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Home/ Questions/Q 3321148
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T23:01:41+00:00 2026-05-17T23:01:41+00:00

This is probably pretty basic, but I’m a beginner in PHP – I have

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This is probably pretty basic, but I’m a beginner in PHP – I have one table for categories (cat_id is the primary key & cat_name is the only other field), and another table for posts which uses the cat_id as a foreign key.

My question is when creating a form for a user to submit a new post, how do I get the list of already created categories to appear in the form (by their category name) by way of a drop down box?

I’m assuming I’ll have to select the table, assign values/variables to the data, or make it an array of some sort and then print the data in a table for the user to select?

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  1. Editorial Team
    Editorial Team
    2026-05-17T23:01:41+00:00Added an answer on May 17, 2026 at 11:01 pm

    There is no restriction on running queries for one table when you’ve already run a query for another. Use a simple SELECT * FROM category, perhaps with an ORDER BY clause. I’m assuming you’re using mysql or mysqli, so use a while loop to go through the results:

    // run the query
    $result = $mysqli->query('SELECT * FROM category ORDER BY cat_name ASC');
    // initialize the html string
    $select = '';
    // loop through all results
    while ($row = $result->fetch_array()) {
      // escape any characters that would break the html
      $row['cat_id'] = htmlspecialchars($row['cat_id']);
      $row['cat_name'] = htmlspecialchars($row['cat_name']);
      // create this item's option element and append it to the html string
      $select .= "<option value=\"$row[cat_id]\">$row[cat_name]</option>";
    }
    // wrap the options in a select
    $select = "<select name=\"category\">$select</select>";
    

    You’ll end up with a html string $select that you can echo wherever you like, as many times as you like.

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