android - EditText 中带有 numberDecimal inputType 的小数分隔符逗号 (',')

Question

inputType numberDecimalin使用EditText点.作为小数分隔符。在欧洲，通常使用逗号,代替。即使我的语言环境设置为德语，小数点分隔符仍然是.

有没有办法将逗号作为小数分隔符？

score 113 · Accepted Answer

一种解决方法（直到 Google 修复此错误）是使用EditTextwithandroid:inputType="numberDecimal"和android:digits="0123456789.,".

然后使用以下 afterTextChanged 将 TextChangedListener 添加到 EditText：

public void afterTextChanged(Editable s) {
    double doubleValue = 0;
    if (s != null) {
        try {
            doubleValue = Double.parseDouble(s.toString().replace(',', '.'));
        } catch (NumberFormatException e) {
            //Error
        }
    }
    //Do something with doubleValue
}

score 34 · Accepted Answer

此处提供的“数字”解决方案的变体：

char separator = DecimalFormatSymbols.getInstance().getDecimalSeparator();
input.setKeyListener(DigitsKeyListener.getInstance("0123456789" + separator));

考虑到语言环境分隔符。

score 19 · Accepted Answer

EditText 的以下代码货币掩码 ($ 123,125.155)

Xml 布局

  <EditText
    android:inputType="numberDecimal"
    android:layout_height="wrap_content"
    android:layout_width="200dp"
    android:digits="0123456789.,$" />

代码

EditText testFilter=...
testFilter.addTextChangedListener( new TextWatcher() {
        boolean isEdiging;
        @Override public void onTextChanged(CharSequence s, int start, int before, int count) { }
        @Override public void beforeTextChanged(CharSequence s, int start, int count, int after) { }

        @Override public void afterTextChanged(Editable s) {
            if(isEdiging) return;
            isEdiging = true;

            String str = s.toString().replaceAll( "[^\\d]", "" );
            double s1 = Double.parseDouble(str);

            NumberFormat nf2 = NumberFormat.getInstance(Locale.ENGLISH);
            ((DecimalFormat)nf2).applyPattern("$ ###,###.###");
            s.replace(0, s.length(), nf2.format(s1));

            isEdiging = false;
        }
    });

score 16 · Accepted Answer

这是 Android SDK 中的一个已知错误。唯一的解决方法是创建自己的软键盘。您可以在此处找到实施示例。

score 8 · Accepted Answer

您可以使用以下解决方法也将逗号作为有效输入：-

通过 XML：

<EditText
    android:inputType="number"
    android:digits="0123456789.," />

以编程方式：

EditText input = new EditText(THE_CONTEXT);
input.setKeyListener(DigitsKeyListener.getInstance("0123456789.,"));

这样安卓系统就会显示数字键盘并允许输入逗号。希望这能回答问题:)

score 8 · Accepted Answer

您可以将以下内容用于不同的语言环境

private void localeDecimalInput(final EditText editText){

    DecimalFormat decFormat = (DecimalFormat) DecimalFormat.getInstance(Locale.getDefault());
    DecimalFormatSymbols symbols=decFormat.getDecimalFormatSymbols();
    final String defaultSeperator=Character.toString(symbols.getDecimalSeparator());

    editText.addTextChangedListener(new TextWatcher() {

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count, int after) {

        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {

        }

        @Override
        public void afterTextChanged(Editable editable) {
            if(editable.toString().contains(defaultSeperator))
                editText.setKeyListener(DigitsKeyListener.getInstance("0123456789"));
            else
                editText.setKeyListener(DigitsKeyListener.getInstance("0123456789" + defaultSeperator));
        }
    });
}

score 6 · Accepted Answer

如果您以编程方式实例化 EditText，Martins 的回答将不起作用。我继续修改DigitsKeyListenerAPI 14 中包含的类，以允许逗号和句点作为小数分隔符。

要使用它，请调用setKeyListener()，EditText例如

// Don't allow for signed input (minus), but allow for decimal points
editText.setKeyListener( new MyDigitsKeyListener( false, true ) );

