Question

我正在使用xmlhttp请求将数据发送到我的php文件，然后使用它从数据库中获取数据，然后在我的原始页面显示该响应。以下是代码：

＆＃13;

<script>

 function func2(str) {
if (str == "") {
    document.getElementById("replaceContent").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 (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            document.getElementById("replaceContent").innerHTML = xmlhttp.responseText;
        }
    } 
    xmlhttp.open("GET","projectdata.php?q="+str,true);
    xmlhttp.send();
}
}

＆＃13;

<!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" />
<title>Untitled Document</title>
</head>

<body>
					<div id="replaceContent">
                    	This should be replaced.
                        <button onclick="func2(10)"> Click ME </button>
                    </div>
					
</body>
</html>

＆＃13;

以下是projectData.php：

＆＃13;

<?php
	include 'connEst.php';
	// get the q parameter from URL
	$q = $_GET['id'];
	echo $q;
	$sql="SELECT * FROM projects WHERE ID = '".$q."'";
	$result = mysqli_query($conn,$sql);
	
	echo "<table>
	<tr>
	<th> $q </th>
	</tr>
	<tr>
	<th>Project Name</th>
	<th>Project Location</th>
	<th>Project Value</th>
	<th>Client Name</th>
	<th>Work Done in Project</th>
	<th>Project Summary</th>
	<th>Start Date</th>
	<th>End Date</th>
	</tr>";
	while($row = mysqli_fetch_array($result)) {
		echo "<tr>";
		echo "<td>" . $row['p_name'] . "</td>";
		echo "<td>" . $row['p_loc'] . "</td>";
		echo "<td>" . $row['p_val'] . "</td>";
		echo "<td>" . $row['client_name'] . "</td>";
		echo "<td>" . $row['p_wdone'] . "</td>";
		echo "<td>" . $row['p_summary'] . "</td>";
		echo "<td>" . $row['start_date'] . "</td>";
		echo "<td>" . $row['end_date'] . "</td>";
		echo "</tr>";
	}
	echo "</table>";
	mysqli_close($conn);
?>

＆＃13;

我使用apache服务器安装在我的mac上运行它。我正在Chrome中对此进行测试。我设置的$ q变量在运行文件时返回一个空值。

请帮忙！

Answer 1

您的Ajax函数发送参数＆＃34; q＆＃34;不是＆＃34; id＆＃34;并且PHP查找$_GET['id']

在projectData.php中，将现有内容更改为：

$q = $_GET['q'];

或者将您的javascript更改为：

xmlhttp.open("GET","projectdata.php?id="+str,true);

XMLHTTP请求不将数据发送到PHP文件

1 个答案: