我想從一個名爲Leave Request的表中更新兩個字段,但是當我點擊提交按鈕時。php sqlsrv - 字段沒有更新
<?php
$stmt = $pdo->prepare('UPDATE dbo.[TABLE$Leave Request] dbo.[KPMG$Leave Request].supAprrove_date=?,dbo.[TABLE$Leave Request].supAprrove=? WHERE dbo.[TABLE$Leave Request].id=?');
$stmt->execute(array($adate, $approve, $id));
$adate = $_POST['supAprrove_date'];
$approve =$_POST['supAprrove'];
$id = $_POST['lid'];
//$affected_rows = $stmt->rowCount();
?>
<form id="form1" name="form1" method="post" action="pending_leave.php">
<label>
<select name="supAprrove" id="supAprrove">
<option value="1">Approve</option>
<option value="2">Reject</option>
</select>
</label>
<input name="supAprrove_date" type="hidden" id="supAprrove_date" value="<?php echo date('Y-m-d'); ?>" />
<input type="hidden" name="lid" id="lid" value="<?php echo $data['id']; ?>"/>
<label>
<input type="submit" name="button" id="button" value="Submit" />
</label>
</form>
也就是說數據庫表如何名稱 – user3315848
什麼是你的錯誤?看起來你不會處理來自'sqlsrv'的任何錯誤。請檢查['sqlsrv_errors']的使用情況(https://msdn.microsoft.com/en-us/library/cc296200(v = sql.105).aspx)。此外,'SET'在'UPDATE dbo。[TABLE $ Leave Request]'後面丟失。 – alalp