[英]Display rows based on sql results
我正在我的WordPress网站中使用破火山口表单来收集有关我们俱乐部的新招生数据。 窗体功能按预期方式。 我正在尝试查看打开的应用程序列表和查看它们的页面。 以下是我的SQL表Tor条目( cf_form_entries
)。
+----+-----------------+--------+
| ID | Form ID | Status |
+----+-----------------+--------+
| 1 | CF5852c23e56b1d | active |
+----+-----------------+--------+
| 2 | CF5852c23e56b1d | active |
+----+-----------------+--------+
| 3 | CF5852c23e56b1d | active |
+----+-----------------+--------+
下表包含表单( cf_form_entry_values
)提交的所有信息;
+----+----------+---------------+--------------------+
| id | entry_id | slug | value |
+----+----------+---------------+--------------------+
| 1 | 1 | branch | Branch A |
| 2 | 1 | full_name | asdasd asdasd |
| 3 | 1 | email_address | b2196363@trbvn.com |
| 4 | 1 | phone | 111111111 |
| 5 | 2 | branch | Branch A |
| 6 | 2 | full_name | Full Name |
| 7 | 2 | email_address | asdasd@asdas.com |
| 8 | 2 | phone | 111111111 |
| 9 | 3 | branch | Branch A |
| 10 | 3 | full_name | Namwe |
| 11 | 3 | email_address | wert@wertwert.com |
| 12 | 3 | phone | 111111111 |
+----+----------+---------------+--------------------+
我可以运行一个简单的select
查询,并获取给定分支,内部联接表的打开的应用程序详细信息。
SELECT cf_form_entries.id,
cf_form_entries.form_id,
cf_form_entries.status,
cf_form_entry_values.slug,
cf_form_entry_values.value
FROM cf_form_entry_values
INNER JOIN cf_form_entries
ON cf_form_entry_values.entry_id = cf_form_entries.id
WHERE cf_form_entry_values.slug = 'branch'
AND cf_form_entry_values.value LIKE '%Branch A%'
上面的查询结果如下表所示;
+----+-----------------+--------+--------+----------+
| id | form_id | status | slug | value |
+----+-----------------+--------+--------+----------+
| 1 | CF5852c23e56b1d | active | branch | Branch A |
| 3 | CF5852c23e56b1d | active | branch | Branch A |
+----+-----------------+--------+--------+----------+
我的问题是, 如何显示(回显)所选表的其他详细信息,例如名称,电子邮件地址等?
因此,我的最终结果将在表格中显示所有打开的应用程序的详细信息。 (不仅是分支名称)
我尝试了一会儿循环。 但是我只能回显分支名称,因为它们是我的内部联接表中唯一选择的数据。
一种方法是条件聚合:
SELECT fe.id, fe.form_id, fe.status,
MAX(CASE WHEN fev.slug = 'branch' THEN fev.value END) as branch,
MAX(CASE WHEN fev.slug = 'full_name' THEN fev.value END) as full_name,
. . .
FROM cf_form_entries fe INNER JOIN
cf_form_entry_values fev
ON fev.entry_id = fe.id
GROUP BY fe.id, fe.form_id, fe.status;
这种方法的好处是,添加新值仅需要在SELECT
添加新表达式。
尝试通过在两个表上使用LEFT JOIN进行以下查询。
SELECT cfev.*,cfm.Form ID as entry_id
FROM cf_form_entry_values cfev
LEFT JOIN cf_form_entries cfm ON cfm.id = cfev.entry_id
WHERE cfev .slug = 'branch' AND vfm.value LIKE '%Branch A%'
这可能会有所帮助。 然后您会得到正确的输出...
这应该给您前两个值,其余的可以使用相同的模式:
SELECT entries.id,
entries.form_id,
entries.status,
(SELECT value FROM cf_form_entry_values as v
WHERE slug='branch' and entry_id=entries.entry_id) as branch,
(SELECT value FROM cf_form_entry_values as v
WHERE slug='full_name' and entry_id=entries.entry_id) as full_name
FROM cf_form_entry_values
INNER JOIN (
SELECT * FROM cf_form_entry_values
INNER JOIN cf_form_entries
ON cf_form_entry_values.entry_id = cf_form_entries.id
WHERE cf_form_entry_values.slug = 'branch'
AND cf_form_entry_values.value LIKE '%Branch A%'
) as entries
ON cf_form_entry_values.entry_id = entries.id
GROUP BY cf_form_entries.id
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