PHP image upload function, save in a dir and then return save image url

Question

I am trying to upload an image to server using PHP and the save inside a dir, and then returning the image url.

HTML:

Answer 1

for file upload try this

<?php if(isset($_POST['submit']))
{
$ImageName = $_FILES['photo']['name'];
$fileElementName = 'photo';
$path = 'images/'; 
$location = $path . $_FILES['photo']['name']; 
move_uploaded_file($_FILES['photo']['tmp_name'], $location); 
} ?>
<form name="form1" id="form1" method="post" action="" enctype="multipart/form-data">
<input type="file" name="photo">
<input type="submit" name="submit">
</form>

Answer 2

First you need a multipart/form-data form for uploading. This is a must :)

<form action="upload_file.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>

The PHP part is fairly simple: This would result your file stored in "upload/{filename}" The main part you want to consider is how to get the filename and back to your write_string_to_database procedure, you could do a simple script after the upload page like

save_string_to_database("upload/" . $_FILES["file"]["name"]);

would do the trick.

<?php
if ($_FILES["file"]["error"] > 0)
  {
  echo "Error: " . $_FILES["file"]["error"] . "<br>";
  }
else
  {
  move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
  }
}

Answer 3

<form method='post' action='' enctype='multipart/form-data'> 
  Name  : <input type="text" name="name"  required=""/><br><br>
  Code  : <input type="text" name="code"  required=""/><br><br>
  Price : <input type="text" name="price" required=""/><br><br>
  Image : <input type="file" name="image" required=""/><br><br>
  <button type='submit' class='buy' name="submit">Add Now</button>
</form>
<!--insert data  -->
<?php
session_start();
include('db.php');

if(isset($_POST["submit"]));
{
  /*echo "<pre>";
  print_r($_POST);
  print_r($_FILES);*/
  $name  = $_POST["name"];
  $code  = $_POST["code"];
  $price = $_POST["price"];
  $image = $_FILES["image"]["name"];

  /* folder image save */

  // $target_dir  = "/var/www/html/shivam/new/upload/";
  // $target_file = $target_dir.basename($_FILES["image"]["name"]);
  // /*echo "1121".$target_file;*/

  // $name = basename($_FILES["image"]["name"]);

  // mysqli_query($con,$qry);

  // /* move file */                                   
  // move_uploaded_file($_FILES['image']['tmp_name'],$target_dir.$name);
  /* move_uploaded_file($tmp_name, "$target_dir/$name");*/
  /* end */

  $uploaddir = '/var/www/html/uploads/';
$uploadfile = $uploaddir . basename($_FILES['image']['name']);
echo '44'.$uploadfile;
echo "<p>";

if (move_uploaded_file($_FILES['image']['tmp_name'], $uploadfile)) {
  echo "File is valid, and was successfully uploaded.\n";
} else {
   echo "Upload failed";
}

echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';enter code here
print_r($_FILES);
print "</pre>";
}
?>

Answer 4

This my function, variable $ten_anh is name of file image in html.

function upload_anh($ten_anh){ //$ten_anh la ten tren html vi du "avatar"
if(isset($_FILES[$ten_anh])){
     $errors= array();
     $file_name = $_FILES[$ten_anh]['name'];
     $file_size =$_FILES[$ten_anh]['size'];
     $file_tmp =$_FILES[$ten_anh]['tmp_name'];
     $file_type=$_FILES[$ten_anh]['type'];
     $file_ext=strtolower(end(explode('.',$_FILES[$ten_anh]['name'])));

     $expensions= array("jpeg","jpg","png");

     if(in_array($file_ext,$expensions)=== false){
            $errors[]="Không chấp nhận định dạng ảnh có đuôi này, mời bạn chọn JPEG hoặc PNG.";
     }

     if($file_size > 2097152){
            $errors[]='Kích cỡ file nên là 2 MB';
     }

     if(empty($errors)==true){
            move_uploaded_file($file_tmp,"../images/".$file_name);
            echo "Thành công!!!";
     }
     else{
            print_r($errors);
     }
}

}

Example: - html code:

<input type="file" id="avatar" name="avatar"accept="image/png, image/jpeg"      required/>

• call function php: upload_anh('avatar');

Answer 5

Here's a simple one.

HTML form to upload image

<form enctype="multipart/form-data" action="upload.php" method="POST">
    <input type="hidden" name="MAX_FILE_SIZE" value="512000" />
    Send this file: <input name="userfile" type="file" />
    <input type="submit" value="Send File" />
</form>

Your PHP file that does the Upload

<?php

$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);

echo "<p>";

if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
  echo "File is valid, and was successfully uploaded.\n";
} else {
   echo "Upload failed";
}

echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";

?> 

Source