My image is not getting displayed after resizing it.
require('connect.php');
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
$imagename = $_FILES["image"]["name"];
// Get new dimensions
list($width, $height) = getimagesize($tmpName);
echo "orginal width".$width."<br/>";
echo "orginal height".$height."<br/>";
$new_width= $width * 0.5;
$new_height = $height * 0.5;
echo "new width".$new_width."<br/>";
echo "new height".$new_height."<br/>"; //Its working fine till here.
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($tmpName);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
echo "My image"."<br>";
echo '<img src="data:image/png|image/jpeg|image/gif;base64,' . base64_encode( $image_p ) . '" style="max-width:400px; max-height:300px;"/>';
/* This line is not able to display the image.
Also tried imagejpeg($image_p, null, 100);. Not working , but anyway i need to
display image alongwith other content so i have used echo. Not sure which
variable i should work with to display the image and how do i do it.*/
}
It seems to me that you are missing one step, encoding the image resource as a jpeg image (for example...).
You could output the result from something like imagejpeg to a file and just serve the file or you could capture its output using output buffering and serve it as you do now.
Related
I am trying to load an image from an external URL, then resize it and show it in a PDF. I am trying to achieve it using a single image for now, but the whole functionality will be inside a foreach loop for handling a lot of very large images.
Firstly, I resized the image, then get the contents of the image, applied base65 encoding, built up a source string from that and added the string to my img src tag. Here's my code -
$filename = 'https://jooinn.com/images/nature-319.jpg'; // URL of the image
$percent = 0.25; // percentage of resize
// Content type
header('Content-type: image/jpeg');
// Get new dimensions
list($width, $height) = getimagesize($filename);
$new_width = $width * $percent;
$new_height = $height * $percent;
// Resample
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// Output
$imageData = base64_encode(file_get_contents($image_p));
// Format the image SRC: data:{mime};base64,{data};
$src = 'data: '.mime_content_type($image_p).';base64,'.$imageData;
// Echo out a sample image
echo '<img src="' . $src . '">';
imagedestroy($image_p);
I think the problem is with this line $imageData = base64_encode(file_get_contents($image_p));, I'm doing it wrong. It works nicely with URL's but how can I make it work for a resized image here? For example, the following works perfectly as long as I don't use the resized image -
$filename = 'https://jooinn.com/images/nature-319.jpg'; // URL of the image
// Output
$imageData = base64_encode(file_get_contents($filename));
// Format the image SRC: data:{mime};base64,{data};
$src = 'data: '.mime_content_type($filename).';base64,'.$imageData;
// Echo out a sample image
echo '<img src="' . $src . '">';
Indeed, as you said, the following line in the code is wrong:
$imageData = base64_encode(file_get_contents($image_p));
The $image_p variable is not a filename but a resource created by imagecreatetruecolor.
You first have to convert it to a jpeg file using imagejpeg().
You can avoid saving an intermediate file before encoding to base64 using the ob_*xxx* functions
ob_start();
imagejpeg($image_p);
$imageContent = ob_get_contents();
$imageData = base64_encode($imageContent);
ob_end_clean();
You also have a problem with this line, again as $image_p is not a filename:
$src = 'data: '.mime_content_type($image_p).';base64,'.$imageData;
As you are creating a jpeg file you should just replace it with:
$src = 'data: image/jpeg;base64,'.$imageData;
For convenience here is the full working script:
$filename = 'https://jooinn.com/images/nature-319.jpg'; // URL of the image
$percent = 0.25; // percentage of resize
// Content type
header('Content-type: image/jpeg');
// Get new dimensions
list($width, $height) = getimagesize($filename);
$new_width = $width * $percent;
$new_height = $height * $percent;
// Resample
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// Output
ob_start();
imagejpeg($image_p);
$imageContent = ob_get_contents();
$imageData = base64_encode($imageContent);
ob_end_clean();
// Format the image SRC: data:{mime};base64,{data};
$src = 'data: image/jpeg;base64,'.$imageData;
// Echo out a sample image
echo '<img src="' . $src . '">';
imagedestroy($image_p);
Is there something I am missing here? If I change imagejpeg($thumb, $newImage);
to imagejpeg($thumb); it echos a load of unreadable characters. The thumb image directory exists.
My ultimate aim is to copy an image at a lower quality.
$filename = $imageDirectory . '/1.jpg';
$percent = 0.5;
$newImage = $imageDirectory . '/thumbs/1.jpeg';
echo "image: <br>";
echo $filename;
list($width, $height) = getimagesize($filename);
$newwidth = $width * $percent;
$newheight = $height * $percent;
// Load
$thumb = imagecreatetruecolor($newwidth, $newheight);
$source = imagecreatefromjpeg($filename);
// Resize
imagecopyresized($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
// Output
imagejpeg($thumb, $newImage);
UPDATE: I now realize that the second parameter must be an image location with the new name. So... I have redefined $newImage. The path is fine... if I upload an image named 1.jpg to that location manually it exists at that path.