TextChangedListener但是，在用句点替换逗号的地方，您仍然必须使用 Martin 的技巧

import android.text.InputType;
import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.NumberKeyListener;
import android.view.KeyEvent;

class MyDigitsKeyListener extends NumberKeyListener {

    /**
     * The characters that are used.
     *
     * @see KeyEvent#getMatch
     * @see #getAcceptedChars
     */
    private static final char[][] CHARACTERS = new char[][] {
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '-' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.', ',' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '-', '.', ',' },
    };

    private char[] mAccepted;
    private boolean mSign;
    private boolean mDecimal;

    private static final int SIGN = 1;
    private static final int DECIMAL = 2;

    private static MyDigitsKeyListener[] sInstance = new MyDigitsKeyListener[4];

    @Override
    protected char[] getAcceptedChars() {
        return mAccepted;
    }

    /**
     * Allocates a DigitsKeyListener that accepts the digits 0 through 9.
     */
    public MyDigitsKeyListener() {
        this(false, false);
    }

    /**
     * Allocates a DigitsKeyListener that accepts the digits 0 through 9,
     * plus the minus sign (only at the beginning) and/or decimal point
     * (only one per field) if specified.
     */
    public MyDigitsKeyListener(boolean sign, boolean decimal) {
        mSign = sign;
        mDecimal = decimal;

        int kind = (sign ? SIGN : 0) | (decimal ? DECIMAL : 0);
        mAccepted = CHARACTERS[kind];
    }

    /**
     * Returns a DigitsKeyListener that accepts the digits 0 through 9.
     */
    public static MyDigitsKeyListener getInstance() {
        return getInstance(false, false);
    }

    /**
     * Returns a DigitsKeyListener that accepts the digits 0 through 9,
     * plus the minus sign (only at the beginning) and/or decimal point
     * (only one per field) if specified.
     */
    public static MyDigitsKeyListener getInstance(boolean sign, boolean decimal) {
        int kind = (sign ? SIGN : 0) | (decimal ? DECIMAL : 0);

        if (sInstance[kind] != null)
            return sInstance[kind];

        sInstance[kind] = new MyDigitsKeyListener(sign, decimal);
        return sInstance[kind];
    }

    /**
     * Returns a DigitsKeyListener that accepts only the characters
     * that appear in the specified String.  Note that not all characters
     * may be available on every keyboard.
     */
    public static MyDigitsKeyListener getInstance(String accepted) {
        // TODO: do we need a cache of these to avoid allocating?

        MyDigitsKeyListener dim = new MyDigitsKeyListener();

        dim.mAccepted = new char[accepted.length()];
        accepted.getChars(0, accepted.length(), dim.mAccepted, 0);

        return dim;
    }

    public int getInputType() {
        int contentType = InputType.TYPE_CLASS_NUMBER;
        if (mSign) {
            contentType |= InputType.TYPE_NUMBER_FLAG_SIGNED;
        }
        if (mDecimal) {
            contentType |= InputType.TYPE_NUMBER_FLAG_DECIMAL;
        }
        return contentType;
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
                               Spanned dest, int dstart, int dend) {
        CharSequence out = super.filter(source, start, end, dest, dstart, dend);

        if (mSign == false && mDecimal == false) {
            return out;
        }

        if (out != null) {
            source = out;
            start = 0;
            end = out.length();
        }

        int sign = -1;
        int decimal = -1;
        int dlen = dest.length();

        /*
         * Find out if the existing text has '-' or '.' characters.
         */

        for (int i = 0; i < dstart; i++) {
            char c = dest.charAt(i);

            if (c == '-') {
                sign = i;
            } else if (c == '.' || c == ',') {
                decimal = i;
            }
        }
        for (int i = dend; i < dlen; i++) {
            char c = dest.charAt(i);

            if (c == '-') {
                return "";    // Nothing can be inserted in front of a '-'.
            } else if (c == '.' ||  c == ',') {
                decimal = i;
            }
        }

        /*
         * If it does, we must strip them out from the source.
         * In addition, '-' must be the very first character,
         * and nothing can be inserted before an existing '-'.
         * Go in reverse order so the offsets are stable.
         */

