PHP mysql_num_rows死亡错误 [英] PHP mysql_num_rows die error

问题描述

我想创建一个页面，用户可以在其中添加信息. 我已经创建了该页面，但是我真正的问题是向下的代码.

I want to create one page, where users add their informations.. I allready have that page created, but my real problem its down code..

我在那部分代码上有问题:

I have some kind of problem, with that part of code:

<?php
//Connect to DB
$db = mysql_connect("localhost","USER","PASS") or die("Database Error");
mysql_select_db("DB",$db);

//Get ID from request
$idstire = isset($_GET['idstire']) ? (int)$_GET['idstire'] : 0;

//Check id is valid
if($idstire > 0)
{
//Query the DB
$resource = mysql_query("SELECT * FROM stiri2 WHERE idstire = " . $idstire);
if($resource === false)
{
    die("Eroare la conectarea cu baza de date");
}

if(mysql_num_rows($resource) == 0)
{
    die("Se pare ca stirea nu mai exista, sau a fost stearsa. <a     href='http://www.wanted-web.ro'>ACASA</a>");
}

$user = mysql_fetch_assoc($resource);

echo "
<div class='main-article-content'>
<h2 class='article-title'>asd</h2>

<div class='article-photo'>
<img src='" . $user['poza'] . "' class='setborder' alt='' />
</div>

<div class='article-controls'>

<div class='date'>
<div class='calendar-date'>" . $user['data'] . "</div>

                            </div>

<div class='right-side'>
<div class='colored'>
<a href='' class='icon-link'><span class='icon-text'></span>Printeaza articol</a>
<a href='#' class='icon-link'><span class='icon-text'></span>Trimite prietenilor</a>
                                </div>

                                <div>
<a href='#' class='icon-link'><span class='icon-text'></span>de Cristian Cosmin D.</a>
<a href='#' class='icon-link'><span class='icon-text'></span>39 comentarii</a>
                                </div>
                            </div>

<div class='clear-float'></div>


                        </div>


<div class='shortcode-content'>
<p>" . $user['nume'] . " , " . $user['prenume'] . " , " . $user['varsta'] . " , " . $user['localitatea'] . "</p>
                        </div>
                    </div>


";
}

$query = "UPDATE stiri2 SET accesari = accesari + 1 WHERE idstire=\"" . $idstire . "\"";
$result = mysql_query($query) OR die(mysql_error());
?>

从这里向我显示错误:

if(mysql_num_rows($resource) == 0)
{
    die("Se pare ca stirea nu mai exista, sau a fost stearsa. <a     href='http://www.wanted-web.ro'>ACASA</a>");
}

我真的不明白为什么！?

I really dont understand why!?

有人可以解释我吗? 谢谢！

Can someone explain me? Thank you!

推荐答案

mysql_query应该具有第二个参数作为连接，在您的情况下为$ db

mysql_query should have the second parameter as the connection which is in your case $db

$resource = mysql_query("SELECT * FROM stiri2 WHERE idstire = " . $idstire,$db);

如果这也行不通，请使用mysql_error知道确切的错误

if this also not works then use mysql_error to know the exact error

$row=mysql_num_rows($resource);
if($row)
{

}
else
{
    mysql_error();
}

这将告诉您mysql_num_rows中是否有问题

this will show you if there is problwm in mysql_num_rows

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