No results when query within foreach

Question

I'm trying to match users with a certain description in one table to a general search description in another table. I have a foreach that contains the general descriptions and created a query within the foreach hoping to pull out the user that best fits that general description.

Problem is I'm not getting any results when I echo out the username. I get results when I echo out result['title'].

if (empty($errors)){
  $results = search_results($keywords);
  $results_num = count($results);

  echo '<p><strong style ="font-size:145%;font-family:georgia;"> Titles: </strong></p> ';

  foreach ($results as $result){
    $query = mysql_query("SELECT username, title from clothing 
    WHERE description LIKE '%$results%'");
    while($row=mysql_fetch_array($query)){
    $username=$row['username'];                     
    $title=$row['title'];
    echo $username;
    }
    echo '<p> <strong>',$result['title'],'</strong> </p>';
  }

}

I would be very grateful to know why my user is not getting echoed out. Also any programming tips would be greatly appreciated for I am always learning and I want to always find ways to improve.

Why don't you see if that query returns anything. As in, run it against the database in an SQL tool. — wesside
– wesside, Commented Oct 25, 2012 at 13:23
In your query, you mean to use the foreach variable $result, not $results as you have: LIKE '%$results%'" — Michael Berkowski
– Michael Berkowski, Commented Oct 25, 2012 at 13:25

CodePB · Accepted Answer · 2012-10-25 13:25:19Z

1

You need to replace $results with $result in your query:

$query = mysql_query("SELECT username, title from clothing WHERE description LIKE '%$result%'");

answered Oct 25, 2012 at 13:25

CodePB

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Comments

Svetoslav · Accepted Answer · 2012-10-25 13:26:41Z

Try to set one check (with error message)if count equals to 0 .. And LIKE clause value must be result not results

if (empty($errors)){
                $results = search_results($keywords);
                $results_num = count($results);

                if($results_num==0) {
                    echo "NO RESULTS MESSAGE !!!"; 
                }
                else{

                    echo '<p><strong style ="font-size:145%;font-family:georgia;"> Titles: </strong></p> ';

                    foreach ($results as $result){
                      $query = mysql_query("SELECT username, title from clothing WHERE description LIKE '%$result%'");
                      while($row=mysql_fetch_array($query)){
                        $username=$row['username'];                     
                        $title=$row['title'];
                        echo $username;
                      }
                      echo '<p> <strong>',$result['title'],'</strong> </p>';
                    }
               }

           }

Barak · Accepted Answer · 2012-10-25 13:32:27Z

0

SELECT username, title from clothing WHERE description LIKE '%$results%'

$results,it seems to be an array.I think you have to use $result here.

edited Oct 25, 2012 at 13:32

Barak

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answered Oct 25, 2012 at 13:28

Nahuelistico

422 bronze badges

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