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Description
PHP version: 8.2.11
guzzlehttp/psr7: 2.6.1
Description
The Uri class is able to parse malformed URIs but this results in Uri::__toString() generating a malformed URI. Attempting to pass that URI back into Uri will yield a MalformedUriException.
How to reproduce
$value = 'http://invalid:%20https://example.com';
$uri = new \GuzzleHttp\Psr7\Uri($value);
$uri2 = new \GuzzleHttp\Psr7\Uri($uri->__toString()); // Exception is thrown here.Line 3 throws a GuzzleHttp\Psr7\Exception\MalformedUriException with the following error message:
Unable to parse URI: http://invalid:%20https//example.com
Notice how the colon after the https was stripped.
Hanashi
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