Loop through database...

**Muk108** · Mar 12th, 2003, 08:18 PM

Ok im back again and i am still trying to loop through my database in a template. is this how i would loop through it? i need to loop through the ID of the item first then while its still doing that print out the data?

so

loop {

$dbid

loop {

$dbtext

}

}

or something like that? do you think you could help me put this in php?

**phpman** · Mar 12th, 2003, 09:28 PM

loop through the id first?? huh you can't do that.

just loop through the whole table do this

PHP Code:


$query=("select * from table")



while ($row= musql_fetch_array($query){

echo $row["id"];

echo $row["field1"];

}

that will loop through the whole db and display every ID and FIELD1

so what are you having trouble with? you said a template?

**Muk108** · Mar 13th, 2003, 03:35 PM

well im trying to use a template engine like fasttemplate so i can have someone do the html and stuff and i just work on the php part. but whenever i do it, it only displayes one item from the database.

**The Hobo** · Mar 13th, 2003, 03:49 PM

Originally posted by Muk108
well im trying to use a template engine like fasttemplate so i can have someone do the html and stuff and i just work on the php part. but whenever i do it, it only displayes one item from the database.

You'd probably have to post your code before we can find the bug...

**Muk108** · Mar 13th, 2003, 05:01 PM

PHP Code:


<?

include("db.php");

include("class.FastTemplate.php3");

if ($complete != 1) 

    $news = mysql_query($query); 

    while ($row = mysql_fetch_array($news)) {

      $title_text = $row['main_title'];

    };

    //

    

          $menu = "SELECT * FROM menu_left ORDER BY menu_order";



          if ($complete != 1) 

          $menu_left = mysql_query($menu); 

          while ($menurow == mysql_fetch_array($menu_left)) {



           if ($menurow['menu_show'] == 1) {

           

            $menu_link[] = $menurow['menu_link'];

            $menu_name[] = $menurow['menu_name'];



           } else {

            echo "";

           };

          }

          

          //

        $newsquery = "SELECT * FROM news";



    if ($complete != 1) 

        $news = mysql_query($newsquery); 

        while ($row = mysql_fetch_array($news)) {

         $news_title_text = $row['news_title'];

         $news_message_text = $row['news_message'];

         $news_submitter_text = $row['news_submitter'];

         $news_date_text = $row['news_date'];

        };

        //



$tpl = new FastTemplate("./templates");

  $tpl->define(array(

        'main'                   => "index.tpl",

      )

  );

  //

  

    $tpl->assign(

        array(

            'title'                  =>  $title_text,

            'news_title_text'        =>  $news_title_text,

             'news_message_text'      =>  $news_message_text,

             'news_submitter_text'    =>  $news_submitter_text,

             'news_date_text'         =>  $news_date_text,

            'menu_link'              =>  $menu_link,

            'menu_name'              =>  $menu_name,

            'message_area'           =>  $message_area,)

        );

          $tpl->parse('MAIN', array("main"));



$tpl->FastPrint();

//$tpl->showDebugInfo();

exit;

?>

thats what i have, i have been working with this for a long time, hope you can help me.

**phpman** · Mar 13th, 2003, 05:44 PM

is this it?

PHP Code:


while ($row = mysql_fetch_array($news)) {

         $news_title_text = $row['news_title'];

         $news_message_text = $row['news_message'];

         $news_submitter_text = $row['news_submitter'];

         $news_date_text = $row['news_date'];

        };

that will only get the last info from teh db, so the last id will only be sent. I thought we fixed this a while ago.

**Muk108** · Mar 13th, 2003, 05:46 PM

no, i dont think we did, or we did and when i tryed putting it in it didnt work. could you help me with this one?

**The Hobo** · Mar 13th, 2003, 06:26 PM

This is a problem right here:

Code:

while ($menurow == mysql_fetch_array($menu_left)) {

Should only be one = sign.

**Muk108** · Mar 13th, 2003, 06:37 PM

that doesnt seem to matter, i still only get one resault printed out on the page.

**The Hobo** · Mar 13th, 2003, 10:03 PM

Originally posted by Muk108
that doesnt seem to matter, i still only get one resault printed out on the page.

It may not fix your problem, but yes, it does matter, as that's not how you do it.

