Displaying 20 checkboxes in three columns

Question

In order to display a database item (table 1) item as check box.
the values of chexk boxes are the results of another table with 20 records (table2), this I will have 20 check box in my form.
I proceed like the following :
I created a region with type is SQL Query ,
the query is :
select x.name, x.code
from (select wwv_flow_item.display_and_save(1, table2.name) as name, htmldb_item.checkbox (2, table2.code, null, decode (table1.code_..., null, 0, table2.code)) as code
from table2 table2 left outer join table1 table1
on table2.code = table1.code_..and table1.code_ = :page_CODE) x
ANYWAY, the I have my result, BUT ,
because I have 20 cheeck boxex, I want to display them
in a way that they appear in three columns.
How can I change my query or how can I do it???????
NOTE : using Lists is not a solution for me!!!!
Thanks for any Help!

Answer 1

First "392451", asking Raj the same question twice within a span of 16 hours will probably not get your question answered any faster. He is only one man and has a lot of questions coming in, not to mention other job responsibilities. Second, he's on the West coast, so is not up yet.
Now on to your answer. I've used this phrase before, and it definitely applies here: This is not an HTML DB 101 answer. You asked a challenging question (and a good one at that), so the answer may challenge you a bit too. I'll use the emp table which has 14 rows for this example, but this should work for any table.
First, lets get the empno and ename columns to use for our checkboxes. I'm going to create 2 additional columns, col_num and grp_by to use in the next step. The text description of "col_num" from the inside-out is: take the rownum, now divide it by 3 and give me the remainder (mod(n,3)) which will return 0,1, or 2, then replace any of the 0's with 3's so it will return 1,2 or 3. We will use these numbers for our 3 columns in the next step.
The grp_by column is simply the rownum divided by 3, rounded up to the next highest integer. The results of rownum divided by 3 for the first 3 rows are .33, .66, and 1. When you apply ceil(n) to these, they all become 1. This will give us a column to group by in the final step.
All of these queries can be run in SQL Workshop, assuming the schema you run it as has access to the emp table.
OK, let's start with the first query:
select empno,ename,
       replace(mod(rownum,3),0,3) col_num /* alias to col_num */,
       ceil(rownum/3) grp_by /* alias to grp_by */
      from(
            select empno,ename
              from emp
             order by ename)
     EMPNO ENAME      COL_NUM                                      GRP_BY
      7876 ADAMS      1                                                 1
      7499 ALLEN      2                                                 1
      7698 BLAKE      3                                                 1
      7782 CLARK      1                                                 2
      7902 FORD       2                                                 2
      7900 JAMES      3                                                 2
      7566 JONES      1                                                 3
      7839 KING       2                                                 3
      7654 MARTIN     3                                                 3
      7934 MILLER     1                                                 4
      7788 SCOTT      2                                                 4
      7369 SMITH      3                                                 4
      7844 TURNER     1                                                 5
      7521 WARD       2                                                 5Now, let's pivot the rows to columns. Any row with a value of col_num = 1 will be in the first column, col_num=2 will be in the second column, etc...
select decode(col_num,1,empno||'-'||ename) col1,
       decode(col_num,2,empno||'-'||ename) col2,
       decode(col_num,3,empno||'-'||ename) col3
  from(
    select empno,ename,
           replace(mod(rownum,3),0,3) col_num /* alias to col_num */,
           ceil(rownum/3) grp_by /* alias to grp_by */
          from(
                select empno,ename
                  from emp
                 order by ename))
COL1         COL2        COL3
7876-ADAMS 
             7499-ALLEN
                         7698-BLAKE
7782-CLARK 
             7902-FORD
                         7900-JAMES
7566-JONES 
             7839-KING
                         7654-MARTIN
7934-MILLER
             7788-SCOTT
                         7369-SMITH
7844-TURNER
             7521-WARDNow, we need to "compress" our results so the empty gaps are filled in. This is where the grp_by column comes in. I'll add a max() around the the_col1 etc column so I can use a group by on the query. I will then group by the grp_by column:
select max(decode(col_num,1,empno||'-'||ename)) col1,
       max(decode(col_num,2,empno||'-'||ename)) col2,
       max(decode(col_num,3,empno||'-'||ename)) col3
  from(
    select empno,ename,
           replace(mod(rownum,3),0,3) col_num /* alias to col_num */,
           ceil(rownum/3) grp_by /* alias to grp_by */
          from(
                select empno,ename
                  from emp
                 order by ename))
group by grp_by
COL1         COL2          COL3
7876-ADAMS   7499-ALLEN    7698-BLAKE
7782-CLARK   7902-FORD     7900-JAMES
7566-JONES   7839-KING     7654-MARTIN
7934-MILLER  7788-SCOTT    7369-SMITH
7844-TURNER  7521-WARD    Now, simply replace the empno concatenated to ename with a call to htmldb_item.checkbox. Use the "col_num" column for the value p_idx,and the empno column for the value of p_value. Then concatenate in initcap(ename) so the checkbox has a label:
select max(decode(col_num,1,htmldb_item.checkbox(col_num,empno)||initcap(ename))) col1,
       max(decode(col_num,2,htmldb_item.checkbox(col_num,empno)||initcap(ename))) col2,
       max(decode(col_num,3,htmldb_item.checkbox(col_num,empno)||initcap(ename))) col3
  from(
    select empno,ename,
           replace(mod(rownum,3),0,3) col_num /* alias to col_num */,
           ceil(rownum/3) grp_by /* alias to grp_by */
          from(
                select empno,ename
                  from emp
                 order by ename))
group by grp_by
(Can't display the results here.  Try in SQL Workshop.)Good luck,
Tyler