Question

Drupal Loop Items From Database

Asked by: walker6o9

I'm trying to grab all of my node ids from a table in my database.  Unfortunately, my code in it's current form returns the node with the lowest item twice (4, 4), rather than returning both of the nodes stored in the database (4, 6).  How do I change this to return all (in this case just 2, but there will be more) items of the database?

$result = db_query('SELECT nid FROM {content_type_agg_source} WHERE nid > %d', 1);
 while ($feed = db_fetch_array($result)) {
 $result2 = db_query('SELECT nid FROM {content_type_agg_source}');
 $items .= db_result($result2);
 $items .= "<br>";
  
  }
 return $items;

                                  
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Asked On
2009-06-03 at 17:20:11ID24462181
Tags

drupal

,

mysql

Topics

Drupal

,

PHP Scripting Language

,

Web-Based CMS

Participating Experts
1
Points
500
Comments
6

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Answers

 

by: pricelinePosted on 2009-06-03 at 20:48:11ID: 24543411

Your logic is little wrong, try this:

$result2 = db_query('SELECT nid FROM {content_type_agg_source}');
while ($feed = db_fetch_array($result2)) {
$items .= db_result($result2);
$items .= "<br>";

}
return $items;

 

by: walker6o9Posted on 2009-06-04 at 11:09:06ID: 24549680

That only outputs one value (the last one).

 

by: pricelinePosted on 2009-06-04 at 18:30:04ID: 24552876

Sorry, can you try this:
$result2 = db_query('SELECT nid FROM {content_type_agg_source}');
while ($feed = db_fetch_array($result2)) {
$items .= db_result($feed);
$items .= "<br>";

}
return $items;

 

by: walker6o9Posted on 2009-06-04 at 23:58:00ID: 24553820

That code threw this error:



    * warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/dolodev.com/httpdocs/andrew/ids/defense/trunk/includes/database.mysql.inc on line 176.
    * warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/dolodev.com/httpdocs/andrew/ids/defense/trunk/includes/database.mysql.inc on line 176.

 

by: pricelinePosted on 2009-06-05 at 21:31:35ID: 24561597

I am not able to try your scenario.. But try this last time:

Sorry, can you try this:
$result2 = db_query('SELECT nid FROM {content_type_agg_source}');
while ($feed = db_fetch_array($result2)) {
$items .= $feed['nid'];
$items .= "<br>";

}
return $items;


also try echo in the loop and see if it is failing for the first time or third time.  Good luck!

 

by: walker6o9Posted on 2009-06-08 at 12:02:06ID: 31588578

That did it.

20120131-EE-VQP-002

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