javascript - 如果语句不起作用，jQuery/AJAX 成功函数？

Question

我有一段代码，用于通过 jQuery AJAX 调用检查所需的用户名是否可用:

function check_username()
{

    // Set user_name to the current value of input #new_user_name
    var user_name = document.getElementById('new_user_name').value;

    // Ajax to check availability of username
    $.ajax({
        url: '_check_username.php',
        type: 'post',
        data: { 'user_name': user_name },
        success: function(data) {   

            var does_it_pass_or_fail = data;

            if ( does_it_pass_or_fail === "fail" )
            {
                alert( "Sorry, the username already exists. does_it_pass_or_fail = " + does_it_pass_or_fail ); );
            }
            else
            {
                alert( "That username is available! does_it_pass_or_fail = " + does_it_pass_or_fail ); );
            }
        }, error: function() {
           // Not sure what to put here...
        }
    });

}

脚本按预期工作——我在 _check_username.php 上检查 MySQL 数据库，如果名称可用，则成功返回“通过”，如果名称已被占用，则成功返回“失败”，并将其存储在 JS 中变量 does_it_pass_or_fail。

但是，无论 does_it_pass_or_fail 是否包含“pass”或“fail”，我只会收到“该用户名可用!”的响应

所以，AJAX/PHP/MySQL 部分似乎工作正常，但是这个简单的 JavaScript if 语句让我难住了!

Answer 1

尝试使用这个:您的代码中存在一些接近alert.try日志错误的语法错误。

function check_username()
{

    // Set user_name to the current value of input #new_user_name
    var user_name = document.getElementById('new_user_name').value;

    // Ajax to check availability of username
    $.ajax({
        url: '_check_username.php',
        type: 'post',
        data: { 'user_name': user_name },
        success: function(data) {   

            var does_it_pass_or_fail = data;
            console.log(data);
            if ( does_it_pass_or_fail === "fail" )
            {
                alert( "Sorry, the username already exists. does_it_pass_or_fail = " + does_it_pass_or_fail );
            }
            else
            {
                alert( "That username is available! does_it_pass_or_fail = " + does_it_pass_or_fail );
            }
        }, error: function(erro) {
          console.log(error);
        }
    });

}