Javascript Ajax搜索未连接到数据库_Javascript_Php_Ajax_Search_Livesearch

Javascript Ajax搜索未连接到数据库

javascript php ajax search

Javascript Ajax搜索未连接到数据库,javascript,php,ajax,search,livesearch,Javascript,Php,Ajax,Search,Livesearch,我正在尝试进行实时ajax搜索。当你输入一个单词时，它会自动显示建议，就像W3学校一样。但由于某种原因，我的索引文件和php没有交换数据值，或者我的数据库由于某种原因没有连接。我总是得到“找不到国家”。你能检查代码有无错误吗？这是php文件： <?php include_once('dbconnect.php'); $q = intval($_GET['q']); $query = mysqli_query($conn,"SELECT * FROM `users` WHERE u

我正在尝试进行实时ajax搜索。当你输入一个单词时，它会自动显示建议，就像W3学校一样。但由于某种原因，我的索引文件和php没有交换数据值，或者我的数据库由于某种原因没有连接。我总是得到“找不到国家”。你能检查代码有无错误吗？这是php文件：

<?php
include_once('dbconnect.php');
$q = intval($_GET['q']);
    $query = mysqli_query($conn,"SELECT * FROM `users` WHERE userCountry LIKE '%".$q."%'"); 
        $count = mysqli_num_rows($query);
//Replace table_name with your table name and `thing_to_search` with the column you want to search
    if($count == "0" || $q == ""){
        $s[] = "No Country found!";
    }else{
                while($row = mysqli_fetch_array($query)){
        $s[] = $row['userCountry']; // Replace column_to_display with the column you want the results from
                 } 
        }
for($x = 0; $x < $count; $x++) {
          echo $s[$x];
          echo "<br>";
}
?>

<head>
      <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
      <link rel="stylesheet" href="//code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css">
      <link rel="stylesheet" href="/resources/demos/style.css">
      <script src="https://code.jquery.com/jquery-1.12.4.js"></script>
      <script src="https://code.jquery.com/ui/1.12.1/jquery-ui.js"></script>
<script>
function showCountry(str) {
    if (str == "") {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else { 
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("txtHint").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET","indexsearchquery.php?q="+str,true);
        xmlhttp.send();
    }
}
</script>
<input id="search-box" name="q" type="text" autocomplete="off" placeholder="Search country..." onchange="showCountry(this.value)" />
       <input type='image' name='search' id="search-icon" value='Submit' src="search-icon.png" >
       <p style="color:white;">Suggestions: <span id="txtHint" ></span></p>

您的国家/地区名称将不会是整数。但是您可以将其转换为intval
将php文件更改为
include_once('dbconnect.php');
$q = $_GET['q']; //<-----remove intval
    $query = mysqli_query($conn,"SELECT * FROM `users` WHERE userCountry LIKE '%".$q."%'"); 
        $count = mysqli_num_rows($query);
//Replace table_name with your table name and `thing_to_search` with the column you want to search
    if($count == "0" || $q == ""){
        $s[] = "No Country found!";
    }else{
                while($row = mysqli_fetch_array($query)){
        $s[] = $row['userCountry']; // Replace column_to_display with the column you want the results from
                 } 
        }
for($x = 0; $x < $count; $x++) {
          echo $s[$x];
          echo "<br>";
}

include_once（'dbconnect.php'）；
$q=$_GET['q']// 了解dbconnect.php文件的内容很重要，因为它是包含数据库连接变量的文件。
检查您的MySQL服务器是否已启动，并且所有变量都已正确设置。
调试时，发送到服务器的值是多少？运行时执行的实际SQL查询是什么？users
中的数据是什么？您希望返回什么记录？可能值得注意的是，这个答案造成了明显的SQL注入漏洞……这起到了作用。现在，我如何保护自己免受SQL注入的影响？使用准备好的语句，您将找到大量教程it@Peslis: