convert Bash command to php

Question

This is the bash command echo -n x && (echo 618cf954-6576-491d-8ac6-a1b888c4705d |xxd -r -p |openssl base64|tr '/+' '_-')

This is my php

<?
$uuid = "618cf954-6576-491d-8ac6-a1b888c4705d";
$voiceid = "x" . $uuid;
$voiceid  = base64_encode($voiceid);
$voiceid = str_replace("+", "-", $voiceid);
$voiceid = str_replace("/", "_", $voiceid);
echo $voiceid;
?>

The bash gives the right output, the php one isn't. I'm not sure what i need to do deferent in php.

The output should be xYYz5VGV2SR2KxqG4iMRwXQ==

A C++ version is at https://gist.github.com/1e096b658097c19cf309

Dennis Williamson · Accepted Answer · 2011-02-07 07:52:24Z

3

Also, in the Bash version, the "x" isn't included in the pipe through xxd and openssl. If you want it to be then do this:

{ echo -n x; echo -n 618cf954-6576-491d-8ac6-a1b888c4705d; } | xxd -r -p | openssl base64 | tr '/+' '_-')

If, on the other hand, you want the PHP to match the existing Bash:

<?
$uuid = "618cf954-6576-491d-8ac6-a1b888c4
$voiceid = str_replace("-", "", $uuid);
$voiceid = pack("H*", $voiceid);
$voiceid  = base64_encode($voiceid);
$voiceid = str_replace("+", "-", $voiceid);
$voiceid = str_replace("/", "_", $voiceid);
$voiceid = "x" . $voiceid;
echo $voiceid;
?>

You will need to use pack to convert the hex string to binary.

Edit: Fixed several of my errors.

edited Feb 7, 2011 at 7:52

answered Feb 7, 2011 at 7:18

Dennis Williamson

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5 Comments

Jander Over a year ago

echo 618cf954 |xxd -r -p |hexdump -C shows that xxd discards the newline.

Dennis Williamson Over a year ago

@Jander: I've revised my answer accordingly. Thanks.

Keverw Over a year ago

Why did you put x at the end? Prepending the x prevents conflicts with the Vivox api.

Dennis Williamson Over a year ago

@Keverw: See the first sentence in my answer. The "x" is in your Bash version is prepended without being processed by the pipeline. There is another problem with your PHP version and that is their is no step that corresponds to the xxd step. By the way, this is the first you've mentioned the context.

Dennis Williamson Over a year ago

@Keverw: Please see my edited answer. Sorry for the confusion.

Jander · Accepted Answer · 2011-02-07 07:25:53Z

3

I think the following line is wrong:

$voiceid = "x" . $uuid;

Adding the "x" onto the string should be the last thing you do before the PHP echo.

answered Feb 7, 2011 at 7:25

Jander

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