Php MYSQL：有条件地向表中添加和更新临时列

我正在写一个标签系统

我有一个标签表，还有一个完整的标签项，由标签项ID和标签ID组成

我正在编写一个获取脚本，该脚本将为数据生成一些复选框，以便我可以轻松地可视化/编辑给定项的标记

我想添加一个名为checked的临时值，然后在标记表中存在一个提供的条目ID的条件下，它有这个标记，然后在我的表中将temp check值设置为true

SELECT t.name AS name, t.id AS id, '0' AS checked  FROM tags t IF( id IN (SELECT DISTINCT tag_id FROM tagged WHERE workshop_id = $id) UPDATE t SET checked='1'

它只是抛出一个一般的sql错误，不起作用。我没有足够的经验来判断这是否真的是一个坏主意，甚至它是否可行。任何输入都将不胜感激。

您不需要更新，您可以使用CASE表达式测试tags.id是否在tagged表中找到，并将其放入选中的临时列中，类似于以下内容：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

请在此处查看它的实际操作：

这将为您提供如下信息：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

更新： 如果要将条件添加到标记表中，如WHERE workshop_id=26，请将其添加到子查询中，而不是左侧联接条件，如下所示：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

您不需要更新，可以使用CASE表达式来测试tags.id是否在taged表中找到，并将其放入选中的临时列中，如下所示：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

请在此处查看它的实际操作：

这将为您提供如下信息：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

更新： 如果要将条件添加到标记表中，如WHERE workshop_id=26，请将其添加到子查询中，而不是左侧联接条件，如下所示：

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
) AS t2 ON t1.id = t2.tag_id;

| ID | NAME | CHECKED |
-----------------------
|  1 |    a |       1 |
|  2 |    b |       0 |
|  3 |    c |       1 |
|  4 |    d |       0 |

SELECT 
  t1.id AS id,
  t1.name AS name, 
  CASE WHEN t2.tag_id IS NULL THEN 0 ELSE 1 END AS Checked
FROM tags AS t1
LEFT JOIN
(
   SELECT * 
   FROM tagged
   WHERE workshop_id = 26
) AS t2 ON t1.id = t2.tag_id;

---

我们可以有两个表的简化示例，每个表中有几行示例数据吗？我们可以有两个表的简化示例，每个表中有几行示例数据吗？选择t1.id作为id，t1.name作为name，当t2.workshop_id为NULL时，则0否则1结束，从标记为t1的标记中选中左连接选择*从标记为t2的26=t2.workshop_id；这与我想要的非常接近，但我需要根据外部id进行筛选…我尝试了硬编码26，但它只生成了一个巨大的表…如果我这样尝试，您的解决方案仍然有效吗？选择t1.id作为id，t1.name作为名称，当t2.workshop_id为NULL时，则0否则1结束，从标记为t1的标记中选中左连接选择*从标记为t2的26=t2.workshop_id；这非常接近我想要的，但我需要根据外部id将其过滤掉…我尝试硬编码26，但它只生成一个巨大的表…如果我这样尝试，您的解决方案仍然有效吗？