How can i view uploaded image in php?

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

its my upload.php

the image is uploading fine. but the image is not displaying. a small box only showing.

this part not working...

/* Displaying Image*/
       $image=$_FILES["file"]["name"]; 
              $img="upload/".$image;
              echo '
user3214024user3214024
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Try this:
header['Content-Type: image/jpeg'];
readfile[$imgPath];
 answered Feb 11, 2014 at 13:28
 
ManuManu
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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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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
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Don't put $img in quotations. When you do it the output is simply $img, not upload/...
echo < img src=" ' ,$img ,' ">;
 

Barnee
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 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
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 answered Feb 21, 2018 at
13:46
 
 

			
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