PHP counter increment in a while loop

Question

Im having a problem incrementing a counter in one of my while loops basically i just want to alternate between two image links that were fetched in my database but my counter wont increase and im not sure why can anyone help?

while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
      $img_link = $row['Image'];
      $img_link_alt = $row['Image_alt'];
      $i = 0;

       echo '<div class="col-xs-6 col-sm-3 placeholder">'; 
       $img = ( $i % 2 == 0 ) ?  $img_link : $img_link_alt; 

            echo $i;
            //'?' . date("h:i:sa").'
            echo '<img style="height:200px; border-radius:0%; width:300px;" src="screenshots/'. $img . '">';          
            echo '<h4>Screenshot</h4><span class="text-muted">Updated Screenshot of the Botting session: <b>' . $row['script_name'] .' </b></span>'; 
            echo '</div>';        
            $i++;        
        }

Ive even tried declaring $i outside of the while loop and still nothing..... any help would be much appreciated

Delete $i = 0; in the while loop. Just above while ($row...), put $i = 0; Try another option. Instead of incrementing and then finding $i % 2, just do $i = ($i === 0) ? 1 : 0; That way, each time you go in the loop, $i will change. — zedfoxus
– zedfoxus, Commented Apr 9, 2016 at 15:12
"Ive even tried declaring $i outside of the while loop and still nothing..... any help would be much appreciated" sorry guys i tried it outside the while loop and still nothing.... — Christian Bartram
– Christian Bartram, Commented Apr 9, 2016 at 15:17
try also to print the condition in your while loop it might help you... — user4386714
– user4386714, Commented Apr 9, 2016 at 15:52

Sanjay Kumar N S · Accepted Answer · 2016-04-09 15:12:35Z

1

Initialize $i outside the loop.

$i = 0;
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
      $img_link = $row['Image'];
      $img_link_alt = $row['Image_alt'];

       echo '<div class="col-xs-6 col-sm-3 placeholder">'; 
       $img = ( $i % 2 == 0 ) ?  $img_link : $img_link_alt; 

            echo $i;
            //'?' . date("h:i:sa").'
            echo '<img style="height:200px; border-radius:0%; width:300px;" src="screenshots/'. $img . '">';          
            echo '<h4>Screenshot</h4><span class="text-muted">Updated Screenshot of the Botting session: <b>' . $row['script_name'] .' </b></span>'; 
            echo '</div>';        
            $i++;        
        }

answered Apr 9, 2016 at 15:12

Sanjay Kumar N S

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Comments

bksi · Accepted Answer · 2016-04-09 15:12:29Z

0

You have to reset $i outside the loop:

$i = 0;
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
  $img_link = $row['Image'];
  $img_link_alt = $row['Image_alt'];
   echo '<div class="col-xs-6 col-sm-3 placeholder">'; 
   $img = ( $i % 2 == 0 ) ?  $img_link : $img_link_alt; 
       echo $i;
        //'?' . date("h:i:sa").'
        echo '<img style="height:200px; border-radius:0%; width:300px;" src="screenshots/'. $img . '">';          
        echo '<h4>Screenshot</h4><span class="text-muted">Updated Screenshot of the Botting session: <b>' . $row['script_name'] .' </b></span>'; 
        echo '</div>';        
        $i++;        
}

answered Apr 9, 2016 at 15:12

bksi

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2 Comments

Christian Bartram Over a year ago

Ive even tried declaring $i outside of the while loop and still nothing..... any help would be much appreciated" sorry guys i tried it outside the while loop and still nothing....

bksi Over a year ago

try to print $row to see if there is actual result. I would print $i, to know if this loop iterates over a data. Try to put print_r($row); inside the loop

ameenulla0007 · Accepted Answer · 2016-04-09 15:13:07Z

0

$i  = 0;
while($i<5) {
    $alternateVal   = $i%2==0 ? "type1" : "type2";
    echo $alternateVal;
    $i++;
}

Above is an example to overview alternate values.

answered Apr 9, 2016 at 15:13

ameenulla0007

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Comments

Community · Accepted Answer · 2017-05-23 10:30:35Z

0

move the

$i

initialization:

 $i=0

outside of the while loop and it worked. you every time renizialize the

$i

variable in your while loop,and your counter don't increase, fix it. An alternative is use the rows counter, like that example:

$count = $db->query("SELECT COUNT(id) FROM users")->fetchColumn();

echo $count; //Returns number of rows

like in that answer

edited May 23, 2017 at 10:30

CommunityBot

11 silver badge

answered Apr 9, 2016 at 15:12

user4386714

Comments

Sᴀᴍ Onᴇᴌᴀ · Accepted Answer · 2017-07-24 17:05:14Z

0

Put the $i variable outside the while block, like this:

$i = 0;

while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
      $img_link = $row['Image'];
      $img_link_alt = $row['Image_alt'];

       echo '<div class="col-xs-6 col-sm-3 placeholder">'; 
       $img = ( $i % 2 == 0 ) ?  $img_link : $img_link_alt; 

            echo $i;
            //'?' . date("h:i:sa").'
            echo '<img style="height:200px; border-radius:0%; width:300px;" src="screenshots/'. $img . '">';          
            echo '<h4>Screenshot</h4><span class="text-muted">Updated Screenshot of the Botting session: <b>' . $row['script_name'] .' </b></span>'; 
            echo '</div>';        
            $i++;        
        }

edited Jul 24, 2017 at 17:05

Sᴀᴍ Onᴇᴌᴀ

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answered Apr 9, 2016 at 15:12

nekiala

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1 Comment

Christian Bartram Over a year ago

Ive even tried declaring $i outside of the while loop and still nothing..... any help would be much appreciated" sorry guys i tried it outside the while loop and still nothing....

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