使用Jquery Ajax从Mysql中检索数据list.php：一个简单的ajax代码，我只想显示Mysql表的记录：<html><head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script></head><body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button></body></html>Records.php是从Mysql获取记录的文件。在数据库中只有两个字段：“名称”，“地址”。<?php //database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);?><tr>
<td>Name: </td>
<td>Address: </td></tr><?php while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
} ?>此代码无效。
3 回答
aluckdog
TA贡献1847条经验 获得超7个赞
您不能返回ajax返回值。存储全局变量存储返回后的返回值。
或者像这样更改你的代码。
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;}添加回答
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