duanlongling5308 2018-07-29 13:51 采纳率: 0%
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php mysql如果两个表之间没有相同的数据,我可以打印其他表数据吗? [关闭]

php mysql If the value in the 'x' table is not in the 'y' table, print the data in the 'x' table?

Mysql Query:

SELECT * FROM andesite_advertisement INNER JOIN andesite_adsrice ON (andesite_adsrice.adsrice_adsase = andesite_advertisement.advertisement_base)

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  • dongqiang2069 2018-07-29 13:53
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    Use CASE or IF syntax:

    SELECT IF(table1.age > 18, table1.user, table2.user)
    FROM ...
    

    It works like so (From the documentation):

    If the first argument is TRUE, IF() returns the second argument. Otherwise, it returns the third argument.

    mysql> SELECT IF(1>2,2,3);
        -> 3
    mysql> SELECT IF(1<2,'yes','no');
        -> 'yes'
    mysql> SELECT IF(STRCMP('test','test1'),'no','yes');
        -> 'no'
    

    Further EDIT (In regards to your updated question). I believe what you're looking for is a LEFT JOIN. If the data doesn't exist in the alternate table, you will still return a row in the database (But any columns in the associated table will return null). Then you can either get the data programattically in PHP, or using something like IF(andesite_adsrice.adsrice_adsase IS NULL, andesite_advertisement.column1, andesite_adsrice.column1).

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