        SpannableStringBuilder stripped = null;

        for (int i = end - 1; i >= start; i--) {
            char c = source.charAt(i);
            boolean strip = false;

            if (c == '-') {
                if (i != start || dstart != 0) {
                    strip = true;
                } else if (sign >= 0) {
                    strip = true;
                } else {
                    sign = i;
                }
            } else if (c == '.' || c == ',') {
                if (decimal >= 0) {
                    strip = true;
                } else {
                    decimal = i;
                }
            }

            if (strip) {
                if (end == start + 1) {
                    return "";  // Only one character, and it was stripped.
                }

                if (stripped == null) {
                    stripped = new SpannableStringBuilder(source, start, end);
                }

                stripped.delete(i - start, i + 1 - start);
            }
        }

        if (stripped != null) {
            return stripped;
        } else if (out != null) {
            return out;
        } else {
            return null;
        }
    }
}

score 3 · Accepted Answer

恕我直言，解决此问题的最佳方法是仅使用 InputFilter。一个很好的要点在这里DecimalDigitsInputFilter。然后你可以：

editText.setInputType(TYPE_NUMBER_FLAG_DECIMAL | TYPE_NUMBER_FLAG_SIGNED | TYPE_CLASS_NUMBER)
editText.setKeyListener(DigitsKeyListener.getInstance("0123456789,.-"))
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

score 2 · Accepted Answer

对于 Mono(Droid) 解决方案：

decimal decimalValue = decimal.Parse(input.Text.Replace(",", ".") , CultureInfo.InvariantCulture);

score 1 · Accepted Answer

您可以执行以下操作：

DecimalFormatSymbols d = DecimalFormatSymbols.getInstance(Locale.getDefault());
input.setFilters(new InputFilter[] { new DecimalDigitsInputFilter(5, 2) });
input.setKeyListener(DigitsKeyListener.getInstance("0123456789" + d.getDecimalSeparator()));

然后你可以使用输入过滤器：

    public class DecimalDigitsInputFilter implements InputFilter {

Pattern mPattern;

public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
    DecimalFormatSymbols d = new DecimalFormatSymbols(Locale.getDefault());
    String s = "\\" + d.getDecimalSeparator();
    mPattern = Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((" + s + "[0-9]{0," + (digitsAfterZero - 1) + "})?)||(" + s + ")?");
}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {

    Matcher matcher = mPattern.matcher(dest);
    if (!matcher.matches())
        return "";
    return null;
}

}

score 1 · Accepted Answer

要本地化您的输入，请使用：

char sep = DecimalFormatSymbols.getInstance().getDecimalSeparator();

然后添加：

textEdit.setKeyListener(DigitsKeyListener.getInstance("0123456789" + sep));

不要忘记将“，”替换为“。” 所以 Float 或 Double 可以毫无错误地解析它。

score 1 · Accepted Answer

我可以确认建议的修复程序不适用于三星 IME（至少在 S6 和 S9 上），也许是 LG。无论语言环境如何，它们仍然显示一个点作为小数点分隔符。切换到 Google 的 IME 可以解决此问题，但对于大多数开发人员来说几乎不是一个选择。

这些键盘的 Oreo 也没有修复它，因为这是三星和/或 LG 必须做的修复，然后甚至推到他们的旧手机上。

相反，我分叉了数字键盘项目并添加了一种模式，它的行为类似于 IME：fork。有关详细信息，请参阅项目示例。这对我来说效果很好，类似于您在银行应用程序中看到的许多“PIN 输入”假 IME。

score 1 · Accepted Answer

这里的所有其他帖子都有很大的漏洞，所以这里有一个解决方案：

根据地区强制使用逗号或句号，不会让您键入相反的。
如果 EditText 以某个值开头，它会根据需要替换正确的分隔符。

在 XML 中：

<EditText
    ...
    android:inputType="numberDecimal" 
    ... />

类变量：

private boolean isDecimalSeparatorComma = false;

