有辦法少做但─ 非常基本的策略是通過查詢像你所說的,以配合我們的表列輸入 -
1. SELECT * FROM table WHERE (name='%name%' or zip='%name%' or city='%name%');
2. SELECT * FROM table WHERE LOCATE(name, GROUP_CONCAT(name,city,zip)) > 0;
3.
SELECT * FROM table WHERE name like '%name%'
UNION
SELECT * FROM table WHERE name like '%name%'
UNION
SELECT * FROM table WHERE name like '%name%';
但假設輸入框具有字符串的情況下,"varun bharti"
但數據庫中的實際名稱是"varun bal bharti"
因此,當您搜索時,您將錯過該記錄。對於這種情況,您應該將字符串按空格分解爲數組元素並將這些查詢用於元素,或者您可以替換名稱列中的空格並進行匹配。
set @var=REPLACE ('varun bharti', ' ', '%');
SELECT * FROM table WHERE name like concat('%',@var,'%') or
zip like concat('%',@var,'%') or
city like concat('%',@var,'%');
您也可以使用regualar表達式。 例如輸入字符串the day boss get
Hitesh> select * from test;
+--------------------+
| name |
+--------------------+
| i am the boss |
| You will get soon |
| Happy birthday bro |
| the beautiful girl |
| oyee its sunday |
+--------------------+
5 rows in set (0.00 sec)
Hitesh> set @var=CONCAT('.*',REPLACE('the day boss get',' ','.*|.*'),'.*');
Query OK, 0 rows affected (0.00 sec)
Hitesh> select @var;
+----------------------------------+
| @var |
+----------------------------------+
| .*the.*|.*day.*|.*boss.*|.*get.* |
+----------------------------------+
1 row in set (0.00 sec)
Hitesh> select * from test where name REGEXP @var;
+--------------------+
| name |
+--------------------+
| i am the boss |
| You will get soon |
| Happy birthday bro |
| the beautiful girl |
| oyee its sunday |
+--------------------+
5 rows in set (0.00 sec)
您可以使用多種方法:例如有「賓」和「費城故事」,你可以這樣做:'SELECT * FROM學校WHERE name =「賓」及城市=「費城」 ;'你可以把幾個條件結合在一起。另一種方法:'SELECT * FROM school s1,school s2加入s1.name = s2.name WHERE s1.name ='penn'AND s1.city ='philadelphia';'這裏你也可以使用多個連接不要去它有一個很好的索引,但有一個好的索引有助於後者的健壯 – Hamed
還有另一種方法,你可以使用幾個嵌套的SELECT(不推薦),如下所示:'SELECT * FROM schools WHERE(SELECT * FROM schools WHERE name =' penn')AS s1 AND city ='philadelphia';'。不要對此太深,因爲它會變得討厭。選擇最適合你的東西 – Hamed
@RyanVincent這不是一個真正的答案,更像是一些提示 – Hamed