Define variable based on concatenated PostgreSQL output in PHP

Question

I am trying to generate a PHP variable that can be used to generate select options on a website.

The php is currently as follows:

$sql_names = pg_query($dbconn, "SELECT id, name external_name FROM names);
    if (!$sql_names) {
      echo "Er\n";
      exit;
    }

    while ($row = pg_fetch_array($sql_names)) foreach($row as $a => $b){ $name_options = '<option value="' . $row[0] . '">' . $row[1] . '</option>';}

This is then called later on as follows:

<select name="NAMES[]">
        <?php echo $name_options; ?>
</select>

However it only returns 1 option when there should be 6.

Any suggestions?

Samuel Teixeira · Accepted Answer · 2017-02-07 14:57:23Z

1

the problem is that you are not append the results in the $name_options variable. This way you are returning only the last row. Try this:

$sql_names = pg_query($dbconn, "SELECT id, name external_name FROM names);
if (!$sql_names) {
    echo "Er\n";
    exit;
}
$name_options = ""; 
while ($row = pg_fetch_array($sql_names)){ 
    $name_options .= '<option value="' . $row[0] . '">' . $row[1] . '</option>';
}

edited Feb 7, 2017 at 14:57

answered Feb 7, 2017 at 14:50

Samuel Teixeira

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2 Comments

user2798841 Over a year ago

This almost worked, but it ended up returning 5 of each row! Solution seems to be:

$sql_names = pg_query($dbconn, "SELECT id, name external_name FROM names); if (!$sql_names) {     echo "Er\n";     exit; } while ($row = pg_fetch_array($sql_names))           $name_options .= '<option value="' . $row[0] . '">' . $row[1] . '</option>';

Just needed to get rid of the foreach!

Samuel Teixeira Over a year ago

Yes, I see it. I edited the answer but I think you could do that. Regards.

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