You'll have to add the correct Content-Type header for the response if you want to output the image to the browser:
header('Content-Type: image/jpeg');
imagejpeg($yourimage);
Please check the first example in the documentation for more information. If you want to decrease the quality before outputting to your browser check the third example.
I found the following script from PHP site to half the size of the image. I use the database to fetch the link of the image. Other things are working correctly which means there isn't any kind of error anywhere except this one.
echo "<img src='".// File and new size
$filename = '$row["image"]';
$percent = 0.5;
// Content type
header('Content-Type: image/jpeg');
// Get new sizes
list($width, $height) = getimagesize($filename);
$newwidth = $width * $percent;
$newheight = $height * $percent;
// Load
$thumb = imagecreatetruecolor($newwidth, $newheight);
$source = imagecreatefromjpeg($filename);
// Resize
imagecopyresized($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
// Output
imagejpeg($thumb);
"'>"
Error:
The whole page is destroyed and there is a broken link of an image.
Hope you guys will help me!
It works after you drop those lines:
echo "<img src='".// File and new size
"'>"
header does it for you, informs that there is jpg coming, no need to echo image tag.
The other solution is to remove this line:
header('Content-Type: image/jpeg');
And create new file, then use it as a source of an image:
// Output
$new_filename = 'new_image.jpg';
imagejpeg($thumb,$new_filename);//saves new image to a file, instead of outputting it to the screen
echo "<img src='$new_filename'>";
Try this it works
<?php
$filename = '$row["image"]';
$percent = 0.5;
// Content type
header('Content-Type: image/jpeg');
// Get new sizes
list($width, $height) = getimagesize($filename);
$newwidth = $width * $percent;
$newheight = $height * $percent;
// Load
$thumb = imagecreatetruecolor($newwidth, $newheight);
$source = imagecreatefromjpeg($filename);
// Resize
imagecopyresized($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
// Output
echo "<img src='".imagejpeg($thumb)."'>";
?>
Scenario : Earlier I was reading the url of image directly and resized the image into 4 different size . But I had execution time out . Now I read the url and copy it into a temporary folder and pass the images in local temporary folder to imagecreatefromjpeg().
protected static function saveImage($row,$url){
$percent = 1.0;
$imagethumbsize = 200;
$db = PJFactory::getDbo();
$details = $db->getImageDetails();
$max = sizeof($details);
$tempfilename = "C:".DS."xampp".DS."htdocs".DS."opg-uat".DS."img".DS."temp".DS.$row['CategoryID'].".jpg";
$tempcopy = copy($url,$tempfilename);
foreach ($details as $array) {
$new_width=$array[2];
$new_height=$array[3];
$newfilename = "C:".DS."xampp".DS."htdocs".DS."opg-uat".DS."img".DS."c".DS.$row['CategoryID']."-".$array[1].".jpg";
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($tempfilename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
imagejpeg($image_p, $newfilename);
}
}
Error : The images are correctly being saved in temp folder . But in the destination folder images of all sizes are created but the image looks only black . (Not getting the actual image) . I guess there is some problem with the file reading from local .
Any idea ?
Thanks in advance.
If you are trying to read files other than JPG files, it will return only black images,
So while reading images, check which filetype your file really has and then call the correct function between the three (jpeg, png, gif).
Do you set $width and $height anywhere?
$width = imagesx($image);
$height= imagesy($image);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
I am working on improving my Facebook app. I need to be able to resize an image, then save it to a directory on the server. This is the code I have to resize:
<?php
// The file
$filename = 'test.jpg';
$percent = 0.5;
// Content type
header('Content-type: image/jpeg');
// Get new dimensions
list($width, $height) = getimagesize($filename);
$new_width = $width * $percent;
$new_height = $height * $percent;
// Resample
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// Output
imagejpeg($image_p, null, 100);
?>
My question is, how would I save this resized image? Would I need to? Is there a way to manipulate the resized image without saving it?
According to the manual on imagejpeg(), the optional second parameter can specify a file name, which it will be written into.
Filename
The path to save the file to. If not set or NULL, the raw image stream will be outputted directly.
To skip this argument in order to provide the quality parameter, use NULL.
It's usually a good idea to write the results to disk for some basic caching, so that not every incoming request leads to a (resource intensive) GD call.
function resize($img){
/*
only if you script on another folder get the file name
$r =explode("/",$img);
$name=end($r);
*/
//new folder
$vdir_upload = "where u want to move";
list($width_orig, $height_orig) = getimagesize($img);
//ne size
$dst_width = 110;
$dst_height = ($dst_width/$width_orig)*$height_orig;
$im = imagecreatetruecolor($dst_width,$dst_height);
$image = imagecreatefromjpeg($img);
imagecopyresampled($im, $image, 0, 0, 0, 0, $dst_width, $dst_height, $width_orig, $height_orig);
//modive the name as u need
imagejpeg($im,$vdir_upload . "small_" . $name);
//save memory
imagedestroy($im);
}
it should be work
http://www.php.net/manual/en/function.imagecopyresampled.php#90038