我正在构建一个简单的网页,允许用户从数据库中删除记录,然后页面将重新加载删除的记录。无法找出用于完成此操作的代码。这是我的主网页:用php删除记录,然后重新加载页面
<html>
<head>
<title>Change Record form</title>
<style type="text/css">
td {font-family: tahoma, arial, verdana; font-size: .875em }
</style>
</head>
<body>
<a href="add_new_form.php">Add new record</a>
<br>
<br>
<?php
$con=mysqli_connect("localhost", "root", "", "customers");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$strSQL = "SELECT * FROM music";
$rs = mysqli_query($con, $strSQL);
while($row = mysqli_fetch_array($rs, MYSQLI_BOTH)) {
// Write the value of the column FirstName (which is now in the array $row)
echo "<br>";
echo "<br>";
echo $row['artist'] . "<br />";
echo $row['title'] . "<br />";
echo $row['format'] . "<br />";
echo $row['notes'] . "<br />";
echo '<FORM METHOD="LINK" ACTION="update_form.php">
<INPUT TYPE="submit" VALUE="Update">
</FORM>';
echo '<FORM METHOD="LINK" ACTION="delete_process.php">
<INPUT TYPE="submit" VALUE="Delete">
</FORM>';
}
?>
然后这是delete_process页:
<?php
$id = $_GET['id'];
$artist = $_GET['artist'];
$title = $_GET['title'];
$format = $_GET['format'];
$notes = $_GET['notes'];
//create connection to DB
$con=mysqli_connect("localhost", "root", "", "customers");
//check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "DELETE from 'MUSIC' WHERE 'id' = $id";
mysql_query($sql);
?>
基本上当用户点击删除按钮,该特定的记录将被删除,主页将刷新记录被删除。真的不知道代码,谢谢!
第一次使用准备好的语句,列引号是反引号而不是单引号 – Ghost 2014-10-30 01:34:27
我很确定'link'不是一个有效的表单方法。您应该使用'POST'并添加一个隐藏字段,其中包含要修改/删除的记录的ID。 – jeroen 2014-10-30 01:36:25
'mysql_query($ sql);'那个不行。在'MUSIC'中加入'$ sql =“DELETE'WHERE'id'= $ id”;' - 点#1:混合API。第二点:错误的标识符。另外,Jeroen说。 – 2014-10-30 01:37:18