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我當前所面臨的問題是,我的MySQL表不顯示當我輸入的網址:www.mydoman.com/search-results.php?town=Blackpool
- 不顯示venuetbl內的結果,並打印錯誤陳述PHP陣列錯誤:資源ID#4
<?php
session_start();
include('dbconnect.php');
if(isset($_GET['town'])) {
$town = $_GET['town'];
mysql_connect("$host", "$username", "$passwd")or die("cannot connect");
mysql_select_db("$db")or die("cannot select DB");
$res=mysql_query("SELECT * from venuetbl where town = '$town'") or die("error");
}
?>
<?php while($row = mysql_fetch_array($res)) { ?>
<div class="search-results">
<div class="result">
<img src="images/<? echo $rows['imageurl']; ?>.jpg" alt="" width="150" height="100" />
<h2><? echo $rows['name']; ?></h2>
<p><? echo $rows['description']; ?></p>
<a class="button" href="http://www.mydomain.co.uk/result.php?id=<? echo $rows['venueid']; ?>">View More</a>
<a class="button green" href="#">Add To My Venues</a>
</div><!--END Result-->
</div><!--END Search Results-->
<?php
}
?>
你取到'$ row'但在循環您正在訪問'$ rows'。 –
您必須至少在'$ town'上調用'mysql_real_escape_string()',然後再將它傳遞給您的查詢。你的腳本很容易受到SQL注入的影響。感謝 –
,顯然缺乏睡眠 – user1867259