我有一个消息传递系统,它工作正常,但我有它,所以当它读取它mysql_querys并将读取设置为1.这样,你可以告诉它,如果它的opend。它不会在这里更新是查看其更新消息的脚本。感谢mysql不更新
<?php
session_start();
require "../scripts/connect_to_mysql.php";
if (isset($_SESSION['id'])){
$touser = $_SESSION['id'];
}
elseif (!isset($_SESSION['id'])){
header('location: http://www.stat-me.com');
}
$id = $_GET['id'];
$memberfirstname = $_SESSION['firstname'];
if(!isset($id)) {
header('location: inbox.php');
}
elseif(isset($id)) {
mysql_query("UPDATE pms SET read='1' WHERE id='$id'");
$grab_pm = mysql_query("SELECT * FROM pms WHERE touser = '$touser' AND id = '$id'");
while($r= mysql_fetch_object($grab_pm)) {
$subject = $r->subject;
$message = $r->message;
$fromuser = $r->fromuser;
$datesent = $r->datesent;
$read = $r->read;
}
}
?>
什么数据类型是id和读? – 2010-06-21 10:39:21
读取数据库时enum defult设置为“0”,并且消息第一次打开时,它假设更新为“1” id是数据库中消息的id(同样在链接中它会是stat-me.com/messaging/view.php?ID=X),它从 – DonJuma 2010-06-21 10:43:03