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Home/ Questions/Q 7490965
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T15:47:51+00:00 2026-05-29T15:47:51+00:00

I have a table called profile in mysql database Table structure: user | hasimg

  • 0

I have a table called “profile” in mysql database
Table structure:

user | hasimg | imgurl

As you can see “hasimg” and “imgurl”, normally if a user does not have an image, hasimg will be set to 0 and imgurl is set to none.

On the other hand, if a user has an image assigned then hasimg is set to 1 and imgurl will contain the link.

  • How can I display ones that have images (hasimg set to 1) with <img src="imgurl html tags?

  • How can I display ones that do not have images (hasimg set to 0) , without <img src?

I want to make a list which display users with images and user that does not have image.

Here’s the code im using to echo out the data row by row

   <?php

      $sql = "SELECT * FROM profile ORDER BY user DSC"; 
      $query = mysql_query($sql)or die(mysql_error());

               $rows = array();
        while($row = mysql_fetch_array( $query )){
          $rows[] = $row;
          echo "User:$row[user]<br>Image:$row[hasimg]\n";
        }

    ?>

P/S mysql connection established.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T15:47:52+00:00Added an answer on May 29, 2026 at 3:47 pm

    If you want to keep your layout fancy way you can use default image.

    if($hasimg == 0)
       $imgurl = "default.jpg";
    
    echo '<img src="' . $imgurl . '">';
    

    EDIT:

    Your PHP code could be like this.

     $sql = "SELECT * FROM profile ORDER BY user DSC"; 
     $result = mysql_query($sql)or die(mysql_error());
    
    
    echo "<table border='1'>
    <tr>
    <th>Name</th>
    <th>Image</th>
    </tr>";
    
    while($row = mysql_fetch_array($result))
      {
        $imgurl = $row['imgurl'];
        if($row['hasimg'] == 0)
             $imgurl = "default.jpg";
    
          echo "<tr>";
          echo "<td>" . $row['user'] . "</td>";
          echo "<td><img src=\"" .$imgurl . "'\" /></td>";
          echo "</tr>";
      }
    echo "</table>";
    
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