Question

I know this is very simple, but I haven’t used PHP/MySQL in a while and I have been reading other threads/php website and can’t seem to get it.

How can I query a single row from a MySQL Table and print out all of the fields that have data in them? I need to exclude the NULL fields, and only add those that have data to an html list.

To clarify, I would like to display the field data without specifying the field names, just for the reason that I have a lot of fields and will not know which ones will be NULL or not.

Answer 1

What you’ve outlined requires 4 basic steps:

1. Connect to the database.
2. Query for a specific row.
3. Remove the null values from the result.
4. Create the html.

Step 1 is quite environment specific, so that we can safely skip here.

Step 2 – SQL

SELECT * from <tablename> WHERE <condition isolating single row>

Step 3 – PHP (assuming that $query represents the executed db query)

//convert the result to an array
$result_array = mysql_fetch_array($query);

//remove null values from the result array
$result_array = array_filter($result_array, 'strlen');

Step 4 – PHP

foreach ($result_array as $key => $value)
{
    echo $value \n;
}