Loop through database...

[Muk108]

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]

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

just loop through the whole table do this


$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]

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]

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]

<?
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]

is this it?

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. :confused:

[Muk108]

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]

This is a problem right here:

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

Should only be one = sign.

[Muk108]

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

[The Hobo]

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]

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

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

<?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]

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.php?s=&threadid=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]

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.php?s=&threadid=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]

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

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]

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

[The Hobo]

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]

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]

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