how to access image from mysql to html table using php

Question

I want to fetch image from database to html table cell. I have tried something but it gives me an error. Let me know what is wrong with this code.

         <tbody>
          <?php
         $result = mysqli_query($conn,$sql);
         while($row = mysqli_fetch_array($result,MYSQLI_BOTH)) {  
         echo " <tr>
         <th>1</th>

         <td>{$row['itemId']}</td>
        **<td><img src="imageView.php?image_id=<?php echo .$row["imageId"].; ?   
         >" /><br/>**
        <td/>
         <br/>
         </td>
         <td>{$row['itemName']}</td>
         </tr>";
         }
        ?>
       <tr>
      <th scope="row">2</th>
      <td>Nicholas</td>
       <td>Sanchez</td>
     <td>username</td>
     </tr>
     <tr>
    <th scope="row">3</th>
      <td>Debra</td>
     <td>Shaw</td>
   <td>username</td>
   </tr>
  <?php 

  mysqli_close($conn);
 ?>
</tbody>

You can't use the double quotes for the HTML encapsulation, you are already using them for the PHP. Also ` — chris85, Jul 31 '17 at 17:25
Using `" />` in double quotes will always give error when you echo with `" "`. Use quotes carefully. Try to use single and double quotes — Harpreet Singh, Jul 31 '17 at 17:28

node_modules · Answer 1 · 2017-07-31T17:36:44.940

2

You are trying to use the PHP tag inside echo "";

Instead of

<?php echo .$row["imageId"].; ?>

Try this

{$row["imageId"]}

Or

".$row["imageId"]."

Also don't forget to escape the image tag. Don't use double-quotes if your echo contains double-quotes too. Use single quote for the image tag instead '. If you wish to use double quotes for your image tag, escape it like this

<img src=\"imageView.php?image_id=".$row["imageId"]."\" />

But to ensure you're safe with the quotes in your image tag, you could use the single quotes only:

<img src='imageView.php?image_id=".$row["imageId"]."' />

edited Jul 31 '17 at 17:36

answered Jul 31 '17 at 17:29

node_modules

4,790
6
21
37

@chris85 Thanks for pointing out, forgot to state that! :) – node_modules Jul 31 '17 at 17:32
1

Or use single quoutes inside a double quoted string. Easier to read for a start than escaping, specially when the html gets a bit complex – RiggsFolly Jul 31 '17 at 17:35
@RiggsFolly Edited in my answer, thank you :) – node_modules Jul 31 '17 at 17:37
@C0dekid Thank You. It helped. One more thing can you tell me what is the best way to use width and height attributes inside this same image tag. – Jul 31 '17 at 18:02
@user8380237 Like you did with the `src` attribute. – node_modules Jul 31 '17 at 18:09
This gives me error. – Jul 31 '17 at 18:12
@user8380237 Because the width and height attributes also need to have single quotes :) – node_modules Jul 31 '17 at 18:31

score 0 · Answer 2 · answered Jul 31 '17 at 17:35

0

<tbody>
<?php
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result, MYSQLI_BOTH))
{
    echo '
<tr>
    <th>1</th>
    <td>' . $row['itemId'] . '</td>
    <td><img src="imageView.php?image_id=' . $row["imageId"] . '>" /></td>
    <td>' . $row['itemName'] . '</td>
</tr>
';
    }
?>
<tr>
    <th scope="row">2</th>
    <td>Nicholas</td>
    <td>Sanchez</td>
    <td>username</td>
</tr>
<tr>
    <th scope="row">3</th>
    <td>Debra</td>
    <td>Shaw</td>
    <td>username</td>
</tr>
<?php mysqli_close($conn); ?>
</tbody>

If you use HTML in a string use single quotes to surround it. You also shouldn't use opening and closing PHP tags within PHP code. In addition to that you had malformed HTML in your code:

<br/>**
    <td/>
     <br/>

You don't need to put a br /> between two cells.

answered Jul 31 '17 at 17:35

Dan

5,140
2
15
30

Thank You. It helped. One more thing can you tell me what is the best way to use width and height attributes inside this same image tag. – Jul 31 '17 at 18:09
You can either use the `height` and `width` attribute or just format the tag with the `style`-attribute and some CSS (`height` and `width` to ignore the image proportions, `max-width` and `max-height` to scale it down proportional). – Dan Jul 31 '17 at 18:14
This gives me error. – Jul 31 '17 at 18:16
Which error message do you get? – Dan Jul 31 '17 at 18:17
Parse error: syntax error, unexpected '90' (T_LNUMBER), expecting ',' or ';' in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC9\data\localweb\drop2\component-tables - Copy.php on line 1378 – Jul 31 '17 at 18:18
You're still mixing `'` and `"` within a string literal. If you want to use the same character you used to specify the string you have to escape it with a backslash, like this: `echo 'this wouldn\'t break the string.';` ... and the closing tag begins with the slash: `` instead of ``. – Dan Jul 31 '17 at 18:21
Thanx @Spingolini Now i got my mistake – Aug 01 '17 at 04:58

how to access image from mysql to html table using php

2 Answers2