php, output to dropdown list

Question

Good day. There is a list (select). The script should parse the names of the php files in the directory, then display them in a drop-down list. And it outputs, but it is crooked. It turns out that the name of each new file is placed in a separate drop-down list, but it is necessary, that in one.

$dir = '../'; $f = scandir($dir); foreach ($f as $file){ if(preg_match('/\.(php)/', $file)){ echo '<select><option>'.$file.'</option></select>'; echo'<br>'; } }

what am I doing wrong ?

rjhdby rjhdby 12.7k four 22 45 · Accepted Answer · 2016-11-29T08:39:45

Well, you take echo '<select> for the limits of the cycle

 $dir = '../'; $f = scandir($dir); echo '<select>'; foreach ($f as $file){ if(preg_match('/\.(php)/', $file)){ echo '<option>'.$file.'</option>'; } } echo '</select>';

php, output to dropdown list

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