Echo images using php according to a particular condition

Question

I have this query :

$q4 = "SELECT TOP 5 b.BadgeName, b.BadgeImage FROM BadgeImageTable AS b
INNER JOIN employee_badge AS e
ON e.badge_id = b.BadgeID
WHERE e.employee_id = 2164
ORDER BY e.earned_on DESC ";
$stmt3=sqlsrv_query($conn,$q4);
if($stmt3==false)
{
echo 'error to retrieve info !! <br/>';
die(print_r(sqlsrv_errors(),TRUE));
}

the query returns this :

Now in my php, i am trying to echo any three random images from the 'BadgeImage' column,if the query gives more than 3 rows of data.Else,if the query gives 2 rows,display only 2 images,if 1 row,display only 1 'badgeImage' and so on.

PHP Code that I tried and which is not giving the correct result :

<div class="badgesize">
               <?php

               if($count = sqlsrv_num_rows($stmt3) > 0){
                 while($recentBadge = sqlsrv_fetch_array($stmt3)){
                   $result[] = $recentBadge;
                 }

                 if($count > 3){
                   foreach(array_rand($result, 3) as $val){
                     $data[] = $result[$val];
                   }
                   $result = $data;
                 }

                 foreach($result as $recentBadge){
                   echo $recentBadge['BadgeName'], '<img src="'.$badgerecent['BadgeImage'].'">';
                 }
               } else {
                 echo 'no results';
               }

               ?>
             </div>

When I run the above PHP it is giving 'no result' inh the UI i.e "it is not running the if statement' and directly going to the else part of the PHP. This shouldnot happen as the query fetches 5 rows of data as shown in above picture.

When I try to echo simply (using below code) without the above condition,I am able to do so without any issue.

<img src="<?php echo "".($recentBadge['BadgeImage']).""; ?>" >

I am using PHP 5.5.38 and sql server 2012

http://stackoverflow.com/questions/971964/limit-10-20-in-sql-server — u_mulder, Dec 14 '16 at 07:39
yeah.. m already using select TOP ... the issue here is the PHP part as the query is returning the desired rows of data. @u_mulder — jane, Dec 14 '16 at 07:44

score 2 · Answer 1 · answered Dec 14 '16 at 07:14

2

Change the following:

if($count = sqlsrv_num_rows($stmt3) > 0){

to

$count = sqlsrv_num_rows($stmt3);    // getting the record count
   if($count > 0){                   // checking if there is some rows present or not
   while($recentBadge = sqlsrv_fetch_array($stmt3)){
   // your code
   }
}

and one more thing do not put full path of the image in badgeImage column as this will create problem when you move code from local to server. Keep only path of image folder and image name.

answered Dec 14 '16 at 07:14

Mayank Pandeyz

25,704
4
40
59

Thanks buddy.will keep ur advice in mind.Coming to the above question, when I change the code as u say ,its not solving the issue.its still going to the 'else' part of the code. – jane Dec 14 '16 at 07:31
Check the number of records by echoing it – Mayank Pandeyz Dec 14 '16 at 07:58
its not giving any number,when I echo the "$count" – jane Dec 14 '16 at 08:52
Means there is no row available in the resultset – Mayank Pandeyz Dec 14 '16 at 08:53
echo "$count" ; -- when I use this. – jane Dec 14 '16 at 08:54
Yeah... but $count = sqlsrv_num_rows($stmt3) ; is a correct syntax I guess – jane Dec 14 '16 at 08:55
Try this buddy, echo sqlsrv_num_rows($stmt3) ; and let me know what it returns – Mayank Pandeyz Dec 14 '16 at 09:02
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/130586/discussion-between-jane-and-mayank-pandeyz). – jane Dec 14 '16 at 09:09

score 1 · Answer 2 · answered Dec 14 '16 at 07:13

1

replace

if($count = sqlsrv_num_rows($stmt3) > 0){

with

 $count = sqlsrv_num_rows($stmt3); 
 if($count > 0){

answered Dec 14 '16 at 07:13

Naga

2,190
3
16
21

thanks for trying.But its not helping. – jane Dec 14 '16 at 07:34

score 1 · Accepted Answer · edited May 23 '17 at 12:24

Apparently sqlsrv_num_rows() does not return an expected result which is explained here

Alternatively you can use that fix or you can use the following code:

if(sqlsrv_has_rows($stmt3)){
  while( $row = sqlsrv_fetch_array($stmt3, SQLSRV_FETCH_ASSOC) ) {
    $recentBadge[] = $row;
  }

  if(count($recentBadge) > 3){
    foreach(array_rand($recentBadge, 3) as $key){
      $data[] = $recentBadge[$key];
    }
    $recentBadge = $data;
  }
} else {
  $recentBadge = [];
}

foreach($result as $recentBadge){
  echo $recentBadge['BadgeName'], '<img src="'.$badgerecent['BadgeImage'].'">';
}

This way we're checking if there are found rows instead, is so, store them all in an array and count the size of the array instead.

Echo images using php according to a particular condition

3 Answers3