Home/ Questions/Q 3850652
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-19T17:02:34+00:00

ex. c1, c2 = > result c1 c1 result 1 1 to 1000001 2

  • 0


 ex. c1, c2 = >  result

 c1       c1       result
 1         1       to  1000001
 2         9       to  2000009
 3         1       to  3000001
 21        34      to  2100034
 22        35      to  2200035
 23        55      to  2300055
 111       1234    to  1111234
 112       8392    to  1128392
 113       2833    to  1132833


 a part of my MySQL SELECT CONCAT() statement with cut out "c1" look like,


 IF(CHAR_LENGTH(`c2`)=1,  concat('00000',`c2`), 
   IF(CHAR_LENGTH(`c2`)=2, concat('0000',`c2`),
    IF(CHAR_LENGTH(`c2`)=3, concat('000',`c2`),
     IF(CHAR_LENGTH(`c2`)=4, concat('00',`c2`),
      IF(CHAR_LENGTH(`c2`)=5, concat('0',`c2`),`c2` )))))

But is there any other way to reduce this code for concat c1 with c1 into result with zero at the middle and with auto calculation of how many zeros have to be added?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    There might be a math function to help you here, although I don’ t know if it is quicker.

    Lets see. you want to multiply a single value for c1 by 1000000 and then add c2.
    For the more general case, you want to multiply by

      1000000 / 10^(char_length(`c1`)-1)

    more:
    Your final value will be 

    (c1 *  (1000000 / 10^(char_length(`c1`)-1))) + c2

    But that LPAD function from @eumiro might be a better answer

    • 0