php mysql_num_rows() 期望参数 1 是资源，布尔值在 C:\wamp\www 中给出
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mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www
提问by user2446626
The code is working fine with few pages and shows some sort warning like "mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www"... it always confuses me because it is working for some page and at some point shows a warning message which frustrates me a lot...help please...
该代码在几页上运行良好，并显示了某种警告，例如“mysql_num_rows() 期望参数 1 为资源，布尔值在 C:\wamp\www 中给出”......它总是让我感到困惑，因为它适用于某些页面并且在某些时候显示一条警告消息，这让我很沮丧......请帮助......
<?php
$emp_id=$_SESSION['username'];
$s=mysql_query("select semester from faculty_advisor where emp_id = '$emp_id' ");
echo $s; //just to check if am getting the correct value
$subject=mysql_query("select * from subject_list where semester=$s");
$numrows=mysql_num_rows($subject);
if($numrows!=0)
{
while($row=mysql_fetch_array($numrows))
{
?>
<td align="center"><?php echo $row['sid']; ?></td>
<td align="center"><?php echo $row['subject_name']; ?></td>
<td align="center"><?php echo "<a class='tooltip' href=\"fac_adv_reg_confrm.php?uid=".$row['sid']."\">Edit<span>Edit User Details</span></a>"?></td>
<?php
}
}
else
{
die('Error: ' . mysql_error());
}
?>
回答by Goutam Pal
$s=mysql_query("select semester from faculty_advisor where emp_id = 'macs410' ");
echo $s;
$subject=mysql_query("select * from subject_list where semester=$s");
Should be
$s=mysql_query("select semester from faculty_advisor where emp_id = 'macs410' ");
$row = mysql_fetch_assoc($s);
$subject=mysql_query("select * from subject_list where semester='".$row['semester']."'");
回答by DevZer0
This is happening because your $subject=mysql_query("select * from subject_list where semester=$s");returned FALSEdue to the query failing, use mysql_error()to find out the error.
发生这种情况是因为您因查询失败而$subject=mysql_query("select * from subject_list where semester=$s");返回FALSE，用于mysql_error()找出错误。
Usually do mysql_query() or die (mysql_error())to ensure the code doesn't continue after an error.
通常这样做是mysql_query() or die (mysql_error())为了确保代码在出错后不会继续。
UPDATE
更新
i also noticed that in your first query result $syour using directly as if its a variable. you need to extract the row using mysql_fetch_assocor mysql_fetch_rowbefore passing it to the second query
我还注意到，在您的第一个查询结果中，$s您直接将其用作变量。您需要使用mysql_fetch_assoc或mysql_fetch_row在将其传递给第二个查询之前提取该行
回答by Goutam Pal
$s=mysql_query("select semester from faculty_advisor where emp_id = 'macs410' ");
echo $s;
$subject=mysql_query("select * from subject_list where semester=$s");
Hear $s is a resource object so
mysql_query("select * from subject_list where semester=$s");
will give an error
hence $subject is not a valid resource object
