php – 在一个更新语句中更新多个MySQL表时出错

嗨,

我有

Query was empty

更新数据时出错.

if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1"))
{
   $updatesql = sprintf("UPDATE client,release SET client.`client_name`=%s, release.`client_name`=%s WHERE client.`client_id`=%s AND release.`client_id`=%s", GetsqlValueString($_POST['newcust'], "text"), GetsqlValueString($_POST['select'], "int"));
   MysqL_select_db($database_trackntrace, $trackntrace);
   $Result1 = MysqL_query($updatesql, $trackntrace) or die(MysqL_error());
}

我认为查询存在问题,任何人都可以确定问题出在哪里.

亲切的问候

解决方法:

问题出在sprintf函数中.
你告诉sprintf你将传递4个字符串,但你只传递2个.

因为您只传递2个参数,并且它需要4,所以它将返回false.

尝试在代码上方添加以下内容：

error_reporting(E_ALL);
ini_set('display_errors', '1');

这是您应该使用上述设置获得的错误：

Warning: sprintf() [<a href='function.sprintf'>function.sprintf</a>]: Too few arguments

你想改变它来解决这个问题：

if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1"))
{
   $updatesql = sprintf("UPDATE client,release SET client.`client_name`=%s, release.`client_name`=%s WHERE client.`client_id`=%s AND release.`client_id`=%s", GetsqlValueString($_POST['newcust'], "text"), GetsqlValueString($_POST['newcust'], "text"), GetsqlValueString($_POST['select'], "int"), GetsqlValueString($_POST['select'], "int"));
   MysqL_select_db($database_trackntrace, $trackntrace);
   $Result1 = MysqL_query($updatesql, $trackntrace) or die(MysqL_error());
}