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此代码尝试捕获 EditText 的字符串并将其与数组的字符串进行比较。代码没有错误，但是当我单击按钮时没有任何反应（甚至没有显示吐司），我做错了什么？谢谢

这是数组数组。列表.xml

<resources>
        <string-array name="array1">
            <item name="number">1</item>
            <item name="name">first</item>
        </string-array>
        <string-array name="array2">
            <item name="number">2</item>
            <item name="name">second</item>
        </string-array>

    <array name="list">
        <item>@array/array1</item>
        <item>@array/array2</item>
    </array>
</resources>

这是主xml文件main.xml的代码

<EditText
            android:id="@+id/etname"
            android:layout_width="match_parent"
            android:layout_height="wrap_content"
            android:layout_weight="1"
            android:ems="10"
            android:inputType="text"
            android:imeOptions="actionDone" >

<Button
            android:id="@+id/button1"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:layout_gravity="center_horizontal"
            android:onClick="searchArray"
            android:text="@string/search" />

搜索数组

public void searchArray(View view){
        Resources res = getResources();
        TypedArray ta = res.obtainTypedArray(R.array.list);
        int n = ta.length();
        String[][] array = new String[n][];
        for (int i = 0; i < n; ++i) {
            int id = ta.getResourceId(i, 0);
            if (id > 0) {
                array[i] = res.getStringArray(id);
            } else {
                // something wrong with the XML
            }
        }
        ta.recycle(); 

        TextView name = (TextView) findViewById(R.id.etname);
        String stringname = name.getText().toString();

        if (stringname != null){
            for (int cont = 0; cont < n; ++cont) {
                if (!(array[cont][1].equals(stringname))){
                    Toast toast2 = Toast.makeText(getApplicationContext(), array[cont][1], Toast.LENGTH_SHORT);
                    toast2.show();
                } else {
                    Toast toast = Toast.makeText(getApplicationContext(), "WRONG", Toast.LENGTH_SHORT);
                    toast.show();
                }
            }
        }

    }
4

2 回答 2

0

您似乎是从索引 1 进行比较，而不是 0。更改代码

 for (int cont = 0; cont < n; ++cont)

至

for (int cont = 0; cont < n; cont++)
于 2013-10-21T02:20:53.270 回答
0

像这样试试

public void searchArray(View view){
        Resources res = getResources();
        TypedArray ta = res.obtainTypedArray(R.array.list);
        int n = ta.length();
        String[][] array = new String[n][];
        for (int i = 0; i < n; i++) {
            int id = ta.getResourceId(i, 0);
            if (id > 0) {
                array[i] = res.getStringArray(id);
                System.out.println(res.getStringArray(id));
            } else {
                // something wrong with the XML
            }
        }
        ta.recycle(); 

        EditText name = (EditText) findViewById(R.id.etname);// I changed TextView to EditText
        String stringname = name.getText().toString();

        if (stringname != null){
            for (int cont = 0; cont < n; ++cont) {
                System.out.println(array[cont][1]+" "+stringname);
                if ((array[cont][1].equals(stringname))){//Edit Here, If it matches, Show Toast
                    Toast toast2 = Toast.makeText(getApplicationContext(), array[cont][1], Toast.LENGTH_SHORT);
                    toast2.show();
                } else {
                    Toast toast = Toast.makeText(getApplicationContext(), "WRONG", Toast.LENGTH_SHORT);
                    toast.show();
                }
            }
        }

    }

希望这对您有所帮助。

于 2013-10-21T03:02:54.163 回答