Question

我无法验证重新接收输入。继承我的代码：

// Validate Recaptcha Input
var challenge = $("#recaptcha_challenge_field").val();
var response = $("#recaptcha_response_field").val();

var dataString = 'recaptcha_challenge_field=' + challenge + '&recaptcha_response_field=' + response;


var html = $.ajax({
type: "POST",
    url: "PHP/recaptchaverify.php",
    data: dataString,
    async: true
}).responseText;

console.log(html);

if(html == 'accept') 
{
    alert("CORRECT");
}

else
{
    alert("The reCAPTCHA wasn't entered correctly. Go back and try it again.");
    $("#recaptcha_response_field").focus();
    Recaptcha.reload();
    return false;
}

现在我将我的变量传递给recpatchaverify.php

require_once('../scripts/recaptcha-php/recaptchalib.php');
$privatekey = "MYKEY";

$resp = recaptcha_check_answer ($privatekey, $_SERVER["REMOTE_ADDR"], $_POST["recaptcha_challenge_field"], $_POST["recaptcha_response_field"]);

if (!$resp->is_valid) 
{
    // What happens when the CAPTCHA was entered incorrectly
    echo "error";
} 

else 
{
    // Your code here to handle a successful verification
    echo "accept";
}

现在我的问题是每当我正确输入Recaptcha时html变量都会显示“accept”，但它在IF语句中不起作用？

Answer 1

您正在向服务器发出异步请求，这意味着当$ .ajax（）行完成并继续执行到console.log（）和if（）语句时，对服务器的实际请求仍处于未决状态还没有完成。在if（）语句执行时，responseText实际上是'undefined'。

相反，您需要使用'success'回调函数，如下所示：

// Validate Recaptcha Input
var challenge = $("#recaptcha_challenge_field").val();
var response = $("#recaptcha_response_field").val();

var dataString = 'recaptcha_challenge_field=' + challenge + '&recaptcha_response_field=' + response;


$.ajax({
    type: "POST",
    url: "PHP/recaptchaverify.php",
    data: dataString,
    success: function(html) {
        if(html == 'accept') 
        {
            alert("CORRECT");
        }

        else
        {
            alert("The reCAPTCHA wasn't entered correctly. Go back and try it again.");
            $("#recaptcha_response_field").focus();
            Recaptcha.reload();
            return false;
        }
    }
});

Recaptcha与Ajax错误

1 个答案: