MYSQL / php基本問題

Question

我想看看表T1是否存在指定的值。 如果值存在，我想在表T2輸入值，

我怎么能用PHP做到這一點？

我想檢查存在，然后通過以下命令插入值

我想要這種格式

if ( table1.id = Exist's )

then 

{insert into  table2 ( values ) }

使用PHP編寫此代碼的正確方法是什么？

Answer 1

嗯，這可能會對你有所幫助。

更新的答案

那么，對於表tb1

+-------+-------------+------+-----+---------+-------+
| Field | Type        | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| fld1  | varchar(20) | YES  |     | NULL    |       |
| fld2  | varchar(20) | YES  |     | NULL    |       |
| fld3  | varchar(20) | YES  |     | NULL    |       |
+-------+-------------+------+-----+---------+-------+

和表tb2

+-------+-------------+------+-----+---------+-------+
| Field | Type        | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| fld1  | varchar(20) | YES  |     | NULL    |       |
| fld2  | varchar(20) | YES  |     | NULL    |       |
| fld3  | varchar(20) | YES  |     | NULL    |       |
+-------+-------------+------+-----+---------+-------+

INSERT INTO tb1 (fld1, fld2, fld3) SELECT tb2.fld1, tb2.fld2, 
    tb2.fld3 FROM tb2;  

適用於version 5.1.49 ，答案是

$sql = "INSERT INTO table1 (fld1, fld2, fld3) SELECT table2.fld1, ".
    table2.fld2, table2.fld3 FROM table2 WHERE table2.id= ".$somevalue;

UPDATE不需要檢查id，它會自動檢查，如果沒有id那么，你會追加null

INSERT INTO table2 (fld1, fld2, fld3) SELECT table1.fld1, table1.fld2, table1.fld3 
FROM table1 where table1.id = someid ;