php generate image?

Question

How can i use php to get an image url from database (based on user id) and display it out as an image. http://mysite.com/img.php?id=338576

The code below shows a php script goes to a database , check whether the specific user(using the id in the link above) exist then get the image url out of that user's row.

 <?php
    //connect to database
    include ("connect.php");

    //Get the values
       $id = mysql_real_escape_string($_GET['id']);


    //get the image url
      $result = mysql_query("SELECT * FROM users WHERE id='$id'")
      or die(mysql_error()); 
      $row = mysql_fetch_array($result);
      if($row)
      {
        $img_url = row['imgurl'];
        mysql_close($connect);

      }
      else{
      }

    ?>

The $Img_url is an actual image link www.asite.com/image.jpg

The problem is how can i use php to make an image from the image link($img_url)? By which http://mysite.com/img.php?id=338576 will turn into an image.

I hope someone could guide me

Thanks!

Check out this question. I think it is very similar to yours: stackoverflow.com/questions/1431875/… — Buttle Butkus
– Buttle Butkus, Commented Dec 5, 2011 at 5:13

Brad · Accepted Answer · 2011-12-05 05:10:31Z

3

The easiest way:

header('Location: url/to/image');

You could also proxy the request, which uses your bandwidth:

echo(file_get_contents('url/to/image'));

edited Dec 5, 2011 at 5:10

Brad

164k57 gold badges380 silver badges559 bronze badges

answered Dec 5, 2011 at 4:54

changx

1,9571 gold badge13 silver badges10 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Buttle Butkus Over a year ago

@micha - No, his image is a file whose location is listed in a database. Big difference.

Buttle Butkus Over a year ago

@micha Yeah, so maybe you can +1 my answer now, instead of -1. By the way, you're a pretty cool guy, deleting all your critical comments after you were proven wrong.

noob Over a year ago

for the second method header("Content-type: image/" + $type); is required

Community · Accepted Answer · 2017-05-23 10:34:23Z

-1

This is pretty basic stuff. Am I understanding you correctly? I would just do this:

<?php

$img_html = '<img src="' . $img_url . '" />';


echo $img_html;

?>

Or check this answer: How do I script a php file to display an image like <img src="/img.php?imageID=32" />?

edited May 23, 2017 at 10:34

CommunityBot

11 silver badge

answered Dec 5, 2011 at 4:54

Buttle Butkus

9,53713 gold badges86 silver badges126 bronze badges

8 Comments

sm21guy Over a year ago

that wont work because i need to use mysite.com/img.php?id=338576 as an image src elsewhere.

Buttle Butkus Over a year ago

Why do you need to do that? In that "elsewhere", can't you just use the id to find the actual image location in the database, and then just show the link to the image directly? @micha - Ok, well let's see your answer then!

Buttle Butkus Over a year ago

I.e. use the same query you did above to get the image location. You could store that query in a function and use it wherever you need an image to match an id.

Brad Over a year ago

@ButtleButkus, There are perfectly good reasons why you may want to serve images directly with PHP, rather than outputting HTML. Not everything you do with PHP ends up on a web page...

sm21guy Over a year ago

Why can't i do elsewhere? i have my own reasons to do so. Btw i found an answer to my problem

|

Collectives™ on Stack Overflow

php generate image?

2 Answers 2

3 Comments

8 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

8 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related