嗨,我有一个MySQL数据库表"points",用户可以点击一个按钮,从他们的帐户中删除一个点,他们按下的按钮有另一个用户的ID,因此他们的帐户必须增加一个.
我让它在jQuery中工作,并在Firebug中判断了变量/帖子,它确实发送了正确的数据,例如:
userid= 1
posterid = 4
我认为问题在于我的PHP页面:
<?php
include ('../functions.php');
$userid=mysql_real_escape_string($_POST['user_id']);
$posterid=mysql_real_escape_string($_POST['poster_id']);
if (loggedin())
{
include ('../connection.php');
$query1 = "UPDATE `points` SET `points` = `points` - 1 WHERE `userID` = '$userid'";
$result1=mysql_query($query1);
$query2 = "UPDATE `points` SET `points` = `points` + 1 WHERE `userID` = '$posterid'";
$result2=mysql_query($query2);
if ($result1 && result2)
{
echo "Successful";
return 1;
}
else
{
echo mysql_error();
return 0;
}
}
?>
有什么 idea 吗?谢谢:)