嗨,
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, $trackntrace) or die(MysqL_error());
}
我认为查询存在问题,任何人都可以确定问题出在哪里.
亲切的问候
解决方法:
问题出在sprintf函数中.
你告诉sprintf你将传递4个字符串,但你只传递2个.
因为您只传递2个参数,并且它需要4,所以它将返回false.
error_reporting(E_ALL);
ini_set('display_errors', '1');
这是您应该使用上述设置获得的错误:
Warning: sprintf() [<a href='function.sprintf'>function.sprintf</a>]: Too few arguments
你想改变它来解决这个问题:
if ((isset($_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());
}