php - 检查是否已选择单选按钮之一

Question

我想检查是否选择了一个单选按钮，如果是，那么我想打印它的值。

<form ....>
     Gender:<br />
 <input type="radio" name="sex" value="male">Male<br />
 <input type="radio" name="sex" value="female">Female<br />
</form>

我正在尝试这个，但是当没有选择任何单选按钮时它会出错。

$selected_radio = $_POST['sex'];
if ($selected_radio == 'male') {
    print "Gender is male <br />";
}
else if ($selected_radio == 'female') {
    print "Gender is female <br />";
}
else {
    print "No gender selected. <br />";
}

Answer 1

您已经在单选按钮的value属性中获得了性别信息。无需将它们再次写入字符串。您的比较运算符不能用空格分隔。

<?php
    if (isset($_POST['sex']))  
        echo "Gender is " . $_POST['sex'];
    else 
        echo "No gender selected.";
?>

或者还有另一种单行解决方案：

<?php echo isset($_POST['sex']) ? "Gender is " . $_POST['sex'] : "No gender selected."; ?>

Answer 2

还有“= =”错误

你输入的名字是“sex”

<input type="radio" name="sex" value="male">

你问的是“性别”：

$_POST['gender']

Answer 3

您的比较操作中有一个sintax错误，应该使用 equal sing==而不是= =please change

if ($selected_radio = = 'male') {

和

if ($selected_radio == 'male') {

你也应该这样做

else if ($selected_radio == 'female') {

此处提供比较运算符文档

您的表单中也有错误，您的输入应该name="gender"与您的代码匹配$_POST['gender']

Answer 4

你改写了= =，==这就是问题所在

   $selected_radio = $_POST['gender'];
if ($selected_radio == 'male') {
    print "Gender is male"."<br/>";
}
else if ($selected_radio == 'female') {
    print "Gender is female"."<br/>";
}
else {
    print "No gender selected."<br/>";
}

Answer 5

如果未设置，则不检查。

if (isset($_POST['gender'])) {
    $selected_radio = $_POST['gender'];
    if ($selected_radio == 'male') {
        print "Gender is male"."<br/>";
    } else if ($selected_radio == 'female') {
        print "Gender is female"."<br/>";
    } else {
        print "form invalid submition. CSRF maybe";
    }
} else {
    print "no gender selected";
}