Array outside while doesn't function

Question

Sorry for the beginner question. I'm searching about an hour, but I can't understand why my $row outside from the second while doesn't function... The name variable run just the $row var doesn't function...

$i = 0;
while($i < 8)
{
    $str = "SELECT * FROM `$name[$i]`";
    $result = mysql_query($str, $connessione);
    $l = mysql_num_rows($result);

    while($l > 1)
    {
        $strs = "SELECT * FROM `$name[$i]` WHERE `Livello` = '$l'";
        $results = mysql_query($strs, $connessione);
        $row[$i][$l] = mysql_fetch_array($results);

            if I put here the echo I can view the mysql variable 
            echo $row[$i][$l]['var'];

        $l--;
    }
        if I put here echo $row[$i][$l]['var']; he send me the error " Undefined offset"
    $i++;
}

Hope you can help me...

what is if mysql_num_rows($result) return 1, then $l > 1 is false. — Baalthasarr
– Baalthasarr, Commented Jul 7, 2014 at 11:03
If $l is 0 to begin with, the $row variable will not be defined. — didierc
– didierc, Commented Jul 7, 2014 at 11:03

Marcin Nabiałek · Accepted Answer · 2014-07-07 11:12:05Z

1

In the place where you put:

echo $row[$i][$l]['var'];

$l value is 0 and you set $row values for $l from 1 to mysql_num_rows($result)

if you put there:

echo $row[$i][1]['var'];

it should work fine assuming mysql_num_rows($result) was more than 1 element.

Probably your code should look like this:

$i = 0;
while($i < 8)
{
    $str = "SELECT * FROM `$name[$i]`";
    $result = mysql_query($str, $connessione);
    $l = mysql_num_rows($result);

    while($l > 0) // changed 1 to 0
    {
        $strs = "SELECT * FROM `$name[$i]` WHERE `Livello` = '$l'";
        $results = mysql_query($strs, $connessione);
        $row[$i][$l] = mysql_fetch_array($results);

            if i put here the echo i can view the mysql variable 
            echo $row[$i][$l]['var'];

        $l--;
    }

    // added extra loop to display array values
    $whileIndex = 0;
    while (true) {
      if (!isset($row[$i][$whileIndex]['var']) {
         break;
      }
      echo $row[$i][$whileIndex]['var']; // should work
      ++$whileIndex;
    }


    $i++;
}

edited Jul 7, 2014 at 11:12

answered Jul 7, 2014 at 11:03

Marcin Nabiałek

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2 Comments

zyzz777 Over a year ago

in the first echo i can view the code is in the second the problem... if i put your code i got the same problem

Marcin Nabiałek Over a year ago

I've added possible solution to my answer

pavel · Accepted Answer · 2014-07-07 11:01:26Z

0

When you try to use echo, $i = 8 and $l = 1. These keys doesn´t exists in your array.

answered Jul 7, 2014 at 11:01

pavel

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2 Comments

zyzz777 Over a year ago

i want the variable for the $i between 0 and 7 (8 vars) and for $l between 1 and mysql num rows (different var)

pavel Over a year ago

You have to use echo in second loop, not behind.

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