我想检查使用AJAX的输入值。每次用户按一个键AJAX发送,然后根据服务器的结果我添加有效或无效的类到输入。jQuery AJAX:意外的输入结束
的.js:
$("input[name=username]").keyup(function(event) {
processAJAXCheck(this,"username");
});
$("input[name=username]").change(function(event) {
processAJAXCheck(this,"username");
});
function processAJAXCheck(element, toCheck) {
contentOfInput = $(element).val();
$.ajax({
url: "ajax.php",
type: "POST",
data: {content: contentOfInput, toCheck: toCheck},
success: afterAJAXResponse
});
}
function afterAJAXResponse(data) {
result = $.parseJSON(data);
if(result['toCheck'] != undefined) {
inputSelector = 'input[name='+result['toCheck']+']';
if(result['result'] == true) {
$(inputSelector).removeClass('invalid');
$(inputSelector).addClass('valid');
formValid = true;
} else {
$(inputSelector).removeClass('valid');
$(inputSelector).addClass('invalid');
formValid = false;
}
}
result = null;
}
PHP代码:
if(isset($_POST['toCheck']) && isset($_POST['content'])
&& $_POST['toCheck'] != "" && $_POST['content'] != "")
{
require_once 'classes/AAA.php';
require_once 'classes/Database.php';
require_once 'etc/settings.php';
$db = new Database($GLOBALS['db_server'],$GLOBALS['db_name'],$GLOBALS['db_user'],$GLOBALS['db_pass']);
$aaa = new AAA($db);
switch($_POST['toCheck'])
{
case 'email':
$output['result'] = $aaa->isUsableEmail($_POST['content']);
$output['toCheck'] = 'email';
break;
case 'username':
$output['result'] = $aaa->isUsableUsername($_POST['content']);
$output['toCheck'] = 'username';
break;
default:
$output['result'] = 'error';
break;
}
echo json_encode($output);
}
奇怪的是,当我按下任意字符键一切正常,没有任何问题。但是,当我按“TAB”切换到另一个输入(它使用相同的功能,但不返回任何错误 - 也当我按下Tab键在那里!)我得到以下错误:
Uncaught SyntaxError: Unexpected end of input b.extend.parseJSON
jquery-1.9.1.min.js:3 afterAJAXResponse
registration.php?school_id=1&department_id=1:189 b.Callbacks.c
jquery-1.9.1.min.js:3 b.Callbacks.p.fireWith jquery-1.9.1.min.js:3 k
jquery-1.9.1.min.js:5 b.ajaxTransport.send.r
感谢您预先回应。
添加数据类型:“JSON” – 2014-11-04 21:26:10
我不认为这是在这段代码引起问题,但你应该用'var',例如声明局部变量'var contentOfInput = ...' – Barmar 2014-11-04 21:27:54
@ADASein如果他这样做,他必须删除显式调用'$ .parseJSON'。 – Barmar 2014-11-04 21:29:05