php 警告：mysql_fetch_array()：在第 40 行中提供的参数不是有效的 MySQL 结果
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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL Result in line 40
提问by TheWhiteHatter
Possible Duplicate:
mysql_fetch_array() expects parameter 1 to be resource, boolean given in select
i keep getting the error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL Result in line 40 on my webpage i dont know why.. please help me . Background: this script is to pull my database tables up named yname,username,password,date,comments,works,dworks the works and dworks are for people who tryed what came up and then they can judge if it works - yes it works dworks - no it dosent work its supposed to be voting sortof well forget about that my problem is on like 40 of this script please help me
我不断收到错误警告：mysql_fetch_array(): 提供的参数不是有效的 MySQL 结果在我的网页第 40 行我不知道为什么..请帮助我。背景：这个脚本是把我的数据库表拉起来命名为 yname、用户名、密码、日期、评论、作品、dworks 作品和 dworks 是为那些尝试过的人准备的，然后他们可以判断它是否有效 - 是的，它有效 dworks - 不，它没有工作它应该投票有点忘了我的问题是这个脚本的 40 请帮助我
<php require "br.htm" ?>
<style>
<?php require "styles.css" ?>
</style>
<?php
$host="host"; // Host name
$username="name"; // Mysql username
$password="pass"; // Mysql password
$db_name="dbname"; // Database name
$tbl_name="passes"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// select record from mysql
$sql="SELECT * FROM $tbl_name order by id desc";
$result=mysql_query($sql);
?>
<table background='https://lh6.ggpht.com/DFABQFuYSXLBiB6qlvDPfOONOUxCiwM6_S-dHnjW82iognQToTkORsy7RVXsAz0Y23w=w705' width='50%'>
<tr>
<th align='center'>Submition By</th><th align='center'>ScreenName</th><th align='center'>Password</th><th align='center'>Does This Work?</th>
</tr>
<tr>
<th align='center'>
<hr color='lime' width='100%'/>
</th>
<th align='center'>
<hr color='lime' width='100%'/>
</th>
<th align='center'>
<hr color='lime' width='100%'/>
</th>
<th align='center'>
<hr color='gold' width='100%'/>
</th>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td background='transparent' align='left'><i><b><? echo $rows['yname']; ?> </b></i></td>
<td background='transparent' align='center'><i><b><? echo $rows['username']; ?></b></i></td>
<td background='transparent' align='right'><i><b><? echo $rows['password']; ?></b></i></td>
<td background='transparent' align='right'><i><b> Yes<? echo $rows['works']; ?></b></i></td>
</tr>
<?php
// close while loop
}
?>
</table>
<?php
// close connection;
mysql_close();
?>
<center>
i dont want to re-style the page but if i have to i will.. if you see what is wrong please make a post below and tell me what line its on and how to correct it please then ill come back and let u know if it works :) thank you
我不想重新设计页面的样式，但如果必须的话，我会的。如果它有效:)谢谢
回答by MrCode
This error means your query failed. mysql_query()returns falseif an error occurred, you are then passing falseto mysql_fetch_array()which is triggering the error message.
此错误意味着您的查询失败。如果发生错误，则mysql_query()返回false，然后您将传递false到mysql_fetch_array()触发错误消息的哪个。
Your query could be failing due to a missing/wrong table or field. To see the detailed error, print out the result of mysql_error().
由于缺少/错误的表或字段，您的查询可能会失败。要查看详细错误，请打印结果mysql_error()。
The mysql_*library is deprecated. It is recommended to upgrade to MySQLi or PDO.
该mysql_*库已弃用。建议升级到 MySQLi 或 PDO。
