我编写了删除 MySQL 表行的代码。但是当我点击删除图标时，没有任何反应。有人可以告诉我代码中还剩下什么吗？
<?php
include_once 'include/DatabaseConnector.php';
$query1="SELECT * FROM MyTable;";
$result1=DatabaseConnector::ExecuteQueryArray($query1);
?>
<script type="text/javascript">
function deleteRow(tableName,colName,id){
$.ajax({
type: "POST",
url: "delete.php",
data: "tableName=tableName&colName=colName&id=id",
success: function(msg){
alert( "Row has been updated: " + msg );
}
});
}
</script>
<table id="newspaper-b" border="0" cellspacing="2" cellpadding="2" width = "100%">
<thead>
<tr>
<th scope="col">Opr</th>
<th scope="col">Flt Num</th>
<th scope="col">From</th>
<th scope="col"></th>
</tr>
</thead>
<tbody>
<?php foreach ($result1 as $row):?>
<tr>
<td><?php echo $row['airlineName'];?></td>
<td><?php echo $row['flightNum'];?></td> <td><?php echo $row['from'];?></td>
<td>
<div title='Delete' onclick='deleteRow(<?php echo 'flightschedule','flightNum',$row['flightNum']; ?>)'>
<img src='images/delete.png' alt='Delete' />
</div>
</td>
</tr>
<?php endforeach;?>
</tbody>
删除.php
<?php
/* Database connection */
include_once 'include/DatabaseConnector.php';
if(isset($_POST['tableName']) && isset($_POST['colName']) && isset($_POST['id'])){
$tableName = $_POST['tableName'];
$colName = $_POST['colName'];
$id = $_POST['id'];
$sql = 'DELETE FROM '.$tableName.' WHERE '.$colName.' ="'.$id.'"';
mysql_query($sql);
} else {
echo '0';
}
?>
最佳答案
- 您检查过您的 PHP 日志是否有错误？
- 什么是
Ajax.Request？如果您正在使用原型(prototype)库，它包含在您的 HTML 代码中的什么位置？ - 最后，你确定你的 PHP 代码被调用了吗？ (检查使用例如 Chrome 浏览器中的 Web 开发者工具，“请求”选项卡)
关于javascript - PHP Ajax MySQL 表行删除，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10737229/