更新數據庫表中多個ID的一個值

Question

$upload_files=implode(' ',$_GET['upload_files']);
$upload_user=",".$_GET['upload_user'];
echo $upload_files;
$sql = "UPDATE {$db_pr}files SET userID = CONCAT(userID,'".$upload_user."') WHERE id IN ('".$upload_files."')";

Answer 1

我相信，IN采用逗號分隔的字符串。

嘗試：

$upload_files=implode("','",$_GET['upload_files']);

Answer 2

Well, I got the solution. I used for loop to achieve the result. 

$upload_files=$_GET['upload_files'];
$upload_user=",".$_GET['upload_user'];
for ($i = 0, $count = count($upload_files); $i <= $count; $i++) {
$sql = "UPDATE {$db_pr}files SET userID = CONCAT(userID,'".$upload_user."') WHERE id = '".$upload_files[$i]."'";
$result = mysqli_query($mysqli,$sql) or die("Error occurred - tried  to update file.");

}
echo "<div class='loginMessage loginSuccess'>Assigned Successfully!!!</div>";