[sos] a dumb question for sur

webtuto · 12-31-2007, 05:51 PM

hi well im trying to do a news script wih categories of course
but for exemple in categorie named => TALKPHP
i did a page for news about this categorie like that

PHP Code:


			
$query = "select * from table_name where group='TALKPHP' ";

group is a colomn or (champ) in frensh i dont know if u call it colomn in english lol
when i test it gaves me an error and when i delete the part "where group='TALKPHP'" the script works very well
so whats the problem

dschreck · 12-31-2007, 06:07 PM

Quote:

Originally Posted by webtuto

...
when i test it gaves me an error and when i delete the part "where group='TALKPHP'" the script works very well
so whats the problem

what's the error that it gives you?

webtuto · 12-31-2007, 06:08 PM

when i go back home ill test and tell u the error thanks for replying

Andrew · 12-31-2007, 07:58 PM

Well, make sure 'TALKPHP' group exists, first. If it does, just double check your names for everything.

Aussi, 'column' est 'champ' en Anglais.

Tanax · 12-31-2007, 09:22 PM

PHP Code:


			
$group = 'TALKPHP';
$sql = sprintf("SELECT * FROM `table_name` WHERE `group` = '%s'", mysql_real_escape_string($group));

$query = mysql_query($sql);

if($query) {
     echo 'Woho! The query worked..';
}

else {
     echo 'DooH! The query did not work..';
}

Try that!

Wildhoney · 01-01-2008, 01:13 AM

...And in addition to Tanax's post. If you're new to sprint, printf, et cetera, then we do have a nice article about them right here. I'm all ears for the error it gives you though.

webtuto · 01-01-2008, 01:08 PM

Quote:

Originally Posted by dschreck

what's the error that it gives you?

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\news\cat_info.php on line 5

webtuto · 01-01-2008, 01:11 PM

Quote:

Originally Posted by Tanax

PHP Code:


			
$group = 'TALKPHP';

$sql = sprintf("SELECT * FROM `table_name` WHERE `group` = '%s'", mysql_real_escape_string($group));



$query = mysql_query($sql);



if($query) {

     echo 'Woho! The query worked..';

}



else {

     echo 'DooH! The query did not work..';

}

Try that!

i try it and here is the answer => DooH! The query did not work..

here is the full code of this page

PHP Code:


			
<?php

include("config.php");

$group = 'TALKPHP';

$sql = sprintf("SELECT * FROM `table_name` WHERE `group` = '%s'", mysql_real_escape_string($group));

$res = mysql_query($sql) ;



if($res) {

     echo 'Woho! The query worked..';

}



else {

     echo 'DooH! The query did not work..';

}  



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

  echo "id : ".$row[id]."<br>" ;

  echo "titre : ".$row[title]."<br>" ;

  echo "resumé : ".$row[resume]."<br>" ;

  echo "note : ".$row[rate]."<br>" ;

  echo "écris le ".$row[date]."<br>" ;

} 

?>

webtuto · 01-01-2008, 01:13 PM

Quote:

Originally Posted by Andrew

Well, make sure 'TALKPHP' group exists, first. If it does, just double check your names for everything.

Aussi, 'column' est 'champ' en Anglais.

you must be good in frensh lol thanks for the word colomn
and group is the exact name of the colomn i even copy it and past it in my code

webtuto · 01-01-2008, 01:20 PM

Quote:

Originally Posted by Wildhoney

...And in addition to Tanax's post. If you're new to sprint, printf, et cetera, then we do have a nice article about them right here. I'm all ears for the error it gives you though.

thanks for the article im gonna read it when i finish this script