CODEWITHSUNDEEP
php
Christian Joy
Christian Joy

Reputation: 5

Variable does not change its value

I have been trying to analyze it for like hours now but I can't understand what was wrong with my code :(

$d = 1; //I declare this variable as 1
$a = 0;
while($d<$arraycount-2){
  while ($a < 40){
    $c = 0;
    $b = 0;

    while($c < $fc){
      $finaloutput = $finaloutput.$database_returned_data[$d][$c].","; //But here, in this loop $d is always 1

      $c++;
    }

    while($b < 5){
      $finaloutput = $finaloutput.$mydata[$a][$b].",";
      $b++;
    }

    $finaloutput = $finaloutput."--END--";
    $a++;
  }
  $d++; //But it increments successfully. Hence, the while loop terminates after it meets the criteria.
}

The variable $d is always 1 inside the other loop but it increments outside the loop. Note there is a while statement inside the while. Is there anything wrong?

I'm using $d for my array:

 $finaloutput = $finaloutput.$database_returned_data[$d][$c].",";

I am a noob poster. Feel free to ask for more details :)

Upvotes: 0

Views: 507

Answers (1)

Kai Qing
Kai Qing

Reputation: 18833

You don't set $a here:

while($d<$arraycount-2){
    while ($a < 40){

So on every iteration besides the first this while condition won't run.

just change it to:

while($d<$arraycount-2){
    $a = 0;
    while ($a < 40){

Upvotes: 1

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