Question

这可能很容易解决，但是我遇到了这段代码的问题 - 我是一个自学新手用PHP，而我提出的代码似乎并不想工作。< / p>

这些页面适用于体育比赛的在线录入系统。评委的详细信息存储在表“club_judges”中，用户可以通过将其复制到“competition_judges”中将其输入竞赛。这是通过表格中的复选框表单完成的。

我想添加一些功能，以便已经添加到竞赛中的评委不会出现在导入表单中，但是我的代码似乎不起作用。我使用“bg_number”（运动的识别号码）的唯一字段来搜索现有的条目。

当前代码：

$existing_judges = mysql_query("SELECT bg_number FROM competition_judges WHERE competition='Test Competition'");
$existing_judges_fetch = mysql_fetch_array($existing_judges);
$existing_judges_array = "('" . implode( "', '", $existing_judges_fetch ) . "');" ; 

$query = "SELECT * FROM club_judges WHERE (`club`='Test Club') AND (`bg_number` NOT IN '$existing_judges_array') ORDER BY name ";
$result = mysql_query($query) or die(mysql_error());

显示错误：

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 ''('1234567', '1234567');') ORDER BY name' at line 1

作为参考，1234567是我的测试裁判的bg_number。

非常感谢任何帮助！

Answer 1

为什么不使用一个查询

SELECT * 
FROM club_judges 
WHERE `club`='Test Club' 
AND `bg_number`
NOT IN (SELECT bg_number FROM competition_judges WHERE competition='Test Competition')
ORDER BY name

Answer 2

使用可以使用单个查询，如下所示：

$query = "SELECT * FROM `club_judges` WHERE `club`='Test Club' 
          AND 
          `bg_number` NOT IN 
              (SELECT `bg_number` FROM `competition_judges` WHERE `competition`='Test Competition')
           ORDER BY `name` ";
$existing_judges_fetch = mysql_fetch_array($existing_judges);
$result = mysql_query($query) or die(mysql_error());

Answer 3

这是一个JOIN版本

SELECT  a.*
FROM    club_judges a
        LEFT JOIN competition_judges b
            ON  a.bg_number = b.bg_number AND
                b.competition='Test Competition'
WHERE   b.bg_number IS NULL AND
        a.club = 'Test Club'
ORDER   BY a.name

要充分了解联接知识，请访问以下链接：

Visual Representation of SQL Joins

PHP MySQL - 从表1中选择不在表2中的值

3 个答案: