Read $_POST value in a variable

[andytan91]

Hey guys, how do i pass the $_POST['argument'] into the $newfile variable successfully?

 

<?php
$file = 'service_pack.bat';
$newfile = 'C:\xampp\htdocs\new\old\$_POST['argument']';

if(isset($_POST['argument']) && $_POST['submit'] == "Update")
{
copy($file, $newfile);
}

?>

<b>Save Template</b>
<p>
<form action="copyfile.php" method="post">
Save file as: <input type="text" name="argument"<br>
<input type="submit" name="submit" value="Update">  <p>
</form>
</p>

[GKWelding]

$newfile = 'C:\xampp\htdocs\new\old\'.$_POST['argument'];

 

OR

 

$newfile = "C:\xampp\htdocs\new\old\$_POST['argument']";

 

notice the use of single and double quotes, basic php.

[andytan91]

I tried both and it doesnt work..the 2nd one  says...

 

Warning: copy(C: mpp\htdocs ew\old$_POST['argument']) [function.copy]: failed to open stream: Invalid argument in C:\xampp\htdocs\new\copyfile.php on line 10

Save Template

[bh]

2nd: you need those brackets

$newfile = "C:\xampp\htdocs\new\old\{$_POST['argument']}";

 

(Anyway this way of creating your newfile is not the best practice.)

[andytan91]

it gives me Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs\new\copyfile.php on line 3...

 

 

what other options do i have?

[bh]

Oh, sure... cuz the '\' -> you have to escape it

 

1st:

$newfile = "C:/xampp/htdocs/newold/{$_POST['argument']}";

 

2nd:

$newfile = "C:\\xampp\\htdocs\\newold\\{$_POST['argument']}";

 

a better way that you create a simple filename from your

$_POST['argument']