SELECT Statement with substr in WHERE Clause

Question

I have here sample table with three fields.

TableA

    FieldA        FieldB        FieldC
    ======================================
    123XYZ456     XYZ           John
    124NNN333     NNN           Jenny
    232XPT124     XPT           Jade
    456XXX345     XXX           James

FieldA has a fixedlength of 9. I didn't design this table and some applications are already using it.

I want to select FieldB and FieldC with conditions against FieldA .

Using this sql statement:

SELECT FieldB, 
       FieldC
FROM   TableA
WHERE  FieldA LIKE Concat(@paramA, '%', @paramB) 

I cannot achieve my desired result. When I try to search with paramA value of 12 and paramB value of '' , I get the 2 results:

    FieldA        FieldB        FieldC
    ======================================
    123XXX456     XXX           John
    124XXX333     XXX           Jenny

because obviously it matches with 12% and that is not what I want. I want that the params should match the correct index of the string.

If I search for paramA = '12' and paramB = '' then it should have no result. To get the fields ( FieldB , FieldC ), I need the correct values of paramA = '123' and paramB = '456' so it will return XYZ and John . If I want to return James then i have to give paramA = '456' and paramB = '345'

How could I correctly build a SQL Statement for this? Any idea? Thanks.

Answer 1

Use LEFT() and RIGHT() :

SELECT FieldB, FieldC
FROM   TableA
WHERE  LEFT(FieldA,3) = @paramA 
       AND RIGHT(FieldA,3) = @paramB;

Answer 2

SELECT Field2,Field3
FROM TABLE
WHERE SUBSTR(Field1,1,3)='123' AND SUBSTR(Field1,7,3)='456'

Answer 3

你还没有说清楚为什么你甚至允许paramB在你的输入验证中是空白的，但为了得到你所描述的确切行为，你可以检查一下这种情况，即：

WHERE  @paramA!='' AND @paramB!='' AND FieldA LIKE Concat(@paramA, '%', @paramB) 

Answer 4

Like so: eg classification table: Homes in America, Lands in America etc..

mysql_query("select classification from properties where LEFT(classification, 5) = 'Homes'");

Result: Homes