PHP & Jquery - Sending data between the client and server

Question

I am trying to write code that sends a file path to the server, uses php to parse out each line into an array and returns the array to the client to have stuff done to it.

When I run my program it doesn't look like it is processing the php file as the echo that i put in for testing purposes is never called.

I know absolutely nothing about PHP so help is greatly appreciated!

Jquery:

$("#codelines").load('ParsePHPForDisplay.php?filename=DBManager.php');

    console.log($('#codelines'));
    pieces = $("#codelines").html().split("\n");

PHP:

<?php

class Parsing
{
    function ParseStuff()
    {
        echo('hi');
        $parsedFile = file($_REQUEST['filename']);
        echo($_REQUEST['filename']);
//        foreach ($parsedFile as $parsedFile_num => $parsedFile) {
//            echo"Line #<b>{$parsedFile_num}</b> : " .htmlspecialchars($parsedFile) . "<br/>\n";
//        }
        return $parsedFile;
    }
}

?>

A piece of my HTML:

<section id="codestuff">
    <h2>Code Lines</h2>
    <pre id="codelines">
    </pre>
</section>

EDIT:

Here is my PHP now:

<?php


        echo('hi');
        $parsedFile = file($_REQUEST['filename']);
        echo('test');
        echo($_REQUEST['filename']);
//        foreach ($parsedFile as $parsedFile_num => $parsedFile) {
//            echo"Line #<b>{$parsedFile_num}</b> : " .htmlspecialchars($parsedFile) . "<br/>\n";
//        }
        return $parsedFile;



?>

I get this error:

Parse error: syntax error, unexpected T_STRING, expecting T_FUNCTION in ParsePHPForDisplay.php on line 5
.

$myParse = new Parsing(); $myParse->ParseStuff(); Though for your purpose, you really don't need to make it in a class. Seems like you came from a java background. However, if you are only going to have one function, you may as well put it outside of a class scope so it runs when the page is loaded without the need to call it. — Jon
– Jon, Commented Feb 24, 2013 at 23:31
Jon's code should be somewhere in there. The class has to be instantiated then the function has to be called. — Matt
– Matt, Commented Feb 24, 2013 at 23:32
There's a bigger problem, Matt and Jon. He says he is trying to upload a file to the server. The PHP I see, is only loading a file that already exists, by its filename. — mavrosxristoforos
– mavrosxristoforos, Commented Feb 24, 2013 at 23:34
@user1950929 Good job paying attention, she's not uploading a file, just passing a filename. — Jon
– Jon, Commented Feb 24, 2013 at 23:56

mavrosxristoforos · Accepted Answer · 2013-02-25 00:02:54Z

1

Alright, I believe the PHP you should use, since you are splitting the file into an array in the Javascript part, is:

<?php
  $file = file_get_contents($_REQUEST['filename']);
  echo $file;
?>

But before publishing this, you should check security issues, like the user loading files from other directories, using ../ in the filename, also not to fetch php files, etc.

answered Feb 25, 2013 at 0:02

mavrosxristoforos

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Comments

Matt Browne · Accepted Answer · 2013-02-24 23:53:17Z

I don't see why you would need PHP to parse the file into an array...just have it send the whole thing as a big string, then you can split it into an array of lines, if you need to, in Javascript once you get it from the server (e.g. myString.split(/\n\r|\n|\r) ).

Then, from PHP, you can just do:

readfile($_REQUEST['filename']); exit;

(note that this is not secure; you should add some conditions to the PHP code so it can only outptut files from a particular directory, for example. See PHP's explode function which will help you split the filename into an array using DIRECTORY_SEPARATOR as the delimiter).

Note that in PHP, a return statement doesn't cause anything to be output. Very often you'll get the value you need via a function that returns a value, but you still need to echo or print it.

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