Question

尝试让这个脚本调用php。当我使密码不相等时，错误吐出密码不匹配，如果我把它留空，我也得到正确的错误信息。但是，如果我尝试正常使用它，我会得到未知的错误。此代码中是否存在无法匹配的操作顺序。我看不出有什么不对。

<script>
function changepass() {
  var u = _("username").value;

  var cp = _("currentPass").value;

  var np = _("newPass").value;

  var cnp = _("confirmNewPass").value;

  if(np != cnp) {
    _("status").innerHTML = "The passwords given do not match!";
  } else if (cp === "" || np === "" || cnp === "") {
    _("status").innerHTML = "Please fill out all of the fields.";
  } else {
    _("changepassbtn").style.display = "none";
    _("status").innerHTML = 'please wait ...';
    var ajax = ajaxObj("POST", "reset_pass.php");
    ajax.onreadystatechange = function() {
        if(ajaxReturn(ajax) == true) {
        var response = ajax.responseText;
        if(response.trim() == "success"){
            _("status").innerHTML = 'Your password change was successful!';
}       else if (response == "no_exist"){
        _("status").innerHTML = "Your current password was entered incorrectly.";
        _("changepassbtn").style.display = "initial";
}       else if(response == "pass_failed"){
        _("status").innerHTML = "Change password function failed to execute!";
        _("changepassbtn").style.display = "initial";
}       else {
        _("status").innerHTML = "An unknown error occurred";
        _("changepassbtn").style.display = "initial";

}
       }
        }
        ajax.send("u="+u+"&cp="+cp+"&np="+np+"&cnp"+cnp);
}
}
</script>

这是php。

<?php
// AJAX CALLS THIS CODE TO EXECUTE
if(isset($_POST['cp'])) {
    include_once("php_includes/db_conx.php");
    $username = '';
    $oldpasshash = '';
    $newpasshash = '';
    $u = mysqli_real_escape_string($db_conx, $_POST['u']);
    $oldpass = $_POST["cp"];
    $newpass = $_POST["cnp"];
    $oldpasshash = md5($oldpass);
    $newpasshash = md5($newpass);
    $sql = "SELECT username, password FROM users WHERE username='$username' LIMIT 1";
    $query = mysqli_query($db_conx, $sql);
    $row = mysqli_fetch_row($query);
    $db_username = $row["0"];
    $db_password = $row["1"];
    if($db_password != $oldpasshash){
        echo "no_exist";
        exit();
    } else {
        $sql = "UPDATE users SET password='$newpashhash' WHERE username='$db_username' LIMIT 1";
        $query = mysqli_query($db_conx, $sql);
        $sql = "SELECT password FROM users WHERE username='$db_username' LIMIT 1";
        $query = mysqli_query($db_conx, $sql);
        $row = mysqli_fetch_row($query);
        $db_newpass = $row[0];
        if($db_newpass == $newpasshash) {
        echo "success";
        exit();
        } else {
            echo "pass_failed";
            exit();
    }
}
}
?>

我也希望杀死所有用户cookie，但我有我的登出页面中的代码，但我不确定将它放在这个页面的哪个位置。任何建议都是好的。注意：我是一个新手，现在只是关注教程，我希望稍后在我理解它时更改它。

Answer 1

您需要在您的网站中添加jquery的ajax库并使用下面的代码，您可以根据需要更改此代码

function checkpass() {
        var password = $("#newpassword").val();
            password = password.trim();

        var confirmpassword = $("#confirmpassword").val();
            confirmpassword = confirmpassword.trim();

        var status = false;    

        if(password !== '' && confirmpassword !== ''){
            if (password != confirmpassword) {
                $("#newpassword").css('border', '1px solid red');
                $("#confirmpassword").css('border', '1px solid red');
                $('#password_error').html('Password and Confirm Password does not match!');

            } else {       
                $("#newpassword").removeAttr('style');
                $("#confirmpassword").removeAttr('style');
                $('#password_error').html('');
                status = true;
            }
            return status;
        }
    }

    function changepass(){
        var checkin = checkpass();
        var url = 'myurl'; // change according to your need 
        if(checkin == true){
            var newpass = $("#newpassword").val();
            $.ajax({
                url: url,
                type: "POST",
                data: $("#passowrdform").serialize(),
                success: function (response){
                    //alert(response);
                    $('#old_pass').val(newpass)
                    $("#newpassword").val('');
                    $("#confirmpassword").val('');
                }, error: function(){

                }
            });
        }
        return false;
    }

AJAX调用php更改密码表单

1 个答案: