Values of generated column not appearing in table

Question

I'm using mysql version 5.1.41-3ubuntu12.10 (Ubuntu).

mysql> show create table tt\G
*************************** 1. row ***************************
       Table: tt
Create Table: CREATE TABLE `tt` (
  `pz` int(8) DEFAULT NULL,
  `os` varchar(8) DEFAULT NULL,
  `uz` int(11) NOT NULL,
  `p` bigint(21) NOT NULL DEFAULT '0',
  `c` decimal(23,0) DEFAULT NULL,
  KEY `pz` (`pz`),
  KEY `uz` (`uz`),
  KEY `os` (`os`),
  KEY `pz_2` (`pz`,`uz`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1
1 row in set (0.00 sec)

mysql> select pz,uz,pz*uz,
    -> if(pz*uz,1,.5),
    -> left(pz,2) pl,left(lpad(uz,5,0),2) ul,
    -> p from tt limit 10;
+-------+----+-------+----------------+--------+----+--------+
|   pz  | uz | pz*uz | if(pz*uz,1,.5) |   pl   | ul |    p   |
+-------+----+-------+----------------+--------+----+--------+
|  NULL |  0 |  NULL |            0.5 | NULL   | 00 |   4080 |
|  NULL |  0 |  NULL |            0.5 | NULL   | 00 | 323754 |
| 89101 |  0 |     0 |            0.5 | 89     | 00 |   6880 |
|     0 |  0 |     0 |            0.5 | 0      | 00 |  11591 |
| 89110 |  0 |     0 |            0.5 | 89     | 00 |     72 |
| 78247 |  0 |     0 |            0.5 | 78     | 00 |     27 |
| 90062 |  0 |     0 |            0.5 | 90     | 00 |      5 |
| 63107 |  0 |     0 |            0.5 | 63     | 00 |      4 |
|  NULL |  0 |  NULL |            0.5 | NULL   | 00 |  54561 |
| 94102 |  0 |     0 |            0.5 | 94     | 00 |  12499 |
+-------+----+-------+----------------+--------+----+--------+

So far so good. As you see, 0.5 appears as a value of if(pz*uz,1,.5). The problem is:

mysql> select os,
    -> if(pz*uz,left(pz,2)<=>left(lpad(uz,5,0),2),.5) uptwo,
    -> if(pz*uz,left(pz,3)<=>left(lpad(uz,5,0),3),.5) upthree,
    -> sum(p) p,sum(c) c
    -> from tt t
    -> group by os,uptwo,upthree order by null;

+----+-------+---------+---------+-------+
| os | uptwo | upthree |    p    |   c   |
+----+-------+---------+---------+-------+
| u  |     1 |       1 |   52852 |   318 |
| i  |     1 |       1 | 7046563 | 21716 |
| m  |     1 |       1 | 1252166 |  7337 |
| i  |     0 |       0 | 1830284 |  4033 |
| m  |     0 |       0 |  294612 |  1714 |
| i  |     1 |       0 |  911486 |  3560 |
| m  |     1 |       0 |  145182 |  1136 |
| u  |     0 |       0 |   12144 |    23 |
| u  |     1 |       0 |    1571 |     8 |
+----+-------+---------+---------+-------+

Although I group by uptwo, 0.5 doesn't appear in that column. What happened to the 0.5 values?

---

Edit: As noted in the comments to Todd Gibson's answer, I also tried it with
if(pz*uz,cast(left(pz,2)<=>left(lpad(uz,5,0),2) as decimal),.5) instead of
if(pz*uz,left(pz,2)<=>left(lpad(uz,5,0),2),.5), but it, too, didn't work.

Answer 1

Instead of .5 as the false condition of the IF(), use 0.5.

if(pz*uz,left(pz,2)<=>left(lpad(uz,5,0),2),0.5) uptwo

if(pz*uz,left(pz,3)<=>left(lpad(uz,5,0),3),0.5) upthree

I believe what is happening is that after the GROUP BY is evaluated, the values in the conditional column must be of the same data type. Since SELECT is evaluated after GROUP BY, the IF() is converting the returned values in favor of an integer (from the boolean expression), and thus your 0.5 gets rounded up to 1 but if you explicitly put a 0 in front of the decimal place, the IF() will treat returned values as a decimal including the result of the boolean expression (i.e. 1.0 or 0.0).

Or you could even put single quotes around the .4 so that the column values will be treated as strings, so some values would appear to be integers and some decimals. The values should be automatically converted when used in numerical contexts (i.e. SELECT ('2.5' * 3.5) AS test #8.75).

Answer 2

I think the uptwo and upthree columns are being implicitly converted to integers. Use a convert() on expression2 to force it to a decimal.

Answer 3

I think you have a problem in the following statements:

-> if(pz*uz,left(pz,2)<=>left(lpad(uz,5,0),2),.5) uptwo,
-> if(pz*uz,left(pz,3)<=>left(lpad(uz,5,0),3),.5) upthree,

In the statement the value of pz*uz must always be greater than or equal to 1.
And hence value of first expression is used to replace over second expression 0.5.

Let us analyse first if function statement:
Syntax: IF(expr1,expr2,expr3)
Meaning if value of expr1 is >= 1 use expr1 else use expr3
Expression1: pz*uz // this value must be '>= 1'.
Expression2: left(pz,2)<=>left(lpad(uz,5,0),2) // this value is either '1' or '0'.
Expression3: .5