Table generator?

marxx · 03-21-2008, 09:17 AM

Hi all and happy eastern! =)

I have on my new project quite many tables where I collect infos and such. I'm totally pissed of to write all over those tables and thought that if I need to changes something (class name etc) I have to go throught those all.. no way!

So, I know that it is possible to make function which generate table but would need some example of that.

Preferences as query and field names.. below is something what I came in my mind but I just can't get it work.. eh..

PHP Code:


			


function maketable($field, $query) {



$fields[] = array($field);



print("<table cellspacing=\"0\" cellpaddign=\"5\" width=\"100%\">\n");



$arr = mysql_query($query) or die(mysql_error());

while($fetch = mysql_fetch_array($arr)) {



  foreach($fields as $f) {



    print("<tr><td>".$fetch[$f[0]]."</td><td>".$fetch[$f[2]]."</td></tr>\n");



  }

}

print("</table>\n\n");

}

Thanks for all help! =)

freenity · 03-21-2008, 11:15 AM

I think this should work:

PHP Code:


			

function maketable($field, $query) {

$fields[] = array($field);

print("<table cellspacing=\"0\" cellpaddign=\"5\" width=\"100%\">\n");

$arr = mysql_query($query) or die(mysql_error());
while($fetch = mysql_fetch_array($arr)) {
  echo "<tr>\n";
  foreach($fields as $f) {
    print("<td>".$fetch[$f]."</td>\n");
  }
  echo "</tr>";
}
print("</table>\n\n");
}

Also, unless you send an array of array to $field, you can't have indices on $f, when you do foreach($fields as $f)

marxx · 03-21-2008, 03:57 PM

Hi Freenity, thanks for answering.

It seems that I'm missed something, I can't get it work..
I even try copy/paste hole php but no.

Then this came up. How should I make calling this function?

Now I have like this:

PHP Code:


			
maketable("'id', 'f1', 'f2', 'f3', 'f4'","SELECT * FROM test");

// edit:

I print_r $fiels and this came up

PHP Code:


			
Array ( [0] => Array ( [0] => id, f1, f2, f3, f4 ) )

So I have to explode $fields.. ok..

Thanks for all help!