[英]Why can't I get value from radio-button?
我在使用JQuery获取一组单选按钮的值时遇到麻烦。 我不确定问题出在.val()函数还是ajax请求中。 我的单选按钮代码为:
<fieldset class="fieldset">
<legend>New User Signup:</legend>
<label class="left">Username: </label><input type="text" name="username" id="username" placeholder="username" size="16"> <br >
<label class="left">Password: </label><input type="password" name="password" id="password" placeholder="password" size="16"> <br >
<label class="left">Gender: </label><input type="radio" name="cc" value="M"> Male <input type="radio" name="cc" value="F" checked="checked"> Female <br >
<label class="left" >Birth Date: </label><input type="text" id="datepicker" name="age" size="10" maxlength="10"> <br >
<input type="submit" id="signup_button" value="Sign Up">
</fieldset>
而我的js代码是:
jQuery(document).ready(function($) {
$("#signup_button").click(function(){
var username = $("#username").val();
var pass = $("#password").val();
var gender = $('input[name=CC]:checked').val();
var birth = $("#datepicker").datepicker("getDate");
$.ajax({
method: "POST",
url: "register.php",
data: {
username: username,
pass: pass,
gender: gender,
birth: birth
},
success: function(data)
{
window.alert("success");
$("#register_output").text(data);
}
});
return false;
});
});
在我的register.php文件中,我只是这样做:
$username = $_POST['username'];
$password = $_POST['pass'];
$gender = $_POST['gender'];
$birth_date = $_POST['birth'];
echo "$gender";
案例很重要:
<label class="left">Gender: </label>
<input type="radio" name="cc" value="M"> Male
<input type="radio" name="cc" value="F" checked="checked"> Female
cc
!= CC
$('input[name="cc"]:checked').val();
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