我正在编写一个网页,该网页从数据库中选择满足条件的某些字段。已建立与数据库的连接,但表中除标题外不显示任何内容。在 Apache2 日志中我看到
[Mon May 07 01:30:21 2012] [error] [client MyIP] PHP Notice: Use of undefined constant localhost - assumed 'localhost' in /var/www/medical.php on line 3
[Mon May 07 01:30:21 2012] [error] [client MyIP] PHP Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in /var/www/medical.php on line 7
这是我正在使用的代码:
<?php
include ("/var/www/medicalalerts-config.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die("Unable to select database");
$query = "SELECT * FROM `1213-rep` WHERE medicalAlert <> \'\' and medicalAlert IS NOT NULL ORDER BY lastName, grade";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
?>
<!--Results Table-->
<table border="1" align="center" cellspacing="2" cellpadding="2">
<tr>
<th><font face="Arial, Helvetica, sans-serif" >Name</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Grade</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Medical Alert</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Parent 1 Name</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Parent 1 Phone</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Parent 2 Name</font></th>
<th><font face="Arial, Helvetica, sans-serif" >Parent 2 Phone</font></th>
</tr>
<?php
$i=0;
while ($i < $num) {
$f1=mysql_result($result,$i,"firstName");
$f2=mysql_result($result,$i,"lastName");
$f3=mysql_result($result,$i,"grade");
$f4=mysql_result($result,$i,"medicalAlert");
$f6=mysql_result($result,$i,"parent1Name");
$f7=mysql_result($result,$i,"parent1Phone");
$f8=mysql_result($result,$i,"parent2Name");
$f9=mysql_result($result,$i,"parent2Phone");
?>
<tr>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f1; echo $f2;?> </font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f3; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f4; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f5; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f6; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f7; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif" ><?php echo $f8; ?></font></td>
</tr>
<?php
$i++;
}
?>
</table>
我需要做什么来修复它?
--凌晨 1:51 更新 - 我添加了 $error_msg = mysql_error();
和<?php echo $error_msg ?>
到我的代码,现在我得到 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '\'\' and medicalAlert IS NOT NULL ORDER BY lastName, grade' at line 1
我需要在查询中更改什么?
--凌晨 1:54 更新 - 我修复了它。 PHPMyAdmin 添加了我不需要的额外反斜杠。 谢谢!
最佳答案
试试这个
mysql_connect('localhost',$username,$password);
$num=mysql_num_rows($result);
关于php - MySQL查询网页显示mysql_numrows()错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10477159/