how to get some values of an array php

Question

I'm running a code to get the columns names of a table but I want all the values except the first and the second.

<select name="TableNum" class="table">
  <?php
  include 'connectDb.php';  #Eisagwgi stoixeiwn gia syndesi me ti vasi
  $result=mysqli_query($con,"SELECT COLUMN_NAME FROM 
   INFORMATION_SCHEMA.Columns where TABLE_NAME = 'available'");
  while($row = mysqli_fetch_array($result)) {
    echo '<option>'.$row[0].'</option>';
    }
  ?>                
</select>

I know from Python how to use an array, and also I read this thread in W3Schools.

Also, with a simple print_r, I get as result this:

Array ( [0] => Date [COLUMN_NAME] => Date ) 
Array ( [0] => Time [COLUMN_NAME] => Time )
Array ( [0] => A1 [COLUMN_NAME] => A1 )
Array ( [0] => A2 [COLUMN_NAME] => A2 )
Array ( [0] => A3 [COLUMN_NAME] => A3 ) 
Array ( [0] => A4 [COLUMN_NAME] => A4 )
Array ( [0] => B1 [COLUMN_NAME] => B1 )
Array ( [0] => B2 [COLUMN_NAME] => B2 )
Array ( [0] => B3 [COLUMN_NAME] => B3 )
Array ( [0] => B4 [COLUMN_NAME] => B4 )
Array ( [0] => B5 [COLUMN_NAME] => B5 )
Array ( [0] => B6 [COLUMN_NAME] => B6 )
Array ( [0] => C1 [COLUMN_NAME] => C1 )
Array ( [0] => C2 [COLUMN_NAME] => C2 )

How could i get only the last 12 values(A1-A1,B1-B6,C1,C2) and put it back to option tag.

Thanks

You could just say that you don't want the Date and Time columns, by adding AND TABLE_NAME NOT IN ('Date', 'Time') after where TABLE_NAME = 'available'. — Qirel
– Qirel, Commented Jun 1, 2017 at 23:06
It's not working either. I get a list of blank values as elements — Haris Papadakis
– Haris Papadakis, Commented Jun 1, 2017 at 23:21
Ah, my mistake. It should be AND COLUMN_NAME NOT IN ('Date', 'Time') (I wrote table instead of column). That should do the trick - then no additional PHP is needed, and everything is done in SQL! — Qirel
– Qirel, Commented Jun 1, 2017 at 23:23

Enstage · Accepted Answer · 2017-06-01 23:22:49Z

1

This logic is probably best done in the SQL itself, but if you want to do it the PHP:

$result = mysqli_query($con,"SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.Columns where TABLE_NAME = 'available'");
$result = $result->fetch_all(MYSQLI_ASSOC);
echo '<option>', implode('</option><option>', array_slice(array_column($result, 'COLUMN_NAME'), 2)), '</option>';

edited Jun 1, 2017 at 23:22

answered Jun 1, 2017 at 23:11

Enstage

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2 Comments

Haris Papadakis Over a year ago

It's not working. I get this :Notice: Array to string conversion in C:\Users\papad\Desktop\WebServer\htdocs\toi\form_reserv.php on line.

Enstage Over a year ago

I've added array_column() to my answer which will hopefully fix the issue.

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