unable to display my images

[tchick]

Hi all,

 

Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else?

Appreciate your help with this.

I have included both of the pages codes i am using.

 

Thanks

Tony

 

image2.php

 

<?php

include("common.php");

error_reporting(E_ALL);

$link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error());

mysql_select_db(db) or die(mysql_error());

$sql = "SELECT id FROM photos";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

?>

<table border="1"><tr><td>id</td><td>image</td></tr>

<?php

while($row=mysql_fetch_assoc($result)){

print '<tr><td>'.$row['id'].'</td><td>';

print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">';

}

echo '</td></tr></table>'

?>

 

image1.php

 

<?php

ob_start();

include("common.php");  

mysql_connect(host,username,password) or die(mysql_error());

mysql_select_db(db) or die(mysql_error()); 

 

$query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}";

$row = mysql_fetch_array($query);

$content = $row['image']; 

header('Content-type: image/jpg');

    echo $content;

}

ob_end_flush();

?>

 

 

[KevinM1]

In image1.php, you're attempting to echo the image data itself, but in image2.php you're trying to use that as the path of the image source.  Raw image data is not the same as the image's path.  You also have syntax errors in several places, including using the wrong value for your GET in image1.php.  It should simply be $_GET['id'].  That said, just so I'm completely clear, fixing those syntax errors won't fix your main problem.

 

I'm not sure why you don't simply use a function to do what you want to do.  It's simpler/cleaner than having a separate script.  Try something like the following (all in one file - not tested):

 

   function displayImage($id){
      $query = mysql_query("SELECT imgage FROM photos WHERE id = $id");
      $row = mysql_fetch_array($query);
      $content = $row['image']; 

      header('Content-type: image/jpg');
      echo $content;
   }

   include("common.php");
   error_reporting(E_ALL);

   $link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error());
   mysql_select_db(db) or die(mysql_error());

   $sql = "SELECT id FROM photos";
   $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
?>

<table border="1"><tr><td>id</td><td>image</td></tr>

<?php
   while($row=mysql_fetch_assoc($result)){
      echo '<tr><td>'.$row['id'].'</td><td>';
      displayImage($row['id']);
   }

   echo '</td></tr></table>';

 

At the very least, that's the approach you should be using.

[tchick]

Hi Nightslyr,

 

Thanks for the reply, errors taken on board, I have tried your function script which as you rightly say is the way forward. I have run the script you supplied (untested) and it results in a file save/download box popping up ?.

 

Thanks Tony

[PFMaBiSmAd]

That's because you cannot output an image directly in the HTML on a web page, primarily because you cannot output a header() in a HTTP request after you have sent any content to the browser. You must use an HTML <img> tag, because it is the browser that requests the image and renders it on the page.

 

Your original concept was correct, but it has some errors in it.

 

In your image2.php code, you have a mis-placed " in your <img> tag. Use the following line of code -

print '<img src="image1.php?id='.$row['id'].'" height="75" width="100">';

 

Your image2.php code has several problems -

 

1) You would need to use $_GET['id'] as that is the name of the GET parameter you are putting on the end of the url ?id=123

 

2) You need to both cast $_GET['id'] as an integer and test that $_GET['id'] has a number in it to prevent sql injection and to prevent sql errors in your query.

 

3) You apparently have typo in the column name in your query. imgage should be image

 

4) You should store the image type image/jpg in a column in your table so that your code will support other image types. You would retrieve that value when you retrieve the image data and output it in the Content-type header().

[tchick]

Hi PFMaBiSmAd,

 

Thanks for your reply, I have corrected the typo errors (sorry) not sure what you mean though regarding the $_GET issues points 1&2.

I will add column to insert the image type into.I have now corrected the typos and included the corrected code below, but still i am getting the id number colum showing ok but the image is still a blank square with a red cross inside?.Is it the $_GET that is causing this ?

I know when i check the properties of the images it shows  image1.php?id=10  etc. so i think this looks ok?.

 

Thanks Tony

 

image2.php

 

<?php

include("common.php");

error_reporting(E_ALL);

$link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error());

mysql_select_db(db) or die(mysql_error());

$sql = "SELECT id FROM photos";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

?>

<table border="1"><tr><td>id</td><td>image</td></tr>

<?php

while($row=mysql_fetch_assoc($result)){

print '<tr><td>'.$row['id'].'</td><td>';

print '<img src="image1.php?id='.$row['id'].'" height="75" width="100">';

}

echo '</td></tr></table>'

?>

 

image1.php

 

<?php

ob_start();

include("common.php");   

  mysql_connect(host,username,password) or die(mysql_error());

  mysql_select_db(db) or die(mysql_error()); 

 

$query = mysql_query("SELECT image FROM photos WHERE id=$_GET['id']";

$row = mysql_fetch_array($query);

$content = $row['image']; 

header('Content-type: image/jpg');

    echo $content;

}

ob_end_flush();

?>

 

 

 

[litebearer]

You haven't closed the parans here...

$query = mysql_query("SELECT image FROM photos WHERE id=$_GET['id']"; 

[PFMaBiSmAd]

And, your image1.php code contains an extra } that would be producing a fatal php parse error and no output at all. Remove that stray } at the third line from the end of the file.

 

 

 

 

[tchick]

Hi PFMaBiSmAd & litebearer

 

Thanks for your replies, I should stop doing several things at the same time, these errors were just my typos (sorry again!)even with these typos corrected i am still getting the id number colum showing ok but the image is still a blank square with a red cross inside?.

 

Thanks Tony

 

image2.php

 

<?php

include("common.php");

error_reporting(E_ALL);

$link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error());

mysql_select_db(db) or die(mysql_error());

$sql = "SELECT id FROM photos";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

?>

<table border="1"><tr><td>id</td><td>image</td></tr>

<?php

while($row=mysql_fetch_assoc($result)){

print '<tr><td>'.$row['id'].'</td><td>';

print '<img src="image1.php?id='.$row['id'].'" height="75" width="100">';

}

echo '</td></tr></table>'

?>

 

image1.php

 

<?php

ob_start();

include("common.php");   

  mysql_connect(host,username,password) or die(mysql_error());

  mysql_select_db(db) or die(mysql_error()); 

 

$query = mysql_query("SELECT image FROM photos WHERE id=$_GET['id']");

$row = mysql_fetch_array($query);

$content = $row['image']; 

header('Content-type: image/jpg');

    echo $content;

ob_end_flush();

?>

 

[PFMaBiSmAd]

Your (current) image1.php code will still be producing a fatal php parse error, due to missing {} around the $_GET['id'] in the query (you had these in your original code.)

 

$query = mysql_query("SELECT image FROM photos WHERE id={$_GET['id']}");

[tchick]

Hi PFMaBiSmAd,

 

Fantastic!!!, it finally works thank you so much for your help with this, I really appreaciate you helping me solve this.

 

Regards

Tony

 

[jcg]

Hi all,

I followed all the steps in this topic, but still getting empty square with red cross.

I did not include the code, because it is the same as this topic.

I've tried everything...  other ways, other foruns, ...

I'm getting crazy!!!!!

will not have to do with the apache server?

if an image is from the / images folder, okay ... but an image from the database ... always gives empty square with red cross...

 

Appreciate your help.

 

Thanks

JCG

 

[Pikachu2000]

Start your own new thread, and post your own code in it, within

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tags. This one is over a year old. Locked.