How to $_POST select value from a PHP category list function?

Question

I have a category list as a PHP function. I called this function in a form to display the options. However I cannot get the form to post the selected option to my database. I have searched SO and online but most questions relate to the values being a select option, which would be easier. However my list is quite long and may change. The other values from the form are sent to the database correctly, apart form the pictures which I am working on, however I cannot find a way to insert the selected value from the drop down list within the form.

This is my simple sql query to send information from the form to the database:

if (isset($_POST['submit']))
{
    $title = $_POST['title'];
    $subtitle = $_POST['subtitle'];
    $description = $_POST['description'];
    $category= $_POST['category'];
    $pic1 = $_POST['pic1'];
    $pic2 = $_POST['pic2'];
    $pic3 = $_POST['pic3'];
    $pic4 = $_POST['pic4'];
    $pic5 = $_POST['pic5'];
    $item_location = $_POST['location'];
    $items_wanted = $_POST['wanted'];
    mysql_query("INSERT INTO items_available (title, subtitle, description, category, image_path1, image_path2, image_path3, image_path4, image_path5, item_location, items_wanted) VALUES ('$title', '$subtitle', '$description', '$category', '$pic1', '$pic2', '$pic3', '$pic4', '$pic5', '$item_location', '$items_wanted')")or die(mysql_error());
    header('Location: prompt.php?x=8');
}

This is the code for the field for category selection from my form which is a drop down list:

<div class="field">
    <label for="category">Category:</label>
    <select id=\"category\" name=\"category\" class=\"searchBox\">
        <?php createCategoryList(); ?>
    </select>
</div>

This is the code for the category list function:

//Creates Category <option>'s for search bar
function createCategoryList(){
if( ctype_digit($_GET['category']) ){ $x = $_GET['category']; }else{ $x = 999; }
echo "<option>All Categories</option>";
$i=0;
while(1){
    if(numberToCategory($i)=="Category Does Not Exist"){
        break;
    }else{
        echo " <option value=\"$i\" ";
        if($i==$x){echo ' SELECTED ';}
        echo " > ";
        echo numberToCategory($i);
        echo "</option>";
    }
    $i++;
}
}

//Category Number to String
function numberToCategory($n){
switch($n){
case 0:
    $cat = "Antiques";
    break;
case 1:
    $cat = "Art";
    break;
case 2:
    $cat = "Automotive";
    break;
case 3:
    $cat = "Baby";
    break;
case 4:
    $cat = "Books";
    break;
case 5:
    $cat = "Business & Industrial";
    break;
case 6:
    $cat = "Cameras & Photo";
    break;
case 7:
    $cat = "Clothing & Accessories";
    break;
case 8:
    $cat = "Collectibles";
    break;
case 9:
    $cat = "Computers";
    break;
case 10:
    $cat = "Crafts";
    break;
case 11:
    $cat = "DVD's & Movies";
    break;
case 12:
    $cat = "Electronics";
    break;
case 13:
    $cat = "Health & Beauty";
    break;
case 14:
    $cat = "Home & Garden";
    break;
case 15:
    $cat = "Jewelry & Watches";
    break;
case 16:
    $cat = "Music";
    break;
case 17:
    $cat = "Pet Supplies";
    break;
case 18:
    $cat = "Services";
    break;
case 19:
    $cat = "Sports & Outdoors";
    break;
case 20:
    $cat = "Sports Memorabilia & Cards";
    break;
case 21:
    $cat = "Tools & Home Improvement";
    break;
case 22:
    $cat = "Toys & Hobbies";
    break;
case 23:
    $cat = "Video Games";
    break;
case 24:
    $cat = "Other";
    break;
default:
    $cat = "Category Does Not Exist";
}

Currently the numerical value of the selected option is passed i.e if I select the baby option from the drop down list, then the number 3 is entered into my database. How do I replace this numerical value with it's corresponding value as it appears on the drop down list?

All and any answers are gratefully appreciated.

I would leave it as numbers. If you deal with relational databases you would setup a database with categories and the number would refer to the category selected. It is better programming this way. Your problem is your not returning the $cat variable (return $cat; )... But I would rethink this idea and use a database really. — Martin E.
You should replace that entire switch statement with a basic array of string values too. Just do a regular for loop over the array for the values. That doesn't answer your question just a suggestion. — Matt Pileggi

ajon ajon · Accepted Answer · 2014-02-28T22:00:13

The value submitted is whatever the value attribute of your option tag is set as. So if you change it to the below it will submit the category name instead of the number:

$catName = numberToCategory($i);
echo "<option value='$catName'>$catName</option>";

However What Martin suggested in his comment would be better. Your table containing categories should have an ID and that is the field you should put in your database and make it a foreign key.

In general it is bad practice to have specially formatted strings in your database, but rather create a table with the strings in them and reference that row id in your other tables.

How to $_POST select value from a PHP category list function?

2 Answers