Devil_Banner
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Thanks alot for helping me out on that one.
Say I wanted to improve and have the user choose in a first dorpdownlist, and by doing so have it populate the second dropdownlist with stuff relevant to the first choice?
Lemme explain :
Say the user chooses "printer" in the first dropdownlist (populated by "ArticleType" column (Article.ArticleType)), then I would want the second dropdownlist (Article.ArticleBrand) to display choices like HP, Epson, and so on...
Is this easy to do ? I mean, dynamically ?
Thanks very much for you guys' time :-) -
Great !!!
I guess that was an easy one, wasn't it ? :)
Lotsa thanks, gterre. It got me stuck for a while.
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Hi,
I'm new at this myself.. ;but maybe you should try a "require" instead of "include" for your variable file
Hope this helps.
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Hi there,
I have this really dumb problem with my query. I have to admit i'm a newbie, hence my stupid problem.
I Use MySQL 5. (Easyphp 1.8 )
I have a table which has a column named ArticleModel.
I Have four entries in this table.
My goal is to select the entries in this table and output them into a dropdown list.
For example, I want the list to contain
LaserJet 4000
Deskjet 1200
Latitude D515
Cisco 2950 series
I can't get further than having 4 times 'array' written in my dropdown list. I have managed to get the first one (Laserjet 4000) to get into the list, but I can't figure out how to loop so it continues to fill my list...
Here's my code:
[color=green]$conn[/color] = [color=blue]mysql_connect[/color]([color=green]$dbHost[/color], [color=green]$dbUser[/color], $dbPassword) or die([color=orange]'DB connexion error'[/color]);
[color=blue]mysql_select_db[/color]($dbName) or die ([color=orange]"DB not reachable"[/color]);
//query
[color=green]$sql[/color] = [color=orange]"SELECT * FROM article "[/color];
[color=green]$query[/color] = [color=blue]mysql_query[/color]([color=green]$sql[/color]);
if (![color=green]$query[/color]) {
// error handler
echo [color=orange]'Query failed. SQL: '[/color], [color=green]$sql[/color], [color=orange]'<br />Error # '[/color], [color=blue]mysql_errno()[/color], [color=orange]' Error msg: '[/color], [color=blue]mysql_error()[/color];
exit;
}
//end of query
[color=green]$row[/color] = [color=blue]mysql_fetch_array[/color]([color=green]$query[/color]);
[color=green]$item[/color] = [color=green]$row[/color][[color=orange]'ModeleArticle'[/color]];
echo "<option value=\"[color=limegreen]$item[/color]\">[color=green]$item[/color]</option>";
Now, I suppose that before the next to last line of code, there should be a loop, but like I said : can't figure out how...
Thanks alot for helping me out.
question po!!
in PHP Coding Help
Posted