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PHP and Database To Generate Report

Posted on 2003-03-27
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I am trying to generate a report that accepts two text entries (start_date and end_date) via a php page and uses these values to query a postgresql database.  My primary objective of this question is to find the most appropriate/accurate way to handle the date range identification and processing so providing thoughts/examples of the necessary loops and arrays is desired. Use any combination of PHP and SQL.

Database table/fields necessary are:

table: tbleventlog
field: id
field: dtdatetime
field: iusername

The input/output (note: users can input any valid start and end dates - no pulldowns, calendars, etc):

Each column of output must represent one week (sunday through saturday mandatory).The trick is to
capture the end_date the user submitted and if it isn't a saturday, find the next calendar saturday and then begin working backward "week-by-week" until we get to the start_date*.  However, the start_date must be a sunday, so if the start_date the user input isn't a sunday, then we need to find the sunday immediately preceeding the start_date supplied by the user.  This functionality must be capable of spanning months and years.

For example: a user submits start-date of 02/18/03 and end_date: 03/07/03.
** 02/18/03 is a tuesday and 03/07/03 is a friday **.  Therefore, the end_date must be redefined as
03/08/03 because that's the next saturday.  Likewise, the start_date  of 02/18/03 will have to be
redefined as 02/16/02 because that's the preceeding saturday.

The report output should resmble the following:

                                       Weekly Logins                
|username                | 02/16/ - 02/22 | 02/23 - 03/01 | 03/02 - 03/08 | total  
|---------------------------------------------------------------------------------------------------|
|johndoe@123.com | 12                    |  0                     |  3                    | 15         |
|janewho@987.com |  3                    |  4                     | 13                    | 20         |
|danwhat@345.com |  1                    | 12                    |  9                    | 22         |
|jimwhen@abc.com |  0                    |  7                     | 15                    | 22         |
|---------------------------------------------------------------------------------------------------|
Thanks and good luck. (Many points for the headache).  Ask for additional details if I haven't been thorough enough.
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Question by:tewald
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11 Comments
 
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Expert Comment

by:drnadeem
ID: 8232930
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by:drnadeem
ID: 8233001
listening
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Expert Comment

by:bljak
ID: 8233457
If i am not wrong, the input with date start-end will be used to create new tables, and the table you have given above is somewhat output. You said they can enter date by themselves. Question is also, do you have valid date check? And where would you love to get those login numbers from?

//bljak
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Author Comment

by:tewald
ID: 8242402
bljak, no tables are created in this example.  The table (tbleventlog) is queried for the information shown in the sample output.  Date validation is not critical but for the sake of this example, let's ensure the user inputs the dates in m/d/Y (03/31/2002) format. Good point about the login numbers. Here the revised tbleventlog:

table: tbleventlog
field: id
field: eventid
field: dtdatetime
field: iusername

ieventid: 4 (Login)

Thanks,

Thomas
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LVL 40

Expert Comment

by:Richard Quadling
ID: 8244314
No code, but hopefully a method to acheive what is required.

Essentially, sub-selects are required.

This is via Access (it is on the pc!).

SELECT DISTINCT
      Actions.name,
      [2003-03-01].CountOfid,
      [2003-03-02].CountOfid,
      [2003-03-03].CountOfid
FROM
      ((Actions
            LEFT JOIN [2003-03-01] ON Actions.name = [2003-03-01].name)
            LEFT JOIN [2003-03-02] ON Actions.name = [2003-03-02].name)
            LEFT JOIN [2003-03-03] ON Actions.name = [2003-03-03].name
ORDER BY Actions.name;

This is the main query.

The queries [2003-03-01], etc are ...

SELECT DISTINCT
      Actions.name,
      Count(Actions.id) AS CountOfid
FROM
      Actions
GROUP BY
      Actions.name,
      Actions.date
HAVING
      Actions.date=#3/1/2003#
ORDER BY
      Actions.name;

And the HAVING clause being different for each day.

So. This is not yet ideal as this requires the queries to already exist and then you pull all of it together in 1 hit.

So, adding them all together is a little messy but ...

SELECT DISTINCT Actions.name, [2003-03-01].CountOfid, [2003-03-02].CountOfid, [2003-03-03].CountOfid
FROM ((Actions
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/1/2003# ORDER BY Actions.name) AS [2003-03-01] ON Actions.name = [2003-03-01].name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/2/2003# ORDER BY Actions.name) AS [2003-03-02] ON Actions.name = [2003-03-02].name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/3/2003# ORDER BY Actions.name) AS [2003-03-03] ON Actions.name = [2003-03-03].name
ORDER BY Actions.name;

So. with my data looking like ...

id      name      state      date
1      name1      email1      01/02/2003
2      name2      inactive      02/02/2002
3      name3      new      23/06/2003
4      name4      active      12/08/2002
5      name2      new      15/03/2003
6      name1      new      16/03/2003
7      name1      inactive      11/10/2002
10      Richard      new      01/03/2003
11      Richard      active      02/03/2003
12      Alan      new      02/03/2003
13      Mike      new      03/03/2003
14      John      new      04/03/2003
15      John      active      06/03/2003
16      Mike      active      07/03/2003
17      Richard      expired      08/03/2003
18      Alan      active      09/03/2003

I get results of ...

name      2003-03-01.CountOfid      2003-03-02.CountOfid      2003-03-03.CountOfid
Alan            1      
John                  
Mike                  1
name1                  
name2                  
name3                  
name4                  
Richard      1      1      

You basically get ALL the users you have in the table and how many entries for the chosen dates.

Ok. This is the query, adding the rest of the code to do all of this is pretty basic, but is this a good enough start for you?

Just to be a little more helpful query, this is the same with the names of the sub-queries being daynames.

SELECT DISTINCT Actions.name, Sun.CountOfid, Mon.CountOfid, Tue.CountOfid, Wed.CountOfid, Thu.CountOfid, Fri.CountOfid, Sat.CountOfid
FROM ((((((Actions
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/2/2003# ORDER BY Actions.name) AS Sun ON Actions.name = Sun.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/3/2003# ORDER BY Actions.name) AS Mon ON Actions.name = Mon.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/4/2003# ORDER BY Actions.name) AS Tue ON Actions.name = Tue.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/5/2003# ORDER BY Actions.name) AS Wed ON Actions.name = Wed.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/6/2003# ORDER BY Actions.name) AS Thu ON Actions.name = Thu.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/7/2003# ORDER BY Actions.name) AS Fri ON Actions.name = Fri.name)
LEFT JOIN (SELECT DISTINCT Actions.name, Count(Actions.id) AS CountOfid FROM Actions GROUP BY Actions.name, Actions.date HAVING Actions.date=#3/8/2003# ORDER BY Actions.name) AS Sat ON Actions.name = Sat.name)
ORDER BY Actions.name;

You can add ...

, SUM(Sun.CountOfid, Mon.CountOfid, Tue.CountOfid, Wed.CountOfid, Thu.CountOfid, Fri.CountOfid, Sat.CountOfid) as WeeklyTotal

to the list of fields, but this won't work in access.

Regards,

Richard Quadling.
0
 
LVL 40

Expert Comment

by:Richard Quadling
ID: 8244359
Slightly better laid out.

SELECT DISTINCT
     Actions.name,
     Sun.CountOfid,
     Mon.CountOfid,
     Tue.CountOfid,
     Wed.CountOfid,
     Thu.CountOfid,
     Fri.CountOfid,
     Sat.CountOfid,
     SUM(
          Sun.CountOfid,
          Mon.CountOfid,
          Tue.CountOfid,
          Wed.CountOfid,
          Thu.CountOfid,
          Fri.CountOfid,
          Sat.CountOfid) as WeeklyTotal
FROM ((((((
     Actions
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/2/2003#
          ORDER BY
               Actions.name) AS Sun
          ON
               Actions.name = Sun.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/3/2003#
          ORDER BY
               Actions.name) AS Mon
          ON
               Actions.name = Mon.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/4/2003#
          ORDER BY
               Actions.name) AS Tue
          ON
               Actions.name = Tue.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/5/2003#
          ORDER BY
               Actions.name) AS Wed
          ON
               Actions.name = Wed.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/6/2003#
          ORDER BY
               Actions.name) AS Thu
          ON
               Actions.name = Thu.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/7/2003#
          ORDER BY
               Actions.name) AS Fri
          ON
               Actions.name = Fri.name)
     LEFT JOIN (
          SELECT DISTINCT
               Actions.name,
               Count(Actions.id) AS CountOfid
          FROM
               Actions
          GROUP BY
               Actions.name,
               Actions.date
          HAVING
               Actions.date=#3/8/2003#
          ORDER BY
               Actions.name) AS Sat
          ON
               Actions.name = Sat.name)
ORDER BY
     Actions.name;

This query can be constructed with different dates for different weeks.
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LVL 14

Accepted Solution

by:
psadac earned 2000 total points
ID: 8249274
If your database is PostgreSQL (good choice), and you have installed pl/pgsql on your database, then run the code below in your psql client. Then :
 
select my_report('view_name','2003-02-18','2003-03-07');

will generate the view 'view_name' and :

select * from view_name;

will display the results.
I've tested it on PostgreSQL 7.3.2 (latest) but it should work with earlier versions.
the php code needed to display this is easy, but if you have any difficulties let me know.

CREATE OR REPLACE FUNCTION my_report(TEXT, DATE, DATE) RETURNS TEXT AS '
    -- create a view with one week per column based on table :
    --      CREATE TABLE tbleventlog (
    --           id SERIAL;
    --          dtdatetime TIMESTAMP;
    --          iusername TEXT;
    --     )
    --
    -- Copyleft 2003, GNU GPL V2, by polo (alias cryptic SQL generator :)
    --
    DECLARE
     _view  ALIAS FOR $1; -- name of the view to create
     _date1 ALIAS FOR $2; -- start date of the report
     _date2 ALIAS FOR $3; -- end date of the report
     datefdow DATE; -- date of the First Day Of Week
     dateldow DATE; -- date Of the Last  Day Of Week
     sql TEXT; -- sql code to be executed
     
    BEGIN
     -- first drop the view if it exists
     EXECUTE ''SELECT viewname FROM pg_views WHERE viewname = '' || quote_literal(_view) || '';'';
     IF FOUND THEN
         EXECUTE ''DROP VIEW '' || quote_ident(_view) || '';'';
     END IF;
     -- initiate the sql string for view creation
     sql:= ''CREATE VIEW '' || quote_ident(_view) || '' AS SELECT iusername,'';
     -- datefdow gets the first day of week
     datefdow := _date1 - CAST(EXTRACT(dow FROM _date1) || '' days'' AS INTERVAL);
     -- iterate until _date2
     WHILE datefdow <= _date2 LOOP
         -- setting the last day for week
         dateldow = datefdow + INTERVAL ''6 days'';
         sql := sql || '' SUM(CASE WHEN CAST(dtdatetime AS DATE) BETWEEN '' || quote_literal(datefdow) || '' AND '' || quote_literal(dateldow) || '' THEN 1 ELSE 0 END) AS "'';
         sql := sql || TO_CHAR(datefdow,''MM/DD - '') || TO_CHAR(dateldow, ''MM/DD'') || ''",'';
         -- next week
         datefdow := datefdow + INTERVAL ''7 days'';
     END LOOP;
     -- add the trailing statements
     sql := sql || '' COUNT(*) AS TOTAL FROM tbleventlog GROUP BY iusername ORDER BY iusername;'';
     -- create the view
     EXECUTE sql;
     -- return the sql generated (for debugging purposes)
     RETURN sql;
    END;
' LANGUAGE 'plpgsql';
0
 
LVL 40

Expert Comment

by:Richard Quadling
ID: 8252189
Server side scripting is a great way to do this sort of thing.

Sub selects don't seem to work quite properly yet in MySQL V4.1 and you can't save the query in Access.

Richard.
0
 
LVL 40

Expert Comment

by:Richard Quadling
ID: 8252190
Server side scripting is a great way to do this sort of thing.

Sub selects don't seem to work quite properly yet in MySQL V4.1 and you can't save the query in Access.

Richard.
0
 
LVL 2

Author Comment

by:tewald
ID: 8253698
Good feedback so far - give me a day or so to try these methods out. Thanks, tewald
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LVL 2

Author Comment

by:tewald
ID: 8299974
Thanks for the tips.
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