问题描述
'SELECT * FROM t1
JOIN t2 ON t1.wid = t2.wid
WHERE t2.wid IS NULL
LIMIT ' . $number;
这段代码没有任何返回给我你能帮我为什么我不取回值吗?
This code nothing returns to me could you help why i do not take values back??
推荐答案
JOIN t2 ON t1.wid = t1.wid
你是那个意思吗?或者你的意思是t1.wid = t2.wid?在这种情况下，您需要左连接.
did you mean that? or do you really mean t1.wid = t2.wid? in which case you'd want a left join.
编辑
好的，你修好了.除非 t2 中的行的 wid 与 t1 中具有相同 wid 的行匹配，否则不会显示任何结果.
Okay, so you fixed it. That won't show up any results unless there are rows in t2 that have a wid that matches a row in t1 with the same wid.
如果你想要结果，把它改成这样:
If you want results, change it to this:
'SELECT * FROM t1
LEFT JOIN t2 ON t1.wid = t2.wid
WHERE t2.wid IS NULL
LIMIT ' . $number;
下一次编辑
如果目标是使用 t1 中尚未在 t2 中的值来更新 t2，那么它将是这样的:
If the goal is to update t2 with values from t1 that aren't ALREADY in t2, then it would be something like this:
'INSERT INTO t2
SELECT t1.* FROM t1
LEFT JOIN t2
ON t1.wid = t2.wid
WHERE t2.wid IS NULL
LIMIT ' . $number;
缺少的步骤只是只返回 t1 的结果，然后将它们插入到 t2 中.
The missing step was simply to return only t1's results, and then insert them into t2.
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