Re: Copy Non-Text file from One Server to Another Issues

Alice Wei wrote: > Hi, > > I have a simple code as shown in the following: > > <?php > > //original file > $file = "http://remote_server/copy/play.txt"; > $file2 = "http://remote_server/copy/test.jpg"; > $file3 = "http://remote_server/copy/sample.pdf"; > > //directory to copy to (must be CHMOD to 777) > $copydir = "."; > > $data = file_get_contents($file); > $file2 = fopen($copydir . "/play.txt", "w+"); > fputs($file2, $data); > fclose($file2); > > $data2 = file_get_contents($file2); > $file3 = fopen($copydir . "/test.jpg", "w+"); > fputs($file3, $data2); > fclose($file3); > > $data3 = file_get_contents($file3); > $file4 = fopen($copydir . "/sample.pdf", "w+"); > fputs($file4, $data3); > fclose($file4); > > $data5= file_get_contents("play.txt"); > echo $data5 . "<br>"; > > $data6= file_get_contents("test.jpg"); > echo $data6 . "<br>"; > > $data7= file_get_contents("sample.pdf"); > echo $data7 . "<br>"; > > ?> > > The first one, $file, has no problems, and I did get the contents copied correctly. However, > the second and third one, I get this error: > > Warning: file_get_contents() expects parameter 1 to be string, resource > given in C:\Inetpub\wwwroot\test\hello.php on line > 18 > ./test.jpg > Warning: file_get_contents() expects parameter 1 to be string, resource > given in C:\Inetpub\wwwroot\test\hello.php on line > 25 > ./sample.pdf > > I notice that the empty file gets created correctly, but I cannot put in the file contents. If > file_get_contents only supports text, could anyone please give me some suggestions on what type of > function I should use to put in the file contents so that it is readable? > > Thanks in advance. > > Alice > _________________________________________________________________ > Use Messenger to talk to your IM friends, even those on Yahoo! > > http://ideas.live.com/programpage.aspx?versionId=7adb59de-a857-45ba-81cc-685ee3e858fe Well, that's because you are probably thinking that $file2 = "http://remote_server/copy/test.jpg"; because you assigned it at the beginning of your code. However, later in your code you do $file2 = fopen($copydir . "/play.txt", "w+"); which assigns a file pointer resource to $file2, which overwrites what you previously assigned. Try not using the same variable name for4 multiple things ;-) Same for $file3. -- Thanks! -Shawn http://www.spidean.com