query to show selected options from multi select field in mysql - mysql

Question

I am trying to get a query working that shows specific info from a mult-select-field that is part of a phpbb forum.
SELECT d.pf_firstname, d.pf_lastname, d.pf_specialism
FROM phpbb_profile_fields_data d
WHERE d.pf_lastname = 'Linssen'
The query above gives the following Result:
-- firstname: Pierre
-- lastname: Linssen
-- pf_specialism: 1;3;8;10;12;15;16 (which are the selected options)
In another table the relation between the available options (0 to 18)
and the related text is given.
result of a "wrong" query, showing all these options
The query I need should show: firstname and lastname of a user and only the specific options that that user (in this case: Linssen) has selected for a specific profile field (in this case: specialism).
So the result of the query should be something like
-- firstname: Pierre
-- lastname: Linssen
-- specialism: (1) Counseling ; (2) Executive Coaching ; (8) Oplossings gericht coachen; (10) Provocatief coachen ; (12) Sales coaching ; (15) Team coaching ; (16) Wandel coaching
How do I do that?
In order to support multi-language, profile fields values (TEXT) are stored in the _lang table, and the actual _data table only holds indexes to the lang table. So that you have, say, 3 stored in the data table, and then you map that to "Project Manager" in English or "Gestor de Proyectos" in Spanish...
2017-01-22: Thanks for input sofar still struggling to get the query to work.
Please find below the three tables with their relevant content.
phpbb_profile_lang has the following relevant fields
-- field_id (41)
-- lang_id (2)
-- lang_name (specialisme)
phpbb_profile_fields_data has the following relevant fields
-- user_id (90)
-- pf_voornaam (Pierre)
-- pf_achternaam (Linssen)
-- pf_specialisme (1;3;8;10;12;15;16)
phpbb_profile_fields_lang has the following relevant fields
-- field_id (41)
-- lang_id (2)
-- option_id (0) lang_value (Business Coaching)
-- option_id (1) lang_value (Counseling)
-- option_id (2) lang_value (Executive Coaching)
-- option_id (3) lang_value (Holistische Coaching)
-- ...............................................
-- ...............................................
-- option_id (17) lang_value (Zingeving)
-- option_id (18) lang_value (Overige)

Answer 1

Your query to join the tables would look like this:
select pf_voornaam, pf_lastname, lang_value
from phpbb_profile_fields_data pfd
inner join phpbb_profile_fields_lang pfl on pfl.feld_id = pfd.pf_specialisme
inner join phpbb_profile_lang pl on pfl.lang_id = pl.lang_id;
If you have control of the schema, I would strongly recommend to change phpbb_profile_lang.lang_id to be the primary key field_id and remove the lang_id - this is redundant based on the data you've shown.
Based on your example, I have used the following query:
select pf_voornaam, pf_achternaam, pf_specialisme, lang_value
from phpbb_profile_fields_data pfd
inner join phpbb_profile_fields_lang pfl on pfl.field_id = pfd.pf_specialisme
inner join phpbb_profile_lang pl on pfl.lang_id = pl.lang_id;
The result is shown in this image

Answer 2

With all the relevant input here and with help of "Javiexin", the following mysqli-based php-script with the various included mysql-queries did the trick. I am open to any suggestions to further improve the queries and script.
Here is a link to a result page.
<?php
/*Open connection to our database. */
$link = mysqli_connect("localhost", "perspec1", "FLeur0798!", "perspec1");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* check if server is alive */
if (mysqli_ping($link)) {
printf ("A: Our connection is OK!<br/><br/>");
} else {
printf ("Error: %s\n", mysqli_error($link));
}
/* First MYSQL query */
$sql = "SELECT d.pf_voornaam, d.pf_achternaam, d.pf_specialisme FROM phpbb_profile_fields_data d WHERE d.pf_achternaam = 'Linssen'";
if (!$sql) { echo "Error: ".mysqli_error(); die();
} else {
printf ("B: First query is successful!<br/><br/>");
}
if ($result = mysqli_query($link, $sql))
{
echo "<table>";
//header
echo "<tr><td>Voornaam</td>";
echo "<td>Achternaam</td>";
echo "<td>Specialisme</td></tr>";
//data
while ($row = mysqli_fetch_array($result))
{
echo "<tr><td>{$row[0]}</td>";
echo "<td>{$row[1]}</td>";
echo "<td>{$row[2]}</td></tr>";
}
echo "</table><br/>";
}
/* Second MYSQL query */
$query2 = mysqli_query($link, $sql );
if (!$query2) { echo "Error: ".mysqli_error(); die();
} else {
printf ("C: Second query is successful!<br/><br/>");
}
while( $row = mysqli_fetch_assoc($query2))
$selected_opts = explode (";", $row['pf_specialisme']);
foreach ($selected_opts as $selected_opt) echo "$selected_opt <br/>";
echo "<br/>";
$selected_opts_string = implode(', ', $selected_opts);
$sql3 = "SELECT fl.option_id, fl.lang_value FROM phpbb_profile_fields f, phpbb_profile_fields_lang fl WHERE f.field_id = fl.field_id AND f.field_name = 'specialisme' AND fl.lang_id = 1 AND fl.option_id IN ($selected_opts_string) ORDER BY fl.option_id ASC";
$query3 = mysqli_query($link, $sql3 );
if (!$sql3) { echo "Error: ".mysqli_error(); die();
} else {
printf ("D: Third query is successful!<br/><br/>");
}
if (!$query3) { echo "Error: ".mysqli_error(); die(); }
echo "<table>";
while( $result = mysqli_fetch_assoc($query3) )
{
echo "<tr><td>$result[option_id]</td>";
echo "<td>$result[lang_value]</td></tr>";
}
echo "</table>";
mysqli_free_result($result);
mysqli_close($link);
?>