dqv84329
2013-08-25 19:58
浏览 834
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您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在''leave'附近使用正确的语法

I always get this error when i run my code

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 ''leave'' at line 1

Here is my coding part

<?php
        $result = mysql_query("select * from 'leave'");
        if ($result == FALSE)
        {
            die(mysql_error());
        }
        while($row = mysql_fetch_assoc($result))
        {
    ?>
    <tr>
        <td><a href = "app_status.php? id = <?php echo $row["Leave_ID"];?>" target = "_blank"></a>Leave ID</td>
        <td><?php echo $row["Emp_ID"];?></td>   
        <td><?php echo $row["Date_Apply"];?></td>
        <td><?php echo $row["Leave_Type"];?></td>
        <td><?php echo $row["Leave_Start"];?></td>
        <td><?php echo $row["Leave_End"];?></td>
        <td><?php echo $row["Status"];?></td>
    </tr>
    <?php
        }

    ?>
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4条回答 默认 最新

  • drevls8138 2013-08-25 19:59
    最佳回答

    Don't use single quaots

    You can try it as

     $result = mysql_query("select * from leave");
    

    Or use ` key

     $result = mysql_query("select * from `leave`");
    
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