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Jessica

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Everything posted by Jessica

  1. Jessica
    You have $Lid=$_POST['Lid']; and you're asking how you would get the posted value of Cid as well? ..... Did you try anything yet? Regardless they'll be the same value. var LocationString = 'Lid='+ $(this).val(); var CityString = 'Cid='+ $(this).val(); .
  2. Jessica
    I was going to ask that, but if it were undefined it should trigger the error, printing out the SQL, making it obvious.
  3. Jessica
    Define not working. Did you get the error?
  4. Jessica
    Does the function return an array which you then put into print_r or does the function actually call print_r?
  5. Jessica
    Well for one thing you didn't attach a photo. I'm confused how you have so many posts on here and have to ask such a poorly worded question? But anyway. Google php explode. That's the only thing I can recommend with the limited specifications.
  6. Jessica
    Only thing left to do is mark is solved, which I just did
  7. Jessica
    I think you guys need to quote some strings, unless you're actually trying to use column names. which I don't think you can do in an IN clause...
  8. Jessica
    Isn't 10 still beta???
  9. Jessica
    And use ONE query. Not three.
  10. Jessica
    But your code does look better. Don't use backticks or quotes on integers.
  11. Jessica
    Google isn't broken for me. I google mysql in and get great tutorials. If you want to actually write an entire application you need to learn the skills to learn.
  12. Jessica
    You have too many ( as well.
  13. Jessica
    Yeah I'm showing you WHERE your problem is. Look at your code. while($rows(mysql_fetch_array($result)) { Look at it.
  14. Jessica
    And did you look up using IN in a MYSQL query?
  15. Jessica
    For the future, uniform indenting would help. This is your problem: while($rows(mysql_fetch_array($result)) { Look at that line and tell me if you see the multiple errors.
  16. Jessica
    Did you try looking up implode in the manual yet? That was quick By using IN in your query you can run only 1 query. You don't need to run an update for each child separately.
  17. Jessica
    Well if you want to write an application, it's time to learn. Arrays are very useful and basic part of programming. If you've never used a function look it up and learn how. Showing us one include statement is useless. This forum is for help with code you write. We don't write it for you, you don't learn that way.
  18. Jessica
    A. Percentage is not the same as a fraction. B. It's a bit easier to go the other way.
  19. Jessica
    Well you already have the checkboxes right? So at the top of your page you need to check if your submit button has been pushed. But first you have to give it a name. In fact all of your form elements need names. The checkboxes should be something like name="children[$cid]". Then you can use run a MySQL UPDATE using IN() and php's implode.
  20. Jessica
    Is there a question? If not, I can move this to critiques.
  21. Jessica
    Try checking for MySQL errors.
  22. Jessica
    It's not hard to avoid the refresh problem - when done processing, use header to send them back to the same page. Just like if you use a separate page. ...
  23. Jessica
    QQ!!! Yeah, darn those people who want you to actually do some work before you get their time for free. I'll fix it for you right now, my consultant fee is $100/hr, min 2 hours.
  24. Jessica
    Well you never try to run the query. mysql_query Awwww.
  25. Jessica
    Google "data normalization". You need to store your data differently, as you've discovered the way you're doing it causes problems as soon as you want to actually use the data.
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