I want to display "No Image" when there's no image 'BLOB' at MySQL database

Question

based on my question above, currently, when there is no image at the database, it will display a broken image while if there is an image, it will display the image. Now how to display "No Image" when there's no image 'BLOB' at MySQL database.

I uses BLOB format at the MySQL Database. Below is the code:

<?php

$query = $conn->query("SELECT * FROM report LEFT JOIN users ON report.badgeid = users.badgeid LEFT JOIN team ON team.team_id = users.team_id WHERE team.team_id = '$team' ORDER BY report.report_date DESC LIMIT 10");
$query -> execute();
$results = $query -> fetchAll(PDO::FETCH_OBJ);
if($query->rowCount() == 0){

  echo "<table class = 'table-bordered' width ='100%'>";
  echo "<thead>";
    echo "<tr>";
        echo "<th>Date</th>
        <th>Task Name</th>
        <th>Before Task</th>
        <th>After Task</th>
        <th>OT From</th>
        <th>OT To</th>
        <th>Status</th>
        <th>Action</th>
      </tr>
    </thead>
    <tbody >
    <tr>
    <td colspan='8'>No report at this moment</td>
    </tr>
    </tbody>
    </table>";
}else{

      echo "<table class = 'table-bordered' width ='100%'>";
      echo "<thead>";
        echo "<tr>";
            echo "<th>Date</th>
            <th>Task Name</th>
            <th>Before Task</th>
            <th>After Task</th>
            <th>OT From</th>
            <th>OT To</th>
            <th>Status</th>
            <th>Action</th>
          </tr>
        </thead>
        <tbody >";

        $query = $conn->query("SELECT * FROM report LEFT JOIN users ON report.badgeid = users.badgeid LEFT JOIN team ON team.team_id = users.team_id WHERE team.team_id = '$team' ORDER BY report.report_date DESC LIMIT 10");
        while($row = $query->fetch(PDO::FETCH_ASSOC)){
        echo "<tr>";

          echo "<td rowspan='2'>". $row['report_date'] . "</td>";
          echo "<td rowspan='2'>". $row['task_name'] . "</td>";
          echo "<td align='center'><img src='data:image/jpeg;base64,".$before."'/></td>";
          echo "<td align='center'><img src='data:image/jpeg;base64,".$row['photo_after']."'/></td>";
          echo "<td rowspan='2' align='center'>". $row['ot_start']. "</td>";
          echo "<td rowspan='2' align='center'>".$row['ot_end']. "</strong></td>";
          echo "<td rowspan='2'><strong>". $row['report_status'] . "</strong></td>";
          echo "<td rowspan='2' align='center'>";
            echo "<a class='btn-view btn-primary btn-sm' href='../view_task/view_task.php?report_id=". $row['report_id'] ."' data-toggle='tooltip'>View</a>";
            echo "</td>";

        echo "</tr>";

        echo "<tr>";

          echo "<td align='center'>". $row['time_photo_before'] . "</td>";
          echo "<td align='center'>". $row['time_photo_after'] . "</td>";

        echo "</tr>";
        }

        echo "</tbody>";
      echo "</table><br>";  
    echo "</div>";
  echo "<div>";
}
?>

Use an `if` statement to check if `$row['photo_after']` contains anything? — brombeer, Nov 25 '19 at 07:58
**Warning:** You are wide open to [SQL Injections](https://php.net/manual/en/security.database.sql-injection.php) and should use parameterized **prepared statements** instead of manually building your queries. They are provided by [PDO](https://php.net/manual/pdo.prepared-statements.php) or by [MySQLi](https://php.net/manual/mysqli.quickstart.prepared-statements.php). Never trust any kind of input! Even when your queries are executed only by trusted users, [you are still in risk of corrupting your data](http://bobby-tables.com/). [Escaping is not enough!](https://stackoverflow.com/q/5741187) — Dharman, Nov 25 '19 at 09:31

score 0 · Accepted Answer · answered Nov 25 '19 at 08:25

Just put a condition to check whether the image exist or not empty in the row.

if(!isset($row['photo_after']) || empty($row['photo_after'])) {
   echo "<td align='center'><img src='default.png'/></td>";
}
else 
{ 
   echo "<td align='center'><img src='data:image/jpeg;base64,".$row['photo_after']."'/></td>"; 
}

Tarun Pandat · Answer 2 · 2019-11-25T10:19:36.707

Try this..., I hope this will work for you. Inside your while loop, I made a condition which checks there is something in your field or it's empty. if it's empty a variable store the text you want to print and if it's not empty it will show your image in variable $image and I echo the variable where you want to show your image.

