<?php
if(isset($_POST['chkStatus'])){
for ($i=0;$i<count($_POST['chkStatus']);$i++) {
$count = count($_POST['chkStatus']);
list($txtClientId, $txStatusValue) = explode('-', $_POST['chkStatus'][$i], 2);
$stmtChkStatus=$con->query("SELECT id,status FROM users WHERE id=$txtClientId");
$SRow = $stmtChkStatus->fetch();
if($SRow['status']!=$txStatusValue) {
if($txStatusValue==0) $stmtStatus=$con->query("UPDATE users SET status='0' WHERE id=$txtClientId");
else $stmtStatus=$con->query("UPDATE users SET status='1' WHERE id=$txtClientId");
}
}
}
?>
<html>
<body>
<dl>
<?php $stmtUsers=$con->query("SELECT id,status FROM users");
while($UsersRow = $stmtUsers->fetch()){
echo "<dt>$UsersRow[name]</dt>
<dd><input type='checkbox' name='chkStatus[]' value='$UsersRow[id]-$UsersRow[status]'";
if($CRow['status']==0) echo " checked";
echo " onchange='this.form.submit()'>
</dd>";
?>
</dl>
</body>
</html>
我的代码的问题是,它只执行选中的复选框。我需要一个代码来改变我单击的复选框(onchange)的SQL DB中的状态。提交表单onchange复选框并更新MySQL分贝
对于未来的贴子请参阅https://meta.stackoverflow.com/questions/251361/how-do-i-format-my-code-blocks,不要对代码格式使用HTML。 – chris85
您的代码容易受到SQL注入攻击。您应该使用[mysqli](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)或[PDO](http://php.net/manual/en/pdo.prepared- statement.php)按照[本文]中描述的准备语句(http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php)。 –
最新问题? – Omi