由于某种原因,这给我一个错误?多行显示在行中
$result = mysql_query("SELECT wpjb_job.*,
wpjb_category.*
FROM wpjb_job ,
wpjb_category
WHERE (is_filled='0' AND is_active='1')
AND wpjb_job.job_category = wpjb_category.id
AND job_country={$countryid}
ORDER BY wpjb_job.job_title") or die(mysql_error());
这是错误:您的SQL语法错误;检查对应于你的MySQL服务器版本的手册,以便在第6行的'ORDER BY wpjb_job.job_title'附近使用正确的语法。至于我可以计算出我在服务器上有MySQL 5.1。该代码仍然有效,但会在表格后出现错误。
我将它改为以下来测试。 现在它运行良好,但它打印表格的标题两次,一次在开始,然后在结束。变化是在$ countryid,我现在把它作为“$ countryid”这没有给出错误
下面表格的打印代码
$joburl = "http://www.x.com/job/view/";
$result = mysql_query("SELECT wpjb_job.*,wpjb_category.*
FROM wpjb_job , wpjb_category
WHERE (is_filled='0' AND is_active='1')
AND wpjb_job.job_category = wpjb_category.id
AND job_country='$countryid'
ORDER BY job_title")
or die(mysql_error());
echo "<table border='1'>";
echo "<tr> <th>Job</th> <th>Company</th> <th>Industry</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array($result)) {
// Print out the contents of each row into a table
echo "<tr><td>";
echo '<a href ="http://www.x.com/job/view/'.$row['job_slug'].'"> '.$row['job_title'].' </a>';
echo "</td><td>";
echo $row['company_name'];
echo "</td><td>";
echo $row['title'];
echo "</td></tr>";
}
echo "</table>";
}
在ORDER BY之前是否有空格 –
您确定{$ countryid}正在被正确评估吗? – ethrbunny
使用'echo $ your_query'来显示正在执行的查询。 – Jocelyn