更新数据库表中多个 ID 的一个值答案

【问题标题】：Update one value at multiple ID's in database table更新数据库表中多个 ID 的一个值
【发布时间】：2016-06-29 22:52:19
【问题描述】：

$upload_files=implode(' ',$_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."')";

【问题讨论】：

标签： php mysql arrays database

【解决方案1】：

我相信IN 采用逗号分隔的字符串。

尝试：

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

【讨论】：

【解决方案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>";

【讨论】：