จะให้แสดงรูปภาพภายในคำสั่ง if แบบ short if ได้อย่างไรคะ

<td>'.( $img != $row["j_pic"]? $img= $row['j_pic'] : '' ).'</td>

<?php
//fetch.php

$output = '';
$img = '';
if(isset($_POST["query"]))
{
 $search = mysqli_real_escape_string($connect, $_POST["query"]);
 $query = "
  SELECT * FROM interdiscip_pillar 
  INNER JOIN interdiscip_detail
  ON interdiscip_detail.j_id = interdiscip_pillar.j_id
  WHERE interdiscip_pillar.j_name LIKE '%".$search."%'
  OR interdiscip_detail.name LIKE '%".$search."%' 
 ";
}
else
{
 $query = "
  SELECT * FROM interdiscip_pillar 
  INNER JOIN interdiscip_detail
  ON interdiscip_detail.j_id = interdiscip_pillar.j_id
  WHERE interdiscip_detail.status= 'Y'
  ORDER BY interdiscip_pillar.j_id DESC,interdiscip_detail.sort ASC
 ";
}
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result) > 0)
{
 $output .= '
  <div class="table-responsive">
   <table class="table table bordered">
    <tr>
     <th>img</th>
     <th>ID</th>
     <th>Name</th>
     <th>Journal</th>
    </tr>
 ';
 while($row = mysqli_fetch_array($result))
 {
  $output .= '
   <tr>
    <td>'.( $img != $row["j_pic"]? $img= $row['j_pic'] : '' ).'</td>
    <td>'.$row["j_id"].'</td>
    <td>'.$row["j_name"].'</td>
    <td>'.$row["name"].'</td>
   </tr>
  ';
 }
 echo $output;
}
else
{
 echo 'Data Not Found';
}
?>