Php Code For Image Upload And Display

[Solved] Php Code For Image Upload And Display | Php - Code Explorer | yomemimo.com
Question : basic code for file upload in php

Answered by : bodhaditya-mukherjee

//This is the minimal code for an image upload for first time learners
//html portion
<!DOCTYPE html>
<html>
<head>	<title>ImageUpload</title>
</head>
<body>	<form action="upload.php" method="post" enctype="multipart/form-data">	<label>Username</label>	<input type="text" name="username">	<br>	<label>UploadImage</label>	<input type="file" name='myfile'>	<br/>	<input type="submit" value="upload">	</form>
</body>
</html> //php portion <?php	$user=$_POST['username'];	$image=$_FILES['myfile'];	echo "Hello $user <br/>";	echo "File Name<b>::</b> ".$image['name'];	move_uploaded_file($image['tmp_name'],"photos/".$image['name']);	//here the "photos" folder is in same folder as the upload.php,	//otherwise complete url has to be mentioned	?>

Source : | Last Update : Fri, 10 Jul 20

Question : Image upload in PHP code with databases

Answered by : sourav-attri

$file = file(1000,100000)."-".$_FILES['file']['name']; $file_loc = $_FILES['file']['tmp_name']; $file_size = $_FILES['file']['size']; $file_type = $_FILES['file']['type']; $folder="uploads/";

Source : | Last Update : Fri, 13 May 22

Question : img upload in php

Answered by : kinjal-suryavanshi

 if(isset($_FILES['image'])) { $img_name = $_FILES['image']['name']; //getting user uploaded name $img_type = $_FILES['image']['type']; //getting user uploaded img type $tmp_name = $_FILES['image']['tmp_name']; //this temporary name is used to save/move file in our folder. // let's explode image and get the last name(extension) like jpg, png $img_explode = explode(".",$img_name); $img_ext = end($img_explode); //here we get the extension of an user uploaded img file $extension= ['png','jpeg','jpg','gif']; //these are some valid img extension and we are store them in array. 

Source : | Last Update : Wed, 01 Sep 21

Question : php photo upload

Answered by : milan-niroula

 if(isset($_FILES['image'])) { $img_name = $_FILES['image']['name']; //getting user uploaded name $img_type = $_FILES['image']['type']; //getting user uploaded img type $tmp_name = $_FILES['image']['tmp_name']; //this temporary name is used to save/move file in our folder. // let's explode image and get the last name(extension) like jpg, png $img_explode = explode(".",$img_name); $img_ext = end($img_explode); //here we get the extension of an user uploaded img file $extension= ['png','jpeg','jpg','gif']; //these are some valid img extension and we are store them in array. 

Source : https://www.codegrepper.com/search.php?q=php%20photo%20upload | Last Update : Wed, 01 Jun 22

Question : how to print image just on side where upload php

Answered by : brainy-baboon-yy4hjtuf4idx

<?php    $galleryPath = "Gallery/";        $descriptions = new DOMDocument("1.0");    $descriptions->appendChild($descriptions->createElement("files"));        while (($uploadedFile = UploadedFiles::fetchNext()) != null){        $sourceFile = $uploadedFile->getSourceFile();        $sourceFileName = $sourceFile->getFileName();            $descriptionsFile = $descriptions->createElement("file");        $descriptionsFile->setAttribute("name", $sourceFileName);        $descriptionsFile->setAttribute("width", $sourceFile->getWidth());        $descriptionsFile->setAttribute("height", $sourceFile->getHeight());        $descriptionsFile->setAttribute("description", $uploadedFile->getDescription());        $descriptions->documentElement->appendChild($descriptionsFile);                    $sourceFile->save($galleryPath. "/" .$sourceFileName);    }            $descriptions->save("Descriptions.xml");?>

Source : https://www.aurigma.com/docs/iu/WritingClient-SideScriptandHandlingUploadusingImageUploaderPHP.htm | Last Update : Sat, 25 Jul 20

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