[SOLVED] Reducing the size of md5 checksums

[PC Nerd]

Hi,

 

I'm working with sending md5checksums of different strings and files to a server to validate they downloaded correctly, and I wanted to (more of an excersise than anythign else) try and reduce what I'm sending.

 

Instead of sending 10 checksums across the net, I wanted to eg. send one checksum which "represents" all the data.

 

Is:

$ck1 = md5("string");

$ck2 = ...

$ck3 = ...

$ck4 = ...

$ck5 = ...

 

md5($ck1+$ck2+$ck3+$ck4+$ck5); // Is this line a "valid" representation of the previous 5 checksums or is there a more standard method?

 

Thanks

 

 

[rtadams89]

When you say reduce what you are sending, do you mean you want to cut down on the actual size (bytes) or the # of variables that have to be passed? If you just want to reduce the number of variables, try putting all the md5 sums in an array. Then you can reference the array and pull out any m5d sum or all of them.

 

I don't think

md5($ck1+$ck2+$ck3+$ck4+$ck5); 

is a very good idea. Manly because this will return a different md5 if the same files are sent, but in a different order. md5($ck1+$ck2+$ck3+$ck4+$ck5) != md5($ck5+$ck4+$ck3+$ck2+$ck1);

 

You could always do something like md5($ck1) + md5($ck2) and have it work for your application, but it's not truly a checksum if you do that.

[aximbigfan]

$md5[] = whatever
$md5[] = whatever2
etc

$big = '';

foreach ($md5 as $value)
{
    $big = $big . $value;
}

$out = md5($big);

 

Chris

[flyhoney]

As long as the md5's are in the same order, your solutions should work just fine.

[PC Nerd]

Ok - thanks all

 

* and to answer the first question - it was both a task of reducing the numebr of variables and the number of bytes.  ultimately - a checksum of all the data is less than individual checksums of each data entry

 

Thanks - Sovled