其中是我的sql代码中的错误？ [英] where is the fault in my sql code?

问题描述

 'SELECT * FROM t1 
 JOIN t2 ON t1.wid = t2.wid 
 WHERE t2.wid IS NULL 
 LIMIT'。 $ number; 
  

此代码没有任何回报给我，您能帮我为什么我不采取价值回来

解决方案

  JOIN t2 ON t1.wid = t1.wid 
  

这是你的意思吗？或者你真的是t1.wid = t2.wid吗？

好的。

，所以你修复它。这不会显示任何结果，除非t2中的行具有与t1中具有相同宽度的行匹配的宽度。

如果您想要结果，请更改它到这：

 'SELECT * FROM t1 
 LEFT JOIN t2 ON t1.wid = t2.wid 
 WHERE t2.wid IS NULL 
 LIMIT'。 $ number; 
  

NEXT EDIT

如果目标是使用t2中不是ALREADY的t1更新t2，那么它将是这样：

 'INSERT INTO t2 
 SELECT t1。* FROM t1 
 LEFT JOIN t2 
 ON t1.wid = t2.wid 
 WHERE t2.wid IS NULL 
 LIMIT'。 $ number; 
  

缺少的步骤只是返回t1的结果，然后将它们插入到t2。

'SELECT * FROM t1
          JOIN t2 ON t1.wid = t2.wid
          WHERE t2.wid IS NULL
          LIMIT ' . $number;

This code nothing returns to me could you help why i do not take values back??

解决方案

JOIN t2 ON t1.wid = t1.wid

did you mean that? or do you really mean t1.wid = t2.wid? in which case you'd want a left join.

EDIT

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.

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;

NEXT EDIT

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;

The missing step was simply to return only t1's results, and then insert them into t2.

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