php copy（）无法正常工作

Question

I want to do a simple copy using a ajax call. this is my code but it does not work . i kepp getting : a copy(../../images/merchant/particulars/208/) failed to open stream: Is a directory in some/filepath on scriptname.php line x , kind of error.

corrected code: 

$dir_to_make = '../../images/merchant/particulars/'.$copytothisstore;
$dir = '../../images/merchant/particulars/'.$copytothisstore.'  /'.$copyvalue;
$image_to_copy = '../../images/merchant/particulars/'.$copyfromthisstore.'/'.$copyvalue;






    if(is_file($image_to_copy)){

        //chk if there is a folder created for this store
        if(!is_dir($dir_to_make)){
              mkdir($dir_to_make, 0755);  
              chmod($dir_to_make, 0755);

              //copy the image 
              if (!copy($image_to_copy,$dir)) {
                  echo "failed to copy $image_to_copy\n";
              } else {
                  echo"all is well!!";
              }

           } else {
               chmod($dir_to_make, 0755);
               if (!copy($image_to_copy,$dir)) {
                  echo "failed to copy $image_to_copy\n";
              } else {
                  echo"all is well!!";
              }

           }


           echo"$image_to_copy does exist!";
        } else{
            echo"$image_to_copy does not exist!";
        }

Answer 1

Please read your error.

copy(../../images/merchant/particulars/208/) failed to open stream: Is a directory in some/filepath

It says that your source file is not a file, but directory.

Simple debugging could always solve your problems:

$image_to_copy ='../../images/merchant/particulars/'.$copyfromthisstore.'/'.$copyvalue;
echo $image_to_copy; // yes, that could give you the answer

It will show you, that $copyvalue in you example is empty.

---

If you wonder why this is returning TRUE ...

if(file_exists($image_to_copy)){

..it's because directory ../../images/merchant/particulars/208/ does exists.

As manual says:

• file_exists() - Returns TRUE if the file or directory specified by filename exists; FALSE otherwise.

You should change it to:

if(is_file($image_to_copy)){

• is_file() - Returns TRUE if the filename exists and is a regular file, FALSE otherwise.

---

Another thing is destination shuld be file as well:

copy($image_to_copy, $dir.'file.jpg');

Manual:

• http://php.net/manual/en/function.copy.php

Answer 2

I think you are confusing a simple thing. Here is a simple example on how to use copy()

$file = 'path/to/filename.ext'; // Example: http://adomain.com/content/image.jpg

$destination_folder = 'path/of/the/destination/folder';
//example: $_SERVER["DOCUMENT_ROOT"]."/content/files/"


// Execute the copy function
copy($file, $destination_folder.'/newFilaname.ext');

Answer 3

Use this function

!move_uploaded_file($image_to_copy, $dir)

http://php.net/manual/en/function.move-uploaded-file.php