PHP - str_replace returns original string

Question

I'm building a simple JS terminal shell emulator which posts its commands via AJAX to PHP.
Please leave security aside, this is only for learning and demo purposes. Now my problem is, str_replace() won't work as expected, in fact, it returns the unchanged input string. It should work like this:
The name of this host is $hostname --> Yes, this string contains a variable --> Replace $hostname with testserver --> return The name of this host is testserver

What am I doing wrong?

This is my respond script for echo and export:

<?
// get environment variables from JSON
$vars = json_decode(file_get_contents('environment.json'), true);

// get request params
$method = $_SERVER['REQUEST_METHOD'];
$action = $_POST['action'];
$data = $_POST['data'];

switch ($action) {

case 'echo':
    $cmd = $data;

        // if the string in question contains a variable, eg. "the time is $time"
        if (strpos($cmd,'$')) {
        $output = '';

        // for each environment variable as variable => value
        foreach ($vars as $var => $val) {

            // replace every variable in the string with its value in the command
            $output = str_replace($var,$val,$cmd);
        }
        echo $output;
    } else { 

        // if it does not contain a variable, answer back the query string 
        // ("echo " gets stripped off in JS)
        echo $cmd; 
    }
break;

case 'export':

    // separate a variable declaration by delimiter "="
    $cmd = explode('=',$data);

    // add a $-sign to the first word which will be our new variable
    $var = '$' . array_shift($cmd);

    // grab our variable value from the array
    $val = array_shift($cmd);

    // now append everything to the $vars-array and save it to the JSON-file 
    $vars[$var] = $val;
    file_put_contents("environment.json",json_encode($vars));
break;
}

The reason is not to be found in your use of str_replace, so please post the combination $var, $val and $cmd you use. — Alexander
– Alexander, Commented Oct 15, 2014 at 8:09
Beware with strpos, if the word you are searching for is in the first place, it will return 0 and not false, so you should use if (strpos($cmd,'$')) !== false — Sal00m
– Sal00m, Commented Oct 15, 2014 at 8:09
IMHO It would be smarter to create a separate object or an array of closures and address the keys as command names instead of using a switch with logic. — php_nub_qq
– php_nub_qq, Commented Oct 15, 2014 at 8:11
@Alexander, Sal00m, hd: environment.json contains {"$hostname":"testserver","$test":"123"} — Moritz Friedrich
– Moritz Friedrich, Commented Oct 15, 2014 at 8:39

blue112 · Accepted Answer · 2014-10-15 08:09:04Z

2

Better using :

if (strpos($cmd,'$') !== false) {

Then, every single replace will take the "first" data as its input data. You should proceed like this :

    $output = $cmd;

    // for each environment variable as variable => value
    foreach ($vars as $var => $val) {

        // replace every variable in the string with its value in the command
        $output = str_replace($var, $val, $output);
    }

answered Oct 15, 2014 at 8:09

blue112

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2 Comments

Moritz Friedrich Over a year ago

Thank you, that solved it. But I don't understand why, could you explain? That'd be great...

blue112 Over a year ago

Each time you're running the loop again, you're taking the "initial" data ($cmd), and replace in it. So, each time you're looping, you're losing the previous changes you've made on the previous loop.

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