PHP Upload image then show - php

Question

I'm trying to figure out how to allow image upload with the image showing up after it is uploaded. I have found this tutorial on uploading images but I'm not sure how to display them afterwards. Would I have to save it in the database then pull it up afterwards somehow?
<form action="upload.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
echo "File is not an image.";
$uploadOk = 0;
}
}
?>

Answer 1

I think you would benefit from an uploading class or function that returns information for your uploaded image. This will help you store the results or display as you are looking to do. Here is one loosely based on what you provided with notation:
Form:
<form action="" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
Script:
<?php
function UploadImage($settings = false)
{
// Input allows you to change where your file is coming from so you can port this code easily
$inputname = (isset($settings['input']) && !empty($settings['input']))? $settings['input'] : "fileToUpload";
// Sets your document root for easy uploading reference
$root_dir = (isset($settings['root']) && !empty($settings['root']))? $settings['root'] : $_SERVER['DOCUMENT_ROOT'];
// Allows you to set a folder where your file will be dropped, good for porting elsewhere
$target_dir = (isset($settings['dir']) && !empty($settings['dir']))? $settings['dir'] : "/uploads/";
// Check the file is not empty (if you want to change the name of the file are uploading)
if(isset($settings['filename']) && !empty($settings['filename']))
$filename = $settings['filename'];
// Use the default upload name
else
$filename = preg_replace('/[^a-zA-Z0-9\.\_\-]/',"",$_FILES[$inputname]["name"]);
// If empty name, just return false and end the process
if(empty($filename))
return false;
// Check if the upload spot is a real folder
if(!is_dir($root_dir.$target_dir))
// If not, create the folder recursively
mkdir($root_dir.$target_dir,0755,true);
// Create a root-based upload path
$target_file = $root_dir.$target_dir.$filename;
// If the file is uploaded successfully...
if(move_uploaded_file($_FILES[$inputname]["tmp_name"],$target_file)) {
// Save out all the stats of the upload
$stats['filename'] = $filename;
$stats['fullpath'] = $target_file;
$stats['localpath'] = $target_dir.$filename;
$stats['filesize'] = filesize($target_file);
// Return the stats
return $stats;
}
// Return false
return false;
}
?>
To use:
<?php
// Make sure the above function is included...
// Check file is uploaded
if(isset($_FILES["fileToUpload"]["name"]) && !empty($_FILES["fileToUpload"]["name"])) {
// Process and return results
$file = UploadImage();
// If success, show image
if($file != false) { ?>
<img src="<?php echo $file['localpath']; ?>" />
<?php
}
}
?>
RAW Feedback:
// This is what the array would look like on return of successful upload:
Array
(
[filename] => animal.png
[fullpath] => /data/19/2/133/150/2948313/user/2524254/htdocs/mydomain/uploads/animal.png
[localpath] => /uploads/animal.png
[filesize] => 35702
)

Answer 2

yes,you would have to save the path to the file in the database and fetch it but for your use case,you can save the path to a $_SESSION variable and then echo the path immediately the script is done.
But you first have to complete the file transfer with the move_uploaded_file function as without that,you would not be able to retrieve the file path as they are stored as temporary files and deleted once the script is interpreted
http://php.net/manual/en/function.move-uploaded-file.php
After this is done,you are to get the path to the file and use the normal img HTML tag

Answer 3

create <img src="" widht="" height="" /> forever u must move the image to directory path and now i get the image name from table after submit the form.. and given the url to img..example.. ur directory name uploads/img . now your file name save in database table as image01.jpg . sample
$img= 'select imagename from table name ';
if(count($img))
{
<img src="<?php echo 'uploads/img/'.$img" widht="10px" height="20px" /></div>
}

Answer 4

if you upload image on data base , data loading will be slow because image siz too large. better method is upload image in folder & save image file path in data base .when you retrieve image call image web root on image tag
example
Saving Filepath Of Uploaded Image To MySQL Database
GET image path
name refers to the filename on the client-side. To get the filename (including the full path) on the server-side, you need to use tmp_name:
$check = fopen($_FILES["UploadFileName"]["tmp_name"], 'r');