php - 如何通过当前记录 ID 从连接表中读取数据？

Question

我有一个选择语句：

SELECT term.name FROM terms as TERM WHERE term.name LIKE ''

现在，我有一个表选项，其中有一个列 options_name，其值为 = trgovine_TERMID_lokacija

因此，例如，我需要编写 sql 语句，将选项表与术语连接起来，并使用当前 id 搜索列 options_name。

我试过这个：

SELECT term.id, term.name FROM terms as term JOIN options AS o ON (o.option_name = trgovine_term.id_lokacija) WHERE term.name LIKE ''

但它不起作用:(

更新：

我什至现在尝试过这个：

SELECT term.term_id, term.name FROM wp_terms as term INNER JOIN wp_options AS o ON o.option_name = 'trgovine_'+term.term_id+'_lokacija' WHERE term.name LIKE '%%adidas%%'

Answer 1

SELECT term.id, term.name FROM terms as term INNER JOIN options AS o ON o.option_name = 'trgovine_TERMID_lokacija' WHERE term.name LIKE ''

Answer 2

我找到了正确的答案。这是将数据合并为字符串的正确方法。CONCAT FTW！:)

o.option_name = CONCAT('trgovine_', term.term_id, '_lokacija')