javascript - jQuery $.post() 在按下按键时将我注销

Question

我有一个动态下拉搜索栏，它可以搜索网页上表单内的数据库成员。为了查看该网页，您必须先登录。当我在我的域上构建网站时，一切正常。然而，当我将文件传输到不同的域并使用相同的数据库对其进行配置时，除了这种形式的动态搜索之外，一切都很完美。如果我在搜索中输入我的名字(有时与工作不同的奇怪名称)，一切都会正常，但如果我输入其他人，它似乎会保留在页面上，但它会将我注销并重新加载顶部的登录表单其他一切，包括我正在打字的表格。我正在使用 jQuery .post() 使搜索动态化。我将在下面提供代码

index.php

  <script>
        // this is the jQuery function used to post to the search document on key up
        function searchUserQ(){
            var searchTxt = $("input[name='userSearch']").val();
            console.log(searchTxt);
            if (searchTxt != '') {

                $.post("includes/search.php", {searchVal:searchTxt},
                    function(output){
                        $("#userResults").html(output);
                    });
            }
        }
    </script>


  <h1 class="editUser">Edit User</h1>
  <form class="editUser" action="index.php" method="post">
    <h1>Search For Employee</h1>
    <input type="text" name="userSearch" id="userSearch" placeholder="Search For Employee By First Name" onkeyup="searchUserQ();" />
    <submit type="submit" />
    <div id="userResults">

    </div>
 </form>

搜索.php

<?php
    // Connect To Secure Login
    $cfgProgDir = '../phpSecurePages/';
    include($cfgProgDir . "secure.php");
    //These are the includes needed to make the php page run
    // this file connects to the database
    include("connect.inc.php");

    if(isset($_POST['searchVal'])){
        // turn that the user searched into a varible
        $searchQ = $_POST['searchVal'];
        // delete any symbols for security
        $searchQ = preg_replace("#[^0-9a-z]#i", "", $searchQ);

        $output      = "";
        $link        = "";
        $searchArray = array();
        $searchIndex = 0;

        // Search through these columns inside the main database
        $userSearchQuery = mysql_query("SELECT * FROM dealerEmployees WHERE 
            firstName   LIKE '%$searchQ%' 
        ");

        // count the number of results
        $userCount = mysql_num_rows($userSearchQuery);
        if($userCount == 0){
            // $output = "There Were No Search Results";
        }else{
            while($row = mysql_fetch_array($userSearchQuery)){
                // define dynamic varibles for each loop iteration
                $id         = $row['id'];
                $firstName  = $row['firstName'];
                $lastName   = $row['lastName'];
                $address    = $row['address'];
                $phone      = $row['phone'];
                $email      = $row['email'];
                $password   = $row['password'];
                $permission = $row['permission'];
                $photo      = "images/" . $row['profilePhoto'];

                $output .= "<li><div class='employeeSearch' style=\"background: url('$photo'); width: 75px; height: 75px\"></div><h6>" . $firstName  . "</h6>" . " " .  "<h6>" . $lastName . "</h6><a href='#' class='employee' data-firstName='$firstName' data-lastName='$lastName' data-address='$address' data-phone='$phone' data-email='$email' data-password='$password' data-permission='$permission' data-id='$id'>Select Employee</a></li>";
            }
        }
    }

    echo $output;

Answer 1

你能试试这个吗？

> "SELECT id, firstName, lastName, address, phone, email, password,
> permission profilePhone  FROM dealerEmployees WHERE 
>             firstName = '".$searchQ."' Limit 1"

也许类似的条件会给出超过 1 个结果。