**Muk108** · Mar 13th, 2003, 10:20 PM

ok well, thank you. i have tryed a for loop at

PHP Code:


    $tpl->assign(

        array(

            'title'                  =>  $title_text,

            'news_title_text'        =>  $news_title_text,

             'news_message_text'      =>  $news_message_text,

             'news_submitter_text'    =>  $news_submitter_text,

             'news_date_text'         =>  $news_date_text,

            'menu_link'              =>  $menu_link,

            'menu_name'              =>  $menu_name,

            'message_area'           =>  $message_area,)

        );

this point. but still no luck.

**The Hobo** · Mar 13th, 2003, 10:41 PM

I think phpman hit the nail on the head. You need to do something like this:

Code:

<?php
    include("db.php");
    include("class.FastTemplate.php3");

    if ($complete != 1) 
        $news = mysql_query($query); 
    while ($row = mysql_fetch_array($news)) {
        $title_text[] = $row['main_title'];
    }
    //
        
    $menu = "SELECT * FROM menu_left ORDER BY menu_order";

    if ($complete != 1) 
       $menu_left = mysql_query($menu); 

    while ($menurow == mysql_fetch_array($menu_left)) {
       if ($menurow['menu_show'] == 1) {
               
            $menu_link[] = $menurow['menu_link'];
            $menu_name[] = $menurow['menu_name'];

        } else {
            echo "";
        }
    }
              
    $newsquery = "SELECT * FROM news";

    if ($complete != 1) 
        $news = mysql_query($newsquery); 
            
    while ($row = mysql_fetch_array($news)) {
        $news_title_text[] = $row['news_title'];
        $news_message_text[] = $row['news_message'];
        $news_submitter_text[] = $row['news_submitter'];
        $news_date_text[] = $row['news_date'];
    }

    $tpl = new FastTemplate("./templates");
    $tpl->define(array('main' => "index.tpl",)); 
    // ^-- what's that comma doing in there?

        
    for ($i = 0; $i < count($title_text); $i++) {
        $tpl->assign(
            array(
            'title'                  =>  $title_text[$i],
            'news_title_text'        =>  $news_title_text[$i],
            'news_message_text'      =>  $news_message_text[$i],
            'news_submitter_text'    =>  $news_submitter_text[$i],
            'news_date_text'         =>  $news_date_text[$i],
            'menu_link'              =>  $menu_link[$i],
            'menu_name'              =>  $menu_name[$i],
            'message_area'           =>  $message_area[$i],)
            //^--what's up with that last comma?
        );

        $tpl->parse('MAIN', array("main"));

        $tpl->FastPrint();
        //$tpl->showDebugInfo();
    }

    exit;
?>

Give it a try. I believe it will work

**phpman** · Mar 14th, 2003, 12:31 AM

exactly, Hobo.

see that is what you needed to do, Muk108. jsu tlike this thread where you just left it hanging

http://www.vbforums.com/showthread.p...hreadid=228334

you never said it didn't work and we told you what the problem was. and again you ask the same question and got the same answer.

**The Hobo** · Mar 14th, 2003, 08:18 AM

Originally posted by phpman
see that is what you needed to do, Muk108. jsu tlike this thread where you just left it hanging

http://www.vbforums.com/showthread.p...hreadid=228334

you never said it didn't work and we told you what the problem was. and again you ask the same question and got the same answer.

I guess when you don't like the answer, you can ask again later and see if you get a different one.

Sadly, for him, this is the answer.

**Muk108** · Mar 15th, 2003, 09:07 PM

im sorry i havent replyed, i was buisy the last two days. and when i execute that i go this error....

Code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\HTTP\www\intra\index.php on line 71

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\HTTP\www\intra\index.php on line 81

and it just loops through that locking up my server.

**Muk108** · Mar 15th, 2003, 09:23 PM

ok i got the page to load, but now for ME, it doesnt want to loop through and print it out again...

**The Hobo** · Mar 17th, 2003, 06:54 PM

Originally posted by Muk108
ok i got the page to load, but now for ME, it doesnt want to loop through and print it out again...

I don't know what to tell you anymore. Try finding some basic tutorials first. Or maybe programming concepts.

Not to overuse a cliche, but we can only lead you to water...

**Muk108** · Mar 17th, 2003, 08:38 PM

i understand i have been looking at tutorials and havnt found anything. ill keep looking and when i find it ill post it. thanks for all your guys help.

**phpman** · Mar 17th, 2003, 10:40 PM

my guess is that you messed up the query when you got the mysql_fetch_arrary errors.

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