select * from test1 left join test2 on (test1.id &＃61; test2.id)&＃xff1b;select * from test1&＃xff0c;test2 where test1.id &＃61; test2.id(&＃43;);
test1 test2 →
* 关键字“(&＃43;)”表示会以NULL值填充不满足相等的连接条件的相关列&＃xff08;NAME_1&＃xff09;&＃xff0c;以(&＃43;)对面的表来作为外连接的驱动表
test2
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