在 onCreate 中，找到当前语言环境中使用的分隔符：

if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
    NumberFormat nf = NumberFormat.getInstance();
    if (nf instanceof DecimalFormat) {
        DecimalFormatSymbols sym = ((DecimalFormat) nf).getDecimalFormatSymbols();
        char decSeparator = sym.getDecimalSeparator();
        isDecimalSeparatorComma = Character.toString(decSeparator).equals(",");
    }
}

同样 onCreate，如果您正在加载当前值，请使用它来更新它：

// Replace editText with commas or periods as needed for viewing
String editTextValue = getEditTextValue(); // load your current value
if (editTextValue.contains(".") && isDecimalSeparatorComma) {
    editTextValue = editTextValue.replaceAll("\\.",",");
} else if (editTextValue.contains(",") && !isDecimalSeparatorComma) {
    editTextValue = editTextValue.replaceAll(",",".");
}
setEditTextValue(editTextValue); // override your current value

同样在创建时，添加监听器

editText.addTextChangedListener(editTextWatcher);

if (isDecimalSeparatorComma) {
    editText.setKeyListener(DigitsKeyListener.getInstance("0123456789,"));
} else {
    editText.setKeyListener(DigitsKeyListener.getInstance("0123456789."));
}

编辑TextWatcher

TextWatcher editTextWatcher = new TextWatcher() {
    @Override
    public void beforeTextChanged(CharSequence s, int start, int count, int after) { }

    @Override
    public void onTextChanged(CharSequence s, int start, int before, int count) { }

    @Override
    public void afterTextChanged(Editable s) {
        String editTextValue = s.toString();

        // Count up the number of commas and periods
        Pattern pattern = Pattern.compile("[,.]");
        Matcher matcher = pattern.matcher(editTextValue);
        int count = 0;
        while (matcher.find()) {
            count++;
        }

        // Don't let it put more than one comma or period
        if (count > 1) {
            s.delete(s.length()-1, s.length());
        } else {
            // If there is a comma or period at the end the value hasn't changed so don't update
            if (!editTextValue.endsWith(",") && !editTextValue.endsWith(".")) {
                doSomething()
            }
        }
    }
};

doSomething() 示例，转换为标准周期进行数据操作

private void doSomething() {
    try {
        String editTextStr = editText.getText().toString();
        if (isDecimalSeparatorComma) {
            editTextStr = editTextStr.replaceAll(",",".");
        }
        float editTextFloatValue = editTextStr.isEmpty() ?
                0.0f :
                Float.valueOf(editTextStr);

        ... use editTextFloatValue
    } catch (NumberFormatException e) {
        Log.e(TAG, "Error converting String to Double");
    }
}

score 0 · Accepted Answer

Android 有一个内置的数字格式化程序。

您可以将其添加到您的EditText以允许小数和逗号： android:inputType="numberDecimal"和android:digits="0123456789.,"

然后在代码中的某个位置，当用户单击保存或输入文本后（使用侦听器）。

// Format the number to the appropriate double
try { 
    Number formatted = NumberFormat.getInstance().parse(editText.getText().toString());
    cost = formatted.doubleValue();
} catch (ParseException e) {
    System.out.println("Error parsing cost string " + editText.getText().toString());
    cost = 0.0;
}

score 0 · Accepted Answer

我决定仅在编辑时将逗号更改为点。这是我的棘手且相对简单的解决方法：

    editText.setOnFocusChangeListener(new View.OnFocusChangeListener() {
        @Override
        public void onFocusChange(View v, boolean hasFocus) {
            EditText editText = (EditText) v; 
            String text = editText.getText().toString();
            if (hasFocus) {
                editText.setText(text.replace(",", "."));
            } else {
                if (!text.isEmpty()) {
                    Double doubleValue = Double.valueOf(text.replace(",", "."));
                    editText.setText(someDecimalFormatter.format(doubleValue));
                }
            }
        }
    });

someDecimalFormatter 将使用逗号或点取决于区域设置

score 0 · Accepted Answer

我不知道你的答案为什么这么复杂。如果 SDK 中存在错误，您必须覆盖它或绕过它。

我选择了第二种方法来解决这个问题。如果您将字符串格式化为Locale.ENGLISH然后将其放入EditText（即使是空字符串）。例子：

String.format(Locale.ENGLISH,"%.6f", yourFloatNumber);

