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

我运行代码时总是遇到这个错误

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

您的SQL语法有错误;检查与MySQL服务器版本对应的手册,以便在第1行的“离开”附近使用正确的语法

Here is my coding part

这是我的编码部分

<?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
        }

    ?>

4 个解决方案

#1

Don't use single quaots

不要使用单一的四元组

You can try it as

你可以尝试一下

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

Or use ` key

或者使用`键

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

#2

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

Use backtick character ` for table name:

使用反引号字符`表名:

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

#3

possible answers:

1) Any values which are not numeric will need to be quoted.

1)任何非数字值都需要引用。

2) Your input data should be cleaned/escaped too before use. You are currently open to SQL injection.

2)您的输入数据也应在使用前清理/转义。您目前对SQL注入持开放态度。

#4

$result = mysql_query("select * from your_db.leave") or die (mysql_error());

#1