Invalid cast of a string in an object with JSON and regex in PHP

March 4, 2022

Here are some sample code that should lead to the same result. Once in JavaScript and Perl where it is understandable and once in PHP which leads to problems.

All of this takes place in a complete scenario in various components of a framework. Therefore the first definition of $a = ['a\b'] is mandatory.

The result $r is just a control example for an external parser process.

#!env node
const a = ['a\b'];
console.debug(a); // [ 'a\b' ]

const j = JSON.stringify(a);
console.debug(j); // ["a\b"]

let s = 'foo';
s = s.replace(/foo/, j);
console.debug(s); // ["a\b"]

const r = JSON.parse(s);
console.debug(r); // [ 'a\b' ]

#!env php
<?php
$a = ['a\b'];
print_r($a); print "\n"; // Array([0] => a\b)

$j = json_encode($a);
print_r($j); print "\n"; // ["a\\b"]

$s = 'foo';
$s =  preg_replace('/foo/', $j, $s);
print_r($s); print "\n"; // ["a\b"] ⚡️

$r = json_decode($s);
print_r($r); print "\n"; // Array ([0] => a)

#!env perl
useJSON;
use Data::Dumper;

my $a = ['a\b'];
print Dumper($a);

my $j = encode_json($a);
print Dumper($j);

my $s = 'foo';
$s =~ s/foo/$j/;
print Dumper($s);

my $r = decode_json($s);
print Dumper($r);

The result is also valid:

$VAR1 = [
           'away'
         ];
$VAR1 = '["a\\\\b"]';
$VAR1 = '["a\\\\b"]';
$VAR1 = [
           'away'
         ];

>Solution :

In simplest terms, you need to escape the \ for use in the second parameter:

$s =  preg_replace('/foo/', str_replace( '\\', '\\\\', $j), $s);

Although, you should also read the details of the replacement parameter from https://www.php.net/manual/en/function.preg-replace.php to understand other pitfalls.

The replacement parameter isn’t just a string it’s an instruction set for regex replacement meaning there are special chars to be considered for regex purposes versus your JSON needs.