我只是想更新数组中的一个元素，但由于某种原因，该信息似乎没有通过。下面是发送数据的javascript:
var xmlhttp = new XMLHttpRequest();
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) {
// Do Nothing
}
}
xmlhttp.open("GET","updateGame.php?q=" + str + "&e=" + recordText,true);
xmlhttp.send();
其中recordText在别处声明。
这是希望接收q和e的PHP:
<?php
$q = intval($_GET['q']);
$e = strval($_GET['e']);
$servername = "localhost";
$username = "*****";
$password = "*****";
$dbname = "*****";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE Games SET entries=$e WHERE PINs=$q";
$result = $conn->query($sql) or die($conn->error);
$conn->close();
?>
我已经检查过了，没有更新的信息
我修复了你的Ajax文件，在这里，它工作得很好，我还修改了你的PHP文件返回数据只是为了测试的目的。用给定的名称创建两个文本文件，复制粘贴next代码并从浏览器运行:
eli_davies_1.php
<html>
<head>
<script type="text/javascript">
function ajax_send ( a )
{ var ajax = false;
if ( window.XMLHttpRequest )
ajax = new XMLHttpRequest();
else if ( window.ActiveXObject )
ajax = new ActiveXObject( "Microsoft.XMLHTTP" );
if ( ajax )
{ var qq = document.getElementById( "txt_q" ).value; // DATA TO SEND.
var ee = document.getElementById( "txt_e" ).value; // DATA TO SEND.
ajax.open("GET","eli_davies_2.php?q=" + qq + "&e=" + ee ); // EXECUTE PHP.
ajax.onreadystatechange =
function ()
{ if ( ( ajax.readyState == 4 ) &&
( ( ajax.status == 0 ) ||
( ajax.status == 200 ) ) )
{ var d = document.getElementById( "div_result" );
d.innerHTML = ajax.responseText; // DISPLAY DATA RETURNED.
}
}
ajax.send( null );
}
}
</script>
</head>
<body>
<input type="text" id="txt_q" value="123" />
<br/>
<input type="text" id="txt_e" value="xyz" />
<br/>
<button onclick="ajax_send()">Send data</button>
<br/>
<div id="div_result"></div>
</body>
</html>
eli_davies_2.php
<?php
$q = intval($_GET['q']);
$e = strval($_GET['e']);
echo "Data received = >" . $q . "< and >" . $e . "<"; // DATA RETURNED (FOR TESTING ONLY).
/*
$servername = "localhost";
$username = "*****";
$password = "*****";
$dbname = "*****";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE Games SET entries=$e WHERE PINs=$q";
$result = $conn->query($sql) or die($conn->error);
$conn->close();
*/
?>