Variable Syntax in PHP

Question

as you can see from my code below I have set a variable ($query) equal to the data posted from an outside form. Under that I tested the variable by echoing it, so the variable seems to be established correctly.

The problem is that near the bottom I'm trying to create another variable, called $str_to_find, where I want it set to output my original variable, $query. However, when I view the output, nothing shows up at all after the code processes this variable near the bottom of my code. I dont' understand why it wouldn't display output.

<?php
$query = $_POST['query']; 

echo "$query"; 

find_files('.');
function find_files($seed) {
    if(! is_dir($seed)) return false;
    $files = array();
    $dirs = array($seed);
    while(NULL !== ($dir = array_pop($dirs)))
    {
        if($dh = opendir($dir))
        {
            while( false !== ($file = readdir($dh)))
            {
                if($file == '.' || $file == '..') continue;
                $path = $dir . '/' . $file;
                if(is_dir($path)) {
                    $dirs[] = $path;
                }
                else {
                    if(preg_match('/^.*\.(php[\d]?|js|txt)$/i', $path)) {
                        check_files($path);
                    }
                }
            }   
            closedir($dh);
        }
    }
}

function check_files($this_file) {
    $str_to_find = $query;
    if(!($content = file_get_contents($this_file))) {
        echo("<p>Could not check $this_file</p>\n");
    }
    else {
        if(stristr($content, $str_to_find)) {
            echo("<p>$this_file -> contains $str_to_find</p>\n");
        }
    }
    unset($content);
}
?>

UPDATED CODE

<?php
 $query = $_POST['query'];



find_files('.');
function find_files($seed) 

{


if(! is_dir($seed)) return false;
$files = array();
$dirs = array($seed);
while(NULL !== ($dir = array_pop($dirs)))
{
  if($dh = opendir($dir))
    {
      while( false !== ($file = readdir($dh)))
        {
          if($file == '.' || $file == '..') continue;
          $path = $dir . '/' . $file;
          if(is_dir($path)) {    $dirs[] = $path; }
          else { if(preg_match('/^.*\.(php[\d]?|js|txt)$/i', $path)) { check_files($path); }}
        }
      closedir($dh);
    }
}
}

function check_files($this_file) 
{

$query = $_POST['query'];

$str_to_find = $query;
if(!($content = file_get_contents($this_file))) { echo("<p>Could not check $this_file</p>\n"); }
else { if(stristr($content, $str_to_find)) { echo("<p>$this_file -> contains
$str_to_find</p>\n"); }}
unset($content);
}

?>

php.net/manual/en/language.variables.scope.php $query is defined outside the function check_files(), so it doesn't exists there: either make it global or, better, pass it as an argument to the function — Damien Pirsy
– Damien Pirsy, Commented Feb 24, 2012 at 20:51
function check_files($this_file) { global $query; $str_to_find = $query; — Cheery
– Cheery, Commented Feb 24, 2012 at 20:52

jstephenson · Accepted Answer · 2012-02-24 20:52:53Z

3

This is an issue of scoping. Your $query variable (and indeed any variables not instantiated directly within the function body) is not available within check_files.

You should pass $query in as a parameter to the function.

function check_files($this_file, $query) {
    // ...
}

Another options exists to make the variables 'global', however this is seldom a sensible idea.

answered Feb 24, 2012 at 20:52

jstephenson

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5 Comments

simshaun Over a year ago

+1, but not for the global suggestion. If you have to use global, you are most likely doing it wrong.

jstephenson Over a year ago

@simshaun I have suggested it not to be a good idea, however it is still a valid option and I think something to be aware of with respect to variable scope.

Rob Myrick Over a year ago

Thanks jstephenson, I've tried all of the ideas for this post, and I'm still not getting output...hmmm?

jstephenson Over a year ago

You will need to pass $query into file_files as well (if you haven't already), as of course this suffers the same problem when attempting to pass it on to check_files.

Rob Myrick Over a year ago

jrstephenson, you can see I posted the code that gave me output. The only problem now is that whenever I type certain variations of characters into the form, it will sometimes show nothing, then for other characters it shows just fine. The best results happen when I use a single digit. This even more absurd because it an input type=text on the original form. Do you have any thoughts on this?

John V. · Accepted Answer · 2012-02-24 20:54:01Z

2

The reason it's not working is because $query is out of the function's scope. If you want to use a variable declared outside of a function inside of it, you either need to pass it through as a parameter, or use

function check_files($this_file) {
    global $query;
    $str_to_find = $query;

Although passing it through as a parameter is preferred to using global.

answered Feb 24, 2012 at 20:54

John V.

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Comments

satoshi · Accepted Answer · 2012-02-24 20:53:43Z

0

The variable $query is declared outside the function check_files() scope. If you want to access it, put global $query; at the beginning of the function.

answered Feb 24, 2012 at 20:53

satoshi

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Comments

Mathias R. Jessen · Accepted Answer · 2012-02-24 20:55:28Z

0

You need to declare the the $query global as per the PHP manual, if not, the parser will assume that $query is a local scope variable (local as in "only within this function")

function check_files($this_file) {
global $query;
$str_to_find = $query;
...

answered Feb 24, 2012 at 20:55

Mathias R. Jessen

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