Question

我正在尝试进行AJAX搜索以从mySQL数据库返回数据，然后使用返回的信息更新页面上的div。我跟着一个教程（根据我的需要定制它），到目前为止，似乎什么也没做，只是将搜索查询附加到地址。非常感谢任何帮助，谢谢！

HTML

<html>
<head>
<script>
function searchDB(str) {
  if (str=="") {
    document.getElementById("txtHint").innerHTML="";
    return;
  } 
  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 (xmlhttp.readyState==4 && xmlhttp.status==200) {
      document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
  xmlhttp.open("GET","search4.php?q="+str,true);
  xmlhttp.send();
}
</script>
</head>
<body>

<form>
<input type="text" name="q" placeholder="Search Database..." onsubmit="searchDB(this.value)">
</form>
<br>
<div id="txtHint"><b>Search will be returned here.</b></div>

</body>
</html>

PHP

<?php
$q = ($_GET['q']);

$con = mysqli_connect('localhost','username','pass','database');
if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"SearchTest");
$sql="SELECT ID, FirstName, LastName FROM SearchTest WHERE FirstName LIKE '%" . $q ."%' OR LastName LIKE '%" . $q . "%'";
$result = mysqli_query($con,$sql);

echo "<table border='1'>
<tr>
<th>First Name</th>
<th>Last Name</th>
</tr>";

while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

Answer 1

问题最有可能出现在这一行mysqli_select_db($con,"SearchTest"); 将其删除，因为您可能会混淆同名的表格 SELECT ... FROM SearchTest。

我接下来100％确定这是问题，因为您已经选择了('localhost','username','pass','database')

的数据库

将错误报告添加到文件顶部

error_reporting(E_ALL);
ini_set('display_errors', 1);

很可能会向您显示数据库（名称为SearchTest）不存在。

修改

TBH，我无法使您现有的代码工作，我不知道为什么。（Nota：见评论，已解决）

然而，这确实有效：

<强>的search.php

<html> <head> <script src="search.js"></script> <title> Search </title></head> <body> <form> Search: <input type="text" name="search" id="search" onmouseup="searchUser(this.value)"> </form> <div id="searchresult" name="searchresult"> Search results ...</div> </body> </html>

<强> search.js

var xmlHttp function searchUser(str) { xmlHttp=GetXmlHttpObject() if (xmlHttp==null) { alert ("Browser does not support HTTP Request") return } var url="searchresult.php" url=url+"?q="+str url=url+"&sid="+Math.random() xmlHttp.onreadystatechange=stateChanged xmlHttp.open("GET",url,true) xmlHttp.send(null) } function stateChanged() { if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete") { document.getElementById("searchresult").innerHTML=xmlHttp.responseText } } function GetXmlHttpObject() { var xmlHttp=null; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { //Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } } return xmlHttp; }

<强> searchresult.php

<?php error_reporting(E_ALL); ini_set('display_errors', 1); $q=$_GET["q"]; $DB_HOST = "xxx"; $DB_NAME = "xxx"; $DB_USER = "xxx"; $DB_PASS = "xxx"; $con = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME); if($con->connect_errno > 0) { die('Connection failed [' . $con->connect_error . ']'); } $sql = "SELECT ID, FirstName, LastName FROM SearchTest WHERE FirstName LIKE '%" . $q ."%' OR LastName LIKE '%" . $q . "%'"; $result = mysqli_query($con,$sql); echo "<table border='1'> <tr> <th>First Name</th> <th>Last Name</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?>

Answer 2

始终使用jQuery（或任何其他js lib / framework）进行ajax调用。 jQuery也非常适合操纵DOM（document.getElementById("txtHint").innerHTML=xmlhttp.responseText;让我想起了90年代。）

您的PHP代码高度不安全。见How can I prevent SQL injection in PHP?

使用Ajax搜索mySQL数据库

HTML

PHP

2 个答案: