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benanamen

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Everything posted by benanamen

  1. benanamen
    First, you are using obsolete Mysql code that will not work at all in the latest version of Php. You need to use PDO with prepared statements. Second, get rid of all the @'s. DO NOT SUPPRESS ERRORS. Errors are your friend, they tell you when something is wrong. And lets not forget about you jumping case all over the place. Always use lower case names. Why is your table name a variable? Are you going to insert those exact column names into more than one table? In your error message there is a missing quote.
  2. benanamen
    Your <tr></tr> needs to be IN the loop. not outside if it.
  3. benanamen
    You only need ONE query and ONE while loop. Select all the columns you want from your table at one time. SELECT m_title, sub_date, author, tutorial, status FROM methods while{ }
  4. benanamen
    dardime, it is seriously time to stop using deprecated Mysql code. It is not going to work at all in the latest version of Php no matter what you do. Those functions simply no longer exist in Php. You need to use PDO with prepared statements or at the very least mysqli.
  5. benanamen
    Just an FYI, you dont have to set both height and width. You can set one or the other. The image will scale proportionally.
  6. benanamen
    I am not really sure what you are trying to do. The easiest way to control the size of a "Large" image is to set the height and width in your img src tag. I dont know what that means. To the left? To the right? Center? Next to text? Above text? Below text?
  7. benanamen
    A bootstrap panel will scale in height automatically. You don't need CSS. The problem must be with your plugin. <html> <head> <meta charset="utf-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <script type="text/javascript" src="http://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> <script type="text/javascript" src="http://netdna.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script> <link href="http://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.3.0/css/font-awesome.min.css" rel="stylesheet" type="text/css"> <link href="http://pingendo.github.io/pingendo-bootstrap/themes/default/bootstrap.css" rel="stylesheet" type="text/css"> </head> <body> <div class="panel panel-default"> <div class="panel-body"> <img src="http://www.planwallpaper.com/static/images/Winter-Tiger-Wild-Cat-Images.jpg" alt=""> </div> </div> </body> </html>
  8. benanamen
    At present you only get weeks for which there is data That is what I was thinking was happening. Ok, now I am back to being completely lost again. For what that means, will that will also suffice for the dried part? Just to be clear, it may not be a year. The time period will be whatever the select start/end date range is which may be a week, months, a year, or cross a year or more. I know I will be studying what you have done for some time to come. You are quite the Mysql Master.
  9. benanamen
    This is not correct. I think we are really close though. Lets forget the count calculations for a moment (They work right). If our start/end date range covers 3 weeks, we should have three rows of data. If our start/end date range covers twenty weeks, we should have 20 rows of data etc, etc.. Pseudo Php but needs in SQL $week_date_range = 5 consecutive week periods that can cross years if needed foreach ($week_date_range AS $week){ // do slab calculation // do dried calculation echo "Slabs=$slab_count, Dried= $dried_count</br>"; // This would give us 5 rows regardless of the calculations and the results } Lets say there is only one record in the lot table and it only has a slab date of Dec 30, 2013. No dried_in date. This record is a slab status from Dec 30, 2013 until today 2016. Our start date per the earliest drop down is is Dec 29, 2013, so we are within the start date using that date. If we were to select the latest end date of Jan 9, 2016, we should get a row for every week starting from Dec 29 2013 with a count of one slab_count per row until the week of Jan 9, 2016. If I counted right that would be 104 rows (weeks) with a count of one slab_count per row. Since it is still a slab to today, it gets counted as a slab every week period up to the end date selected. What they want to know is, every week period that a lot is a slab status it should be counted in that week period and displayed in that week period So if a lot is a slab status during three start/end week periods, we would have three rows with a count of 1 per row. If it was a slab status for 10 start/end week periods we would have 10 rows with a count of one per row. Same thing applies for the dried count per the conditions we have already worked out. Right now with one record with the earliest drop down start date and the latest drop down end date, there is only one row returned for the first week. There should be a row for every week up to the end week selected. The good news is the calculations for slab and dried are good. What we are missing is a per week result row for the selected date ranges. I know it is confusing. I am still barely grasping it.
  10. benanamen
    While Barand did trash my example, I structured it for the OP's apparently limited understanding. I wouldn't call it trashing. Barand told the exact truth. It had nothing to do with it being you in particular that posted it. I firmly believe we should not dumb down an answer that is incorrect due to an OP's lack of understanding, but rather provide the correct answer and dumb that down so they can understand the correct way however long it may take them. Better for them to learn may things from doing the right thing than the wrong thing. Compared to us, ALL the OP's have a limited understanding on the posted subject or they wouldn't be posting for help.
  11. benanamen
    Do you mean i cannot store those check box value with comma in a same row? You can if you want to do it wrong and go against proper database design and create headaches for yourself and any other developers that has to deal with it down the road as well as anyone offering you help. I gave you a starting point. Look up database normalization and learn it. It is a very important foundation to database design and is one of the simplest things you can master in as little as 10 minutes. Come back with your normalized structure if you are still stuck and we will be happy to help you. * You also didn't post your code, just the form. We cant review what we cant see.
  12. benanamen
    Your doing it wrong. You dont store the data like that in the same column. You need another table with a foreign key to the first table and do a JOIN. Look up database normalization.
  13. benanamen
    It should work with just slab dates only. If the 3 records are slab status, their would not be dried_in dates or complete dates for those 3 records so, I should have a count of 3 for slabs and a count of 0 for dried ins. Dried count is zero as expected, but I get 4 slabs and there are only 3 slabs (three records with a slab date only.)
  14. benanamen
    Not sure what was going on before. I am getting a result with the entire query I posted with changes in #8 (Minus the comments), but, I am getting a slab count of 4. There are only three records in the DB so it should not be possible to count more slab dates than exist. Even still, running the query from post #9 does give a correct count of 3. By the way, why do you put the commas from the previous line at the start of the next lines after it. I find that very confusing to work with.
  15. benanamen
    Ok, so if I take the query apart and run the slab count portion I get the expected result of 3, but it doesnt work as a whole. SELECT @day1 := '2014-01-04' as init1, @day2 := '2016-01-16' as init6, @slab := (SELECT count(lot_id) FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE slab_date BETWEEN @day1 AND @day2 AND IFNULL(dried_in_date,'9999-12-31')>=@day1) as init2
  16. benanamen
    Here is my guess which does not work. I get no results. I expect to get a slab count of 3 crossing three years. (Not in code tags so I can highlight my changes) My slab dates are 2014-01-15, 2015-01-15, and 2016-01-15 SELECT s.week , CONCAT(DATE_FORMAT(@day1 + INTERVAL (s.week-1)*7 DAY, '%e %b'),' - ', DATE_FORMAT(@day1 + INTERVAL (s.week-1)*7 + 6 DAY, '%e %b')) as wkcomm , @slab:=@slab+ifnull(slabcount,0)-@driedcf as slab_count , @dried:=@dried+ifnull(driedcount,0)-@tiledcf as dried_count , @driedcf:=ifnull(driedcount,0) as x1 , @tiledcf:=ifnull(tiledcount,0) as x2 FROM ( SELECT @wk := 0 as init0 , @day1 := '2014-01-04' as init1 // Start Date , @day2 := '2016-01-16' as init6 // End Date , @slab := (SELECT count(lot_id) FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE slab_date BETWEEN @day1 AND @day2 AND IFNULL(dried_in_date,'9999-12-31')>=@day1) as init2 , @dried := (SELECT COUNT(lot.lot_id) FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) LEFT JOIN schedule s ON lot.lot_id=s.lot_id AND work_type_id=1 AND complete_date < @day1 WHERE dried_in_date < @day1 AND s.lot_id IS NULL) as init3 , @driedcf := 0 as init4 , @tiledcf := 0 as init5 ) init INNER JOIN ( SELECT COUNT(lot_id) as slabcount , DATE_FORMAT(slab_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE DATE_FORMAT(slab_date, '%X') = DATE_FORMAT(@day1, '%X') GROUP BY week ) s LEFT JOIN ( SELECT COUNT(lot_id) as driedcount , DATE_FORMAT(dried_in_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE DATE_FORMAT(dried_in_date, '%X') = DATE_FORMAT(@day1, '%X') AND dried_in_date is not null GROUP BY week ) d using (week) LEFT JOIN ( SELECT COUNT(s.lot_id) as tiledcount , DATE_FORMAT(complete_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) INNER JOIN schedule s ON lot.lot_id=s.lot_id AND work_type_id=1 AND complete_date IS NOT NULL WHERE DATE_FORMAT(complete_date, '%X') = DATE_FORMAT(@day1, '%X') GROUP BY week ) t using (week) ;
  17. benanamen
    Are we looking at the same query? There is NO @date2. Do I need to add it? And if so, I have no idea where it would be added or how it would be referenced to read the date as an end date.
  18. benanamen
    First thing you need is a form for the user to select the start and end dates. This is where you would preset the start and end with the defaults of 3 months ago and now. I would recommend using date pickers, or date select dropdown menus, so you have control over the format of the dates that are submitted. As you are charting in weeks you would also want to ensure that start dates fall on Sundays and end dates fall on Saturdays. (Alternatively, enter a start date and the number of weeks to be plotted in the form instead of two dates). I already understood every bit of that already but thanks for telling me. You would then change the query to accept @date2 as the end date (in the same way that @date1 is set to the start date). When checking dates in the query, it currently checks that dates are in the required year. You would change this to check that the dates fall between @date1 and @date2. The only problem is there is no @date2. For initial testing I shouldn't need a form yet. I should just be able to hard code the start/end dates in the query and only use the console. If I need to add the @date2 somehow, somewhere, I am completely clueless about doing that.
  19. benanamen
    For starters I am just trying to get the query to read across years. This query is way beyond my current knowledge of Mysql. The ONLY line that has any hard coded date to change is: , @day1 := '2015-01-04' as init1 I am testing with three lot records that are slabs only. One each of Jan 2014,2015, and 2016. If I change the only hard coded date to the start of week one in 2014 (2015-01-05) I expect to get a slab count of three. I only get a slab count of one which is the one record in 2014. Same thing if I change it to 2015, I should get a count of two, but I get one (Expect the one slab in 2015 and the one slab in 2016) I have no idea whatsoever of what to do to be able to set a start week to start counting from, let alone an end week. As is the query only accurately counts for a specific year. I also noticed that it doesnt care about the day or month. I can put the day and month after the slab date and it still gets counted. I also noticed it requires a complete mysql date to function even though the query seems to only be looking at the year portion.
  20. benanamen
    @Psycho, why did you start with an empty $output?
  21. benanamen
    Weekly Chart Part Two Need default view to be last 3 months based on start date of current week start date Need to be able to dynamically select a start and end week maximum 52 week range from drop-down select Will need to be able to cross years. Current query is for one particular full year. The following query will currently give correct results for one particular full year. What I tried: The following query Results: Correct result for year 2015 New Expected results: See above SELECT s.week , CONCAT(DATE_FORMAT(@day1 + INTERVAL (s.week-1)*7 DAY, '%e %b'),' - ', DATE_FORMAT(@day1 + INTERVAL (s.week-1)*7 + 6 DAY, '%e %b')) as wkcomm , @slab:=@slab+ifnull(slabcount,0)-@driedcf as slab_count , @dried:=@dried+ifnull(driedcount,0)-@tiledcf as dried_count , @driedcf:=ifnull(driedcount,0) as x1 , @tiledcf:=ifnull(tiledcount,0) as x2 FROM ( SELECT @wk := 0 as init0 , @day1 := '2015-01-04' as init1 , @slab := (SELECT count(lot_id) FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE slab_date < @day1 AND IFNULL(dried_in_date,'9999-12-31')>=@day1) as init2 , @dried := (SELECT COUNT(lot.lot_id) FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) LEFT JOIN schedule s ON lot.lot_id=s.lot_id AND work_type_id=1 AND complete_date < @day1 WHERE dried_in_date < @day1 AND s.lot_id IS NULL) as init3 , @driedcf := 0 as init4 , @tiledcf := 0 as init5 ) init INNER JOIN ( SELECT COUNT(lot_id) as slabcount , DATE_FORMAT(slab_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE DATE_FORMAT(slab_date, '%X') = DATE_FORMAT(@day1, '%X') GROUP BY week ) s LEFT JOIN ( SELECT COUNT(lot_id) as driedcount , DATE_FORMAT(dried_in_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) WHERE DATE_FORMAT(dried_in_date, '%X') = DATE_FORMAT(@day1, '%X') AND dried_in_date is not null GROUP BY week ) d using (week) LEFT JOIN ( SELECT COUNT(s.lot_id) as tiledcount , DATE_FORMAT(complete_date, '%V') as week FROM lot INNER JOIN block b USING (block_id) INNER JOIN community c on b.community_id = c.community_id AND c.contract_type_id IN (1,3) INNER JOIN schedule s ON lot.lot_id=s.lot_id AND work_type_id=1 AND complete_date IS NOT NULL WHERE DATE_FORMAT(complete_date, '%X') = DATE_FORMAT(@day1, '%X') GROUP BY week ) t using (week) ;
  22. benanamen
    echo "<tr><td>{$row['username']} {$row['COUNT(*)'] }</td></tr>"; @Jaques1 will tell you about escaping the output. I am just giving you the quick and dirty.
  23. benanamen
    Where you are requiring the fg_member site you need two dots instead of one. It is doing that because you are requiring an include from a subfolder of root that includes an include. I usually put my config file in the root directory and then include the include files to that main config file. Then you just need to include the config from root as needed.
  24. benanamen
    Response I got: Thanks man! You are awesome!
  25. benanamen
    It would be easier if you could post a dump of your DB. As far as the names, just make them unique. states.state_name, recruiter.recruiter_name, counties.county_name Also, you are mixing case around. Always use lower case for your column names, table names and variable names. It will save you trouble in the long run. Always be consistent in your naming You said But your WHERE condition is based on a state id.
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