<?php

$query = $conn->query("SELECT * FROM report LEFT JOIN users ON report.badgeid = users.badgeid LEFT JOIN team ON team.team_id = users.team_id WHERE team.team_id = '$team' ORDER BY report.report_date DESC LIMIT 10");
$query -> execute();
$results = $query -> fetchAll(PDO::FETCH_OBJ);
if($query->rowCount() == 0){

  echo "<table class = 'table-bordered' width ='100%'>";
  echo "<thead>";
    echo "<tr>";
        echo "<th>Date</th>
        <th>Task Name</th>
        <th>Before Task</th>
        <th>After Task</th>
        <th>OT From</th>
        <th>OT To</th>
        <th>Status</th>
        <th>Action</th>
      </tr>
    </thead>
    <tbody >
    <tr>
    <td colspan='8'>No report at this moment</td>
    </tr>
    </tbody>
    </table>";
}else{

      echo "<table class = 'table-bordered' width ='100%'>";
      echo "<thead>";
        echo "<tr>";
            echo "<th>Date</th>
            <th>Task Name</th>
            <th>Before Task</th>
            <th>After Task</th>
            <th>OT From</th>
            <th>OT To</th>
            <th>Status</th>
            <th>Action</th>
          </tr>
        </thead>
        <tbody >";

        $query = $conn->query("SELECT * FROM report LEFT JOIN users ON report.badgeid = users.badgeid LEFT JOIN team ON team.team_id = users.team_id WHERE team.team_id = '$team' ORDER BY report.report_date DESC LIMIT 10");
        while($row = $query->fetch(PDO::FETCH_ASSOC)){


      // Here is my code...



          if(empty($row['photo_after'])) {
               $image = "NO IMAGE FOUND..."; //anything you want to diaplay if there is no image
             }
             else{ $image = "<img src='data:image/jpeg;base64,".$row['photo_after']."'/>"; }





     // Check the $image variable in your table




        echo "<tr>";

          echo "<td rowspan='2'>". $row['report_date'] . "</td>";
          echo "<td rowspan='2'>". $row['task_name'] . "</td>";
          echo "<td align='center'><img src='data:image/jpeg;base64,".$before."'/></td>";
          echo "<td align='center'>".$image."</td>";   // $image store your data
          echo "<td rowspan='2' align='center'>". $row['ot_start']. "</td>";
          echo "<td rowspan='2' align='center'>".$row['ot_end']. "</strong></td>";
          echo "<td rowspan='2'><strong>". $row['report_status'] . "</strong></td>";
          echo "<td rowspan='2' align='center'>";
            echo "<a class='btn-view btn-primary btn-sm' href='../view_task/view_task.php?report_id=". $row['report_id'] ."' data-toggle='tooltip'>View</a>";
            echo "</td>";

        echo "</tr>";

        echo "<tr>";

          echo "<td align='center'>". $row['time_photo_before'] . "</td>";
          echo "<td align='center'>". $row['time_photo_after'] . "</td>";

        echo "</tr>";
        }

        echo "</tbody>";
      echo "</table><br>";  
    echo "</div>";
  echo "<div>";
}
?>

While this code may solve the question, [including an explanation](//meta.stackexchange.com/q/114762) of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please [edit] your answer to add explanations and give an indication of what limitations and assumptions apply. — Dharman, Nov 25 '19 at 09:31

I want to display "No Image" when there's no image 'BLOB' at MySQL database

2 Answers2