查询就是这个:
SELECT resume.user_id
FROM ".$wpdb->prefix."wpjb_meta as meta
JOIN ".$wpdb->prefix."wpjb_meta_value as value_von
ON meta.id = value_von.meta_id AND meta.meta_object = 'resume' AND meta.name = 'arbeitsregion_von'
JOIN ".$wpdb->prefix."wpjb_meta_value as value_bis
ON meta.id = value_bis.meta_id AND meta.meta_object = 'resume' AND meta.name = 'arbeitsregion_bis'
JOIN ".$wpdb->prefix."wpjb_resume as resume
ON value_bis.object_id = resume.id
WHERE value_von.value <= 222222 AND value_bis.value >= 222222
我的问题是第二次加入:它再次加入同一个表,添加一个内容为“arbeitsregion_bis”的列。结果应该是这样的:
resume.user_id
-------4-------
-------6-------
但是查询的结果总是没有条目。为什么呢?
编辑: 我不需要这些值,我只需要在WHERE条件下的那些,sry。
答案 0 :(得分:2)
通过你正在进行的连接,你假设meta.name
同时有两个值,'arbeitsregion_von'和'arbeitsregion_bis',这是不可能的。
此查询应该为您提供所需的内容(每个用户只有两行)。
SELECT resume.user_id, meta.name as von_bis, my_values.value
FROM ".$wpdb->prefix."wpjb_meta as meta
JOIN ".$wpdb->prefix."wpjb_meta_value as my_values
ON meta.id = my_values.meta_id AND meta.meta_object = 'resume'
JOIN ".$wpdb->prefix."wpjb_resume as resume
ON my_values.object_id = resume.id
WHERE
meta.name IN ('arbeitsregion_von', 'arbeitsregion_bis')
要将它放在一列中,您必须转动:
SELECT resume.user_id,
MAX(CASE WHEN meta.name = 'arbeitsregion_von' THEN my_values.value END) AS von,
MAX(CASE WHEN meta.name = 'arbeitsregion_bis' THEN my_values.value END) AS bis
FROM ".$wpdb->prefix."wpjb_meta as meta
JOIN ".$wpdb->prefix."wpjb_meta_value as my_values
ON meta.id = my_values.meta_id AND meta.meta_object = 'resume'
JOIN ".$wpdb->prefix."wpjb_resume as resume
ON my_values.object_id = resume.id
WHERE
meta.name IN ('arbeitsregion_von', 'arbeitsregion_bis')
GROUP BY resume.user_id
编辑:为了满足您的新要求,最简单的方法
SELECT resume.user_id,
MAX(CASE WHEN meta.name = 'arbeitsregion_von' THEN my_values.value END) AS von,
MAX(CASE WHEN meta.name = 'arbeitsregion_bis' THEN my_values.value END) AS bis
FROM ".$wpdb->prefix."wpjb_meta as meta
JOIN ".$wpdb->prefix."wpjb_meta_value as my_values
ON meta.id = my_values.meta_id AND meta.meta_object = 'resume'
JOIN ".$wpdb->prefix."wpjb_resume as resume
ON my_values.object_id = resume.id
WHERE
meta.name IN ('arbeitsregion_von', 'arbeitsregion_bis')
GROUP BY resume.user_id
HAVING von <= 222222 AND bis >= 222222