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Home/ Questions/Q 916569
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T18:04:25+00:00 2026-05-15T18:04:25+00:00

I’m make a little game in php with mysql. Now I have a problem

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I’m make a little game in php with mysql. Now I have a problem with one of the sql query’s I created. The idea is that the query checks if the user has enough materials.

I have a query that if I use it like this it works:

SELECT
(
  SELECT COUNT(*)
  FROM building_requirements
  WHERE building_id = '1'
) as building_requirements_count,
(
  SELECT COUNT(*)
  FROM user_materials, building_requirements
  WHERE user_materials.material_id = building_requirements.material_id
    AND user_id = '27'
    AND building_id = '1'
    AND (user_material_amount >= building_material_amount) = 1
) as user_materials_count;

But when I add one column that use the result of those subquery’s it fails:

SELECT
(
  SELECT COUNT(*)
  FROM building_requirements
  WHERE building_id = '1'
) as building_requirements_count,
(
  SELECT COUNT(*)
  FROM user_materials, building_requirements
  WHERE user_materials.material_id = building_requirements.material_id
    AND user_id = '27'
    AND building_id = '1'
    AND (user_material_amount >= building_material_amount) = 1
) as user_materials_count, 
building_requirements_count = user_materials_count as enough_materials;

I get the error:

#1054 - Unknown column 'building_requirements_count' in 'field list'

Can someone explain to me why I can’t use the results of the subquery here? And how I can fix this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T18:04:25+00:00Added an answer on May 15, 2026 at 6:04 pm

    Cause the is no field called “building_requirements_count” in your table definition.
    You are not allowed to use self-defined fields here except for the WHERE part.

    Why don’t you use your self-defined fields in the WHERE section of your query?

    EDIT:
    It would be easier for you to get each value seperate out of the DB and do the calculating stuff in PHP.

    $result = mysql_query("SELECT COUNT(*) FROM building_requirements WHERE building_id = '1'");
    if ($result) {
      $row = mysql_fetch_row($result);
      $building_requirements_count  = $row[0];
    }
    else {
      $building_requirements_count = 0;
    }
    
    $query = "  SELECT COUNT(*)
      FROM user_materials, building_requirements
      WHERE user_materials.material_id = building_requirements.material_id
        AND user_id = '27'
        AND building_id = '1'
        AND (user_material_amount >= building_material_amount) = 1";
    
    $result2 = mysql_query($query);
    if ($result2) {
    
      $row = mysql_fetch_row($result2);
      $user_material_count  = $row[0];
    }
    else {
      $user_material_count = 0;
    }
    
    $enough_materials = ( $user_material_count >= $building_requirements_count) ? true : false;
    
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