PHP Copy function

[Don E]

Hello everyone, Does the copy function only take literal strings for parameters, and not variables that are strings? is there a way to copy a file where the path and name are stored in a variable?http://php.net/manual/en/function.copy.php Thanks.

[jeffman]

Example 1 shows copy being used with variables. So, yes.

[justsomeguy]

For all intents and purposes there is no difference between a literal value and a variable holding the value (unless you're in a situation that requires a variable, like passing by reference).

[Don E]

I meant more along the lines of the following example. $origImg = "uploads/$this->pic_name";$copyImg = "uploads/$this->pic_name" . "orig";copy($origImg, $copyImg);When using the above without quotes I thought why the copy failed but doing the above with double quotes fails as well.

[justsomeguy]

Do one of these: $origImg = "uploads/" . $this->pic_name; $origImg = "uploads/{$this->pic_name}";

[jeffman]

Echo the value of those strings to be sure they are what you think. Then make sure the relative paths are correct. Then make sure the write permissions in the directory are correct.

[jeffman]

I wondered about interpolating the object element inside the quotes, too, but FWIW, this works for me:

$o->file = "file";echo "str/$o->file";

[Don E]

Thanks for the input guys. Unfortunately the following examples didn't work either: $origImg = "uploads/" . $this->pic_name;$origImg = "uploads/{$this->pic_name}"; I echoed out the variables to see if they contain the proper values and they do as well. I even tried to cast them to strings like the following and still get fail on copy: $imgToCopy = (string)"uploads/$this->pic_name";$setAsOrig = (string)"uploads/$this->pic_name" . ".orig"; It works though when setting the variables like the following: $imgToCopy = "uploads/sfsGDFDmyimage.jpg";$setAsOrig = "uploads/sfsGDFDmyimage.jpg" . ".orig";

Edited January 19, 2013 by Don E

[jeffman]

You might try trim() on the strings, in case they contain non-printing characters or newlines at the end. Before you do that, maybe try strlen() to see if the lengths match.

[Don E]

I managed to figure out what the problem was. There was an issue with the upload. Thanks everyone for their input.Nice to see DD around again!