PHP imagejpeg doesn't work - php

Question

So I looked almost every question and tried to work to show my image using the php script below. But it doesn't work. I also tried to see if GD library is enabled with PHPInfo() and it is working as well. I am very new to PHP but can't seem to get this working. Thanks for your help!
EDIT: I get a broken image icon after I click on upload a doc button.
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_FILES['photo'])
&& is_uploaded_file($_FILES['photo']['tmp_name'])
&& $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
foreach ($_FILES['photo'] as $key => $value) {
echo "$key : $value<br />";
}
if($_FILES['photo']['type'] == 'image/jpeg') {
$tmp_img = $_FILES['photo']['tmp_name'];
$image = imagecreatefromjpeg($tmp_img);
header('Content-Type: image/jpeg');
imagejpeg($image,NULL);
imagedestroy($image);
} else {
echo "Uploaded file ewas not a jpg image.";
}
echo "no photo uploaded.";
}
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>

Answer 1

You can try add die(); after imagedestroy($image); to prevent any other output except image data. Also change imagejpeg($image,NULL); to imagejpeg($image);.

Answer 2

You cannot access to image direcly with tmp_name. Firstly you should save the image to the server and after that you can use with imagecreatefromjpeg
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_FILES['photo'])
&& is_uploaded_file($_FILES['photo']['tmp_name'])
&& $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
if($_FILES['photo']['type'] == 'image/jpeg') {
$tmp_img = $_FILES['photo']['tmp_name'];
$uniq_name = uniqid().".".explode(".",$tmp_img)[1];
move_uploaded_file($_FILES['photo']['tmp_name'],$uniq_name);
$image = imagecreatefromjpeg($uniq_name);
header('Content-Type: image/jpeg');
imagejpeg($image);
imagedestroy($image);
} else {
echo "Uploaded file ewas not a jpg image.";
}
echo "no photo uploaded.";
}
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>

Answer 3

I think the problem lies with the echo and the output strings after the PHP section (HTML code). Remove the foreach loop and output HTML only if it is a new page or if something is wrong.
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_FILES['photo'])
&& is_uploaded_file($_FILES['photo']['tmp_name'])
&& $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
if($_FILES['photo']['type'] == 'image/jpeg') {
$tmp_img = $_FILES['photo']['tmp_name'];
$image = imagecreatefromjpeg($tmp_img);
header('Content-Type: image/jpeg');
imagejpeg($image,NULL);
imagedestroy($image);
exit; // Add this to stop the program here.
} else {
echo "Uploaded file was not a jpg image.";
}
} else {
echo 'No file was sent.';
}
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>
And one other thing, if you code using Notepad, be sure that there are no hidden characters at the beginning of your file.