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PHP mysql_num_rows死错误

转载 作者:太空宇宙 更新时间:2023-11-03 12:24:08 25 4
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我想创建一个页面,用户可以在其中添加他们的信息..我已经创建了那个页面,但我真正的问题是它的代码..

我有一些问题,那部分代码:

<?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());
?>

它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!?

有人可以解释一下吗?谢谢!

最佳答案

mysql_query 应该有第二个参数作为连接,在你的例子中是 $db

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

如果这也不起作用,那么使用 mysql_error 来知道确切的错误

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

}
else
{
mysql_error();
}

这将显示 mysql_num_rows 中是否存在问题

关于PHP mysql_num_rows死错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18418504/

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