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没有说出ID应该是什么。这将删除每行!
您应该对此查询进行更新,指定要删除的记录。
$result ="DELETE FROM saldo WHERE id";
更大的问题是,在代码图片中您没有'$result ="DELETE FROM saldo WHERE id = $id";
'列,因此您需要将查询绑定到唯一标识每一行的内容。您可能需要更新表格,或者可能不会向我们显示整个表格。
警告
Little Bobby说 your script is at risk for SQL Injection Attacks. 了解有关prepared的MySQLi语句。甚至escaping the string都不安全!
,
如果您需要删除特定的行,则需要使用WHERE
中的ID
DELETE FROM saldo WHERE id=?
由于您需要将变量传递给查询,因此mysqli_query()
是不合适的。您必须使用准备好的语句。然后,您可以绑定值并执行。
要将值发送到PHP,可以使用链接或表单。
链接:
<a href="delete.php?id=123">Delete item with ID 123</a>
或采用以下格式:
<form method="get">
<input type="text" name="id" />
<input type="submit" value="Delete" />
</form>
您的固定代码应如下所示。请注意,我已启用错误报告,您应该始终这样做。我还添加了if
语句,因为您只想在特定ID已传递到脚本的情况下执行查询。
<?php
// include database connection file
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = mysqli_connect('localhost','root','','ceksaldo');
$conn->set_charset('utf8mb4'); // always set the charset
// Get id from URL to delete that saldo
if (isset($_GET['id'])) {
$id = $_GET['id'];
// Delete saldo row from table based on given id
$stmt = $conn->prepare('DELETE FROM saldo WHERE id=?');
$stmt->bind_param('s',$id);
$stmt->execute();
echo "Record deleted successfully";
}
echo 'No ID selected. No record deleted!';
// After delete redirect to Home,so that latest saldo list will be displayed.
echo "<a href=index.php>Home</a>";
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