How can i view uploaded image in php?

after uploading the image to a folder. how to display the image..

its my upload.php

 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "
";
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "
";
    echo "Type: " . $_FILES["file"]["type"] . "
";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB
";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "
";

    if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"]."
";

$image=$_FILES["file"]["name"]; /* Displaying Image*/
      $img="upload/".$image;
      echo '
user3214024user3214024
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2
Try this:
header('Content-Type: image/jpeg');
readfile($imgPath);
 answered Feb 11, 2014 at 13:28
 
ManuManu
9125 silver badges16 bronze badges
As you've already put "upload" in $img



$img="upload/".$image;
You don't need to put it in src anymore
echo '
Mario RadomananaMario Radomanana
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0
I have tried your code and its working finely just need to change the line
echo'';
 answered Apr 9, 2015 at 16:22
 
You need:
echo '';

answered Feb 11, 2014 at 13:28
 
You mixed up some quotes. It should be :
$image = $_FILES["file"]["name"]; 
$img = "upload/".$image;
echo "
putvandeputvande
14.9k3 gold badges32 silver badges50 bronze badges
Don't put $img in quotations. When you do it the output is simply $img, not upload/...
echo < img src=" ' ,$img ,' ">;
 

Barnee
3,0248 gold badges41 silver badges48 bronze badges
 answered Aug 13, 2014 at 10:21
 
We have to Put enctype = "multipart/form-data"
inside the form



  
  
  

     
     
      
     

        
Sent file:  echo $_FILES['image']['name'];  
        
File size:  echo $_FILES['image']['size'];  
        
File type:  echo $_FILES['image']['type'] 
        
 
     
      
  
       
We Have
  //file_upload.php

  if(isset($_FILES['image'])){$errors= array();

  $file_name = $_FILES['image']['name'];

  $file_size = $_FILES['image']['size'];

  $file_tmp = $_FILES['image']['tmp_name'];

  $file_type = $_FILES['image']['type'];

  $file_ext=strtolower(end(explode('.',$_FILES['image']['name'])));
  
  $extensions= array("jpeg","jpg","png");
  
  if(in_array($file_ext,$extensions)=== false){
     $errors[]="extension not allowed, please choose a JPEG or PNG file.";
  }
  
  if($file_size > 2097152) {
     $errors[]='File size must be excately 2 MB';
  }
  
  if(empty($errors)==true) {
     move_uploaded_file($file_tmp,"images/".$file_name);
     echo "Success";
  }else{
     print_r($errors);
  }    } ?>
 
 answered Jun 19, 2019 at 17:20

Just do this
echo '
We have to take the $file_name variable because.. "" (double quotes) can only return text in the variable - it cannot return binary number to show image!
 
Wai Ha Lee
8,29772 gold badges59 silver badges88 bronze badges
 answered Feb 21, 2018 at
13:46
 
 

How can I retrieve uploaded image in php?
STORE & RETRIEVE IMAGE IN DATABASE.
 
STEP 1) CREATE A DATABASE TABLE. 1-database.sql. ... .
 
STEP 2) PHP CONNECT TO DATABASE. 2-connect-db.php. ... .
 
STEP 3) UPLOAD IMAGE TO DATABASE. 3-upload.php. ... .
 
STEP 4) RETRIEVE & SHOW THE IMAGE. DIRECT IMAGE OUTPUT..
 
How do I find my uploaded files in php?
The is_uploaded_file() function in PHP is an inbuilt function which is used to check whether the specified file uploaded via HTTP POST or not. The name of the file is sent as a parameter to the is_uploaded_file() function and it returns True if the file is uploaded via HTTP POST.
How do I view an uploaded image in HTML?
How To Display Uploaded Image In Html Using Javascript ?.
 
Hide file upload button from HTML page and replace it with a text or icon link. ... .
 
Create a label for the file input field. ... .
 
Javascript to display uploaded image in html. ... .
 
Entire code block as a whole required to display uploaded image in html using javascript..
 
How display all images from database in php?
Explanation of PHP code:
We are first selecting the records from the table in the $query variable. Then the $result will execute the query. While loop is used to fetch all the records in the $data to fetch the image from the database. And finally, the fetched images are displayed with the help of the  tag.