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通过使用选择连接查询将2个列值相加来更新列值

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我正在寻找一种方法来更新第3列,通过将其他2列的值相加 . 我遇到的问题是UPDATE语句似乎需要指定一个表,但我通过SELECT和JOIN语句使用“虚拟”表 . 这是我目前的代码:

SELECT *
     FROM wp_posts      AS p
LEFT JOIN (
            SELECT tr.object_id AS id, 
                   t.name       AS physical
              FROM wp_term_relationships AS tr  
        INNER JOIN wp_term_taxonomy AS x 
                     ON (x.taxonomy='pa_physical-inventory' 
                    AND x.term_taxonomy_id=tr.term_taxonomy_id)
        INNER JOIN wp_terms AS t 
                     ON t.term_id=x.term_id    
          ) AS mo ON p.id = mo.id
LEFT JOIN (
            SELECT tr.object_id AS id, 
                   t.name       AS murphy
              FROM wp_term_relationships AS tr  
        INNER JOIN wp_term_taxonomy AS x 
                     ON (x.taxonomy='pa_murphy-inventory' 
                    AND x.term_taxonomy_id=tr.term_taxonomy_id)
        INNER JOIN wp_terms AS t 
                     ON t.term_id=x.term_id 
          ) AS pa ON p.id = pa.id
LEFT JOIN (
           SELECT post_id AS id, meta_value AS totalinventory
             FROM wp_postmeta
            WHERE meta_key = '_stock'
          ) AS totalinventory ON p.id = totalinventory.id
    WHERE p.post_status = 'publish' 
      AND p.post_type = 'product'

我希望将“murphy”和“physical”添加到一起,并为返回的每一行插入“totalinventory” . 其中一些行返回“null”(如果“murphy”没有库存集它显示为null)所以我希望在将值一起添加时将其视为“0” .

任何帮助或指导都会非常感激,因为我长期以来一直在摸不着头脑 .

编辑:我在解决方案中对PHP开放,因为最终将使用cron作业运行 .

1 回答

  • 0

    在更简单的情况下,SELECT通常可以像这样转换为UPDATE ...

    SELECT:

    SELECT * 
    FROM [tables and joins]
    WHERE [conditions]
    ;
    

    更新:

    UPDATE [tables and joins (as above)]
    SET tableA.fieldB = tableC.fieldD + tableE.fieldF
    WHERE [conditions (same as above)]
    ;
    

    当然,如果你有LEFT JOIN,你首先需要SET, before 转换为UPDATE查询 .

    免责声明:正如我在前言所述,根据情况和数据的细节,这并不总是有效 .

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