mysql_fetch_数组,无效-在一个php页面中合并表
我有以下代码mysql_fetch_数组,无效-在一个php页面中合并表,php,mysql,join,Php,Mysql,Join,我有以下代码 $data = mysql_query( "SELECT md.*, qr.* FROM moduleDetails md LEFT JOIN qResponses qr ON qr.userno = md.userno"); print "<table border cellpadding=3>"; while($info = mysql_fetch_array( $data )) { print "<
$data = mysql_query(
"SELECT md.*, qr.*
FROM moduleDetails md
LEFT JOIN qResponses qr
ON qr.userno = md.userno");
print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
print "<tr>";
print "<th>Faculty</th> <td>".$info['faculty'] . "</td>";
print "<th>Module Code</th> <td>".$info['moduleCode'] . "</td>";
print "<th>Date Started</th> <td>".$info['dateStarted'] . "</td>";
print "<th>Module Title</th> <td>".$info['moduleTitle'] . "</td>";
print "<th>School</th> <td>".$info['school'] . "</td></tr>";
}
print "</table>";
它在第31行给了我一个错误,那就是$info=mysql\u fetch。。。。等
我做错了什么?。。我看不出有什么解决办法,在我涉及两张桌子之前,它工作得并不好。。任何帮助都很好-看不出有什么问题-加入表的新功能。您可能无法从查询中获得任何结果。执行查询以验证它是否给出了结果,并在尝试获取结果之前检查是否有结果
if (mysql_num_rows($data) == 0)
die("No Records");
elseif (mysql_num_rows($data) > 0)
{
while($info = mysql_fetch_array( $data ))
{
...............
}
}
else { die("Something else went wrong"); }
或者最好用try/catch来包装它,你能在$data=mysql\u query之前发布这行吗?