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伙计们，我正在尝试使用 Ajax 和 PHP 制作登录表单。每次提交表单时，我都会得到正确的响应，但这里的问题是我的 Javascript 文件中有一个“IF CONDITION”在我得到“成功”响应时没有执行。我不知道为什么。这是我的代码，请任何人帮助。

阿贾克斯文件

function ajaxObj( meth, url ) {
    var x = new XMLHttpRequest();
    x.open( meth, url, true );
    x.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    return x;
}

function ajaxReturn(x){
    if(x.readyState == 4 && x.status == 200){
    return true;
    }
}

Javascript 文件处理登录- 问题出在此文件中

function signin(){
    var logstat = _("logstat");
    var loge = _("SignInEmail").value;
    var logp = _("SignInPasswordActive").value;
    if(loge == "" || loge == "Email address" || logp == ""){
    logstat.innerHTML = "Fill out all of the form data.";
    } else {
        _("loginbtn").style.display = "none";
        logstat.innerHTML = 'please wait ...';
        var ax = ajaxObj("POST", "signin.php");
        ax.onreadystatechange = function() {
            if(ajaxReturn(ax) == true) {
                if(ax.responseText != "success"){
                    logstat.innerHTML = ax.responseText;
                    _("loginbtn").style.display = "block";              
                } else {
                    //Here is the problem I get success in Ajax Response, but this condition did not run
                    logstat.innerHTML = "login success";
                    _("loginbtn").style.display = "block";
                }
            }
        }
        ax.send("le="+loge+"&lp="+logp);
    }   
}

PHP 文件

<?php
    if(isset($_POST['le']) && isset($_POST['lp'])){
        $p = $_POST['lp'];
        require_once('includes/database.php');
        include_once('randStrGen.php');
        $e = $_POST['le'];              
        $p_hash = md5($p);      
        $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR'));
        if($e == "" || $e == "Email address" || $p == ""){
            echo "Fill out all of the form data.";
            exit();
        } else {
            $sql = "SELECT id, email, password FROM users WHERE email='$e' AND activated='1' LIMIT 1";
            $query = mysqli_query($db->connection, $sql);
            $numrows = mysqli_num_rows($query); 
            if($numrows == 0){
                echo "This account is not exist";
                exit();
            }else{
                $row = mysqli_fetch_array($query);
                $db_id = $row['id'];
                $db_em = mysqli_real_escape_string($db->connection, $row['email']);
                $db_ps = substr($row['password'],10,-10);   
                if($p_hash != $db_ps){
                    echo "Email or Password combination failed";
                    exit();
                }else{
                    //This is the successful Log in condition
                    echo "success";
                    exit();
                }
            }
            exit();
        }
        exit();
    }
?>
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1 回答 1

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尝试交换条件，以检查它是否等于（甚至 ===）为“成功”，然后在 else 子句中完成其余部分

于 2013-07-01T20:04:06.077 回答