追逐该解决方案，您的结果与显示的键盘兼容。然后浮点数和双精度数以点而不是逗号的编程语言方式以典型的方式工作。

score 0 · Accepted Answer

我的解决方案是：

在主要活动中：

char separator =DecimalFormatSymbols.getInstance().getDecimalSeparator(); textViewPitchDeadZone.setKeyListener(DigitsKeyListener.getInstance("0123456789" + separator));
在 xml 文件中： android:imeOptions="flagNoFullscreen" android:inputType="numberDecimal"

我将editText中的double作为字符串。

score 0 · Accepted Answer

8 年多过去了，我很惊讶，这个问题还没有解决......
我在这个简单的问题上苦苦挣扎，因为@Martin 最赞成的答案允许输入多个分隔符，即用户可以输入“12,,,, ,,,12,1,,21,2,"
另外，第二个问题是在某些设备上，数字键盘上不显示逗号（或需要多次按下点按钮）

这是我的解决方案，它解决了上述问题并让用户输入“。” 和 ','，但在 EditText 中，他将看到与当前语言环境相对应的唯一小数分隔符：

editText.apply { addTextChangedListener(DoubleTextChangedListener(this)) }

和文本观察者：

  open class DoubleTextChangedListener(private val et: EditText) : TextWatcher {

    init {
        et.inputType = InputType.TYPE_CLASS_NUMBER or InputType.TYPE_NUMBER_FLAG_DECIMAL
        et.keyListener = DigitsKeyListener.getInstance("0123456789.,")
    }

    private val separator = DecimalFormatSymbols.getInstance().decimalSeparator

    override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) {
        //empty
    }

    @CallSuper
    override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
        et.run {
            removeTextChangedListener(this@DoubleTextChangedListener)
            val formatted = toLocalizedDecimal(s.toString(), separator)
            setText(formatted)
            setSelection(formatted.length)
            addTextChangedListener(this@DoubleTextChangedListener)
        }
    }

    override fun afterTextChanged(s: Editable?) {
        // empty
    }

    /**
     * Formats input to a decimal. Leaves the only separator (or none), which matches [separator].
     * Examples:
     * 1. [s]="12.12", [separator]=',' -> result= "12,12"
     * 2. [s]="12.12", [separator]='.' -> result= "12.12"
     * 4. [s]="12,12", [separator]='.' -> result= "12.12"
     * 5. [s]="12,12,,..,,,,,34..,", [separator]=',' -> result= "12,1234"
     * 6. [s]="12.12,,..,,,,,34..,", [separator]='.' -> result= "12.1234"
     * 7. [s]="5" -> result= "5"
     */
    private fun toLocalizedDecimal(s: String, separator: Char): String {
        val cleared = s.replace(",", ".")
        val splitted = cleared.split('.').filter { it.isNotBlank() }
        return when (splitted.size) {
            0 -> s
            1 -> cleared.replace('.', separator).replaceAfter(separator, "")
            2 -> splitted.joinToString(separator.toString())
            else -> splitted[0]
                    .plus(separator)
                    .plus(splitted.subList(1, splitted.size - 1).joinToString(""))
        }
    }
}

score 0 · Accepted Answer

简单的解决方案，制作自定义控件。（这是在 Xamarin android 中制作的，但应该很容易移植到 java）

public class EditTextDecimalNumber:EditText
{
    readonly string _numberFormatDecimalSeparator;

    public EditTextDecimalNumber(Context context, IAttributeSet attrs) : base(context, attrs)
    {
        InputType = InputTypes.NumberFlagDecimal;
        TextChanged += EditTextDecimalNumber_TextChanged;
        _numberFormatDecimalSeparator = System.Threading.Thread.CurrentThread.CurrentUICulture.NumberFormat.NumberDecimalSeparator;

