我使用 PHP、AJAX 和 MYSQLI 的组合来加载用户输入内容的建议(每次按键时触发加载)。这一切在 chrome、ie10、firefox、safari 等中都运行得很好。然后我在 IE9 上尝试了它，它只是显示“无建议”(如果没有找到匹配项，我会编写代码来执行此操作)。我唯一的想法是，这可能是我在谷歌上读到的缓存问题，但是在尝试所有建议后..我最终回到了开始的地方..“没有建议”。所以，我很困惑。这是我的 AJAX 代码:
function activelyLoad(passedInfo, insertInto, urlInLib)
{
if (passedInfo.length==0)
{
document.getElementById(insertInto).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(insertInto).innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","../lib/"+urlInLib+"?q="+passedInfo,true);
xmlhttp.send();
}
这是我正在加载的 PHP 文件之一:
<?php
include('config.inc.php');
header( 'Cache-Control: no-store, no-cache, must-revalidate' );
header( 'Pragma: no-cache' );
$full = $_GET['q'];
$fullAr = explode(" ", $full);
$count = count($fullAr);
$var = $fullAr[0];
$query = "SELECT * FROM StudentInfo WHERE (StudentFirstName LIKE '%$var%' OR StudentPrefferedName LIKE '%$var%' OR StudentLastName LIKE '%$var%')";
$response = "";
if ($count >= 2) {
$var2 = $fullAr[1];
$query = "SELECT * FROM StudentInfo WHERE ((StudentFirstName LIKE '%$var%' OR StudentPrefferedName LIKE '%$var%') AND StudentLastName LIKE '%$var2%') OR ((StudentFirstName LIKE '%$var2%' OR StudentPrefferedName LIKE '%$var2%') AND StudentLastName LIKE '%$var%')";
}
$result = mysqli_query($connection, $query);
$num_results = mysqli_num_rows($result);
if ($num_results > 0) {
while ($row = mysqli_fetch_array($result)) {
$id = $row['StudentID'];
$functionList = "\"activelyLoad($id, 'detDiv', 'addtable.php'),showElem('#details')\"";
$functionList2 = "\"logLoad($id), showElem('#contactDets')\"";
if(strcmp($row['StudentPrefferedName'], '') != 0){
$response .= ("<tr><td>" . $row['StudentFirstName'] . " \"" . $row['StudentPrefferedName'] . "\" " . $row['StudentLastName'] . "</td><td><button type='button' onmousedown=" . $functionList . " onclick=" . $functionList2 . ">Get details</button></td></tr>");
}
else{
$response .= ("<tr><td>" . $row['StudentFirstName'] . " " . $row['StudentLastName'] . "</td><td><button type='button' onmousedown=" . $functionList . " onclick=" . $functionList2 . ">Get details</button></td></tr>");
}
}
}
else $response = "No suggestions";
echo($response);
mysqli_close($connection);
?>
以下是该页面在 IE9 中的相关 HTML 输出:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<META HTTP-EQUIV=”Pragma” CONTENT=”no-cache”>
<title>XXXXXXX</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<script src="../lib/searchScripts.js"></script>
</head>
<body>
<h1>XXXXXXXX</h1>
<p>Please select one of the following options:</p>
<ul>
<li><a id="searchLink" href="#" onclick="showElem('#search')">Search for a student</a></li>
<div id="search">
<p><b>To find a student, start typing a name in the input field below:</b></p>
<form>
Student Name: <input type="text" onkeyup="activelyLoad(this.value, 'txtHint', 'gethint.php'), showElem('#suggestions')">
</form>
<div id="suggestions"><h2>Suggestions:</h2> <div id="suggestionsBox"><table id="txtHint"><tr><td>No Suggestions</td></tr></table></div></div>
唯一让我认为这一定是缓存问题的是，这适用于所有现代的最新浏览器。非常感谢任何帮助。
编辑 IE9 的控制台返回
SCRIPT600: Invalid target element for this operation. for 'document.getElementById(insertInto).innerHTML=xmlhttp.responseText;'
这是我的问题吗？怎么无效呢？如果此处出现错误，浏览器如何能够访问 AJAX 来加载“无建议”？
最佳答案
在许多其他问题中，您试图将 AJAX 结果转储到 <table> 中元素为.innerHTML属性。
Internet Explorer 有一个错误，不允许您设置 table 的内容标签通过 .innerHTML直到 IE10 中修复该错误
错误报告:http://webbugtrack.blogspot.ca/2007/12/bug-210-no-innerhtml-support-on-tables.html
您最好创建一个 <div>并设置.innerHTML相反。
关于PHP、AJAX、MYSQLI 和 IE9 未按预期输出，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/17576741/