        KeyListener = DigitsKeyListener.GetInstance($"0123456789{_numberFormatDecimalSeparator}");
    }

    private void EditTextDecimalNumber_TextChanged(object sender, TextChangedEventArgs e)
    {
        int noOfOccurence = this.Text.Count(x => x.ToString() == _numberFormatDecimalSeparator);
        if (noOfOccurence >=2)
        {
            int lastIndexOf = this.Text.LastIndexOf(_numberFormatDecimalSeparator,StringComparison.CurrentCulture);
            if (lastIndexOf!=-1)
            {
                this.Text = this.Text.Substring(0, lastIndexOf);
                this.SetSelection(this.Text.Length);
            }

        }
    }
}

score 0 · Accepted Answer

您可以使用inputType="phone"，但在这种情况下，您将不得不处理多个,或.存在，因此需要额外的验证。

score 0 · Accepted Answer

我对 KOTLIN 的修复

我遇到了同样的错误，我修复了：

val separator = DecimalFormatSymbols.getInstance().decimalSeparator
mEditText.keyListener = DigitsKeyListener.getInstance("0123456789$separator")

这很好用。！但！在三星键盘上，不显示分隔符，因此您不能输入十进制数字。

所以我必须通过检查来解决这个问题，如果使用三星键盘：

    val x = Settings.Secure.getString(getContentResolver(), Settings.Secure.DEFAULT_INPUT_METHOD);
    if (x.toLowerCase().contains("samsung")) {}

但是你仍然有“。” 作为小数分隔符。因此，如果分隔符是逗号，则必须用逗号替换点：

val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
 
  if (separator == ',') {
     mEditText.addTextChangedListener(object : TextWatcher {
                        override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) = Unit
    
                        override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) = Unit
    
                        override fun afterTextChanged(s: Editable?) {
    
                            if (!s.isNullOrEmpty()) {
                                
                                    if (s.toString().contains(".")) {
                                        val replaced = s.toString().replace('.', separator)
                                        mEditText.setText(replaced)
                                        mEditText.setSelection(replaced.length)
                                    }
    
                                
                            }
                        }
                    })
                }

但是你必须检查没有人在 EditTextfield 中输入更多的“，”。这可以通过正则表达式来完成。

我的整个解决方案：

val x = Settings.Secure.getString(getContentResolver(), Settings.Secure.DEFAULT_INPUT_METHOD);
  if (x.toLowerCase().contains("samsung")) {
         val Number_REGEX: Pattern =  Pattern.compile("^([1-9])*([.,]{1}[0-9]{0,10})?$")
         val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
         if (separator == ',') {
            mEditText.addTextChangedListener(object : TextWatcher {
                override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) = Unit

                override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) = Unit

                override fun afterTextChanged(s: Editable?) {

                    if (!s.isNullOrEmpty()) {
                        val matcherMail = Number_REGEX.matcher(s.toString())
                        if (!matcherMail.matches()) {
                            val length: Int = s.length

                            s.delete(length - 1, length);
                        } else {
                            if (s.toString().contains(".")) {
                                val replaced = s.toString().replace('.', separator)
                                mEditText.setText(replaced)
                                mEditText.setSelection(replaced.length)
                            }

                        }
                    }
                }
            })
        }
    } else {
        val separator = DecimalFormatSymbols.getInstance().decimalSeparator
        mEditText.keyListener = DigitsKeyListener.getInstance("0123456789$separator")
       
    }

xml文件：

 <com.google.android.material.textfield.TextInputEditText
        android:id="@+id/tEditText"
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:hint="Input"
        android:inputType="numberDecimal"
        android:imeOptions="actionDone"/>

如果要使用数字，请确保获取正确的格式：

val x = NumberFormat.getInstance().parse(mEditText.text.toString()).toDouble()

score 0 · Accepted Answer

我有一个解决方案，允许用户同时输入点和逗号（如果在键盘上可用），但只显示区域设置默认分隔符。此外，它不允许用户输入超过 1 个分隔符。EditText引用或无限循环没有问题。它是这个线程中几个答案的组合，适合我的需要。

与接受的答案一样，EditText相应地配置：

android:inputType="numberDecimal"
android:digits="0123456789.,"

然后在 EditText 上设置一个自定义 TextWatcher：

myEditText.addTextChangedListener(FlexibleDecimalSeparatorTextWatcher())

并包括自定义 TextWatcher：

import android.text.Editable
import android.text.SpannableStringBuilder
import android.text.TextWatcher
import android.widget.EditText
import java.text.DecimalFormatSymbols

/**
 * The [FlexibleDecimalSeparatorTextWatcher] allows the user to input both the comma (,) and dot (.) as a decimal separator,
 * and will then automatically convert each entered separator into the locale default separator.
 * If the user were to enter multiple separators - every separator but the first will be removed.
 *
 * To provide comma and dot support, set the [EditText] inputType to 'numberDecimal' and its digits to '0123456789.,'.
 */
class FlexibleDecimalSeparatorTextWatcher : TextWatcher {
companion object {
    private val DECIMAL_SEPARATORS = listOf('.', ',')
    private val LOCALE_DEFAULT_DECIMAL_SEPARATOR = DecimalFormatSymbols.getInstance().decimalSeparator
}

override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) {}
override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) {}

override fun afterTextChanged(s: Editable?) {
    if (s != null) {
        val textWithConvertedSeparators = convertSeparatorsToLocaleDefault(s.toString())
        val textWithoutMultipleSeparators = removeAdditionalSeparators(textWithConvertedSeparators)

        // Make the change if required. This only triggers one additional afterTextChanged call if there were changes.
        if(s.toString() != textWithoutMultipleSeparators) {
            s.replace(0, s.length, SpannableStringBuilder(textWithoutMultipleSeparators))
        }
    }
}

/**
 * This function converts all entered separators (in [DECIMAL_SEPARATORS]) to the [LOCALE_DEFAULT_DECIMAL_SEPARATOR].
 */
private fun convertSeparatorsToLocaleDefault(original: String): String {
    var result = original
    DECIMAL_SEPARATORS.forEach { separator ->
        if (separator != LOCALE_DEFAULT_DECIMAL_SEPARATOR && result.contains(separator)) {
            result = result.replace(separator, LOCALE_DEFAULT_DECIMAL_SEPARATOR)
        }
    }
    return result
}

/**
 * Strip out all separators but the first.
 * In this function we assume all separators are already converted to the locale default.
 */
private fun removeAdditionalSeparators(original: String): String {
    var result = original
    var separatorCount = result.count { c -> c == LOCALE_DEFAULT_DECIMAL_SEPARATOR }
    if(separatorCount > 1) {
        // We will reverse the text so we can keep stripping the last (first in reverse) separator off.
        var textReversed = result.reversed()
        val separatorRegex = Regex.fromLiteral(LOCALE_DEFAULT_DECIMAL_SEPARATOR.toString())
        while (separatorCount > 1) {
            textReversed = textReversed.replaceFirst(separatorRegex, "")
            separatorCount--
        }
        // And finally we reverse it back to the original order.
        result = textReversed.reversed()
    }
    return result
}

}

score 0 · Accepted Answer

我不得不想出一个解决方法，它由不同的答案组成。这将允许“，”或“。” 最多一位小数

这是我的编辑文本：

val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
editTextBox.filters = arrayOf<InputFilter>(DecimalDigitsInputFilter(5, 1, separator))
editTextBox.keyListener = DigitsKeyListener.getInstance("0123456789$separator")

和我的班级来处理特定的正则表达式：

class DecimalDigitsInputFilter(
    digitsBeforeZero: Int, 
    digitsAfterZero: Int, 
    separator: Char
) : InputFilter {
    private val mPattern: Pattern =
        Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((\\$separator[0-9]{0," + (digitsAfterZero - 1) + "})?)||(\\$separator)?")

    override fun filter(source: CharSequence, start: Int, end: Int, dest: Spanned, dstart: Int, dend: Int): CharSequence? {
        val matcher = mPattern.matcher(dest)
        return if (!matcher.matches()) "" else null
    }
}

score -4 · Accepted Answer

我认为这个解决方案比这里写的其他解决方案简单：

<EditText
    android:inputType="numberDecimal"
    android:digits="0123456789," />

这样当你按下'。在软键盘中没有任何反应；只允许使用数字和逗号。

android - EditText 中带有 numberDecimal inputType 的小数分隔符逗号 (',')

24 回答 24

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