Since this was removed in PHP7. You can achieve the same output with the following code.
function convertBinToMSSQLGuid($binguid)
{
$unpacked = unpack('Va/v2b/n2c/Nd', $binguid);
return sprintf('%08X-%04X-%04X-%04X-%04X%08X', $unpacked['a'], $unpacked['b1'], $unpacked['b2'], $unpacked['c1'], $unpacked['c2'], $unpacked['d']);
}
Enjoy.
mssql_guid_string
(PHP 4 >= 4.0.7, PHP 5, PECL odbtp >= 1.1.1)
mssql_guid_string — Convertit le GUID binaire de 16 octets en une chaîne de caractères
Avertissement
Cette fonction a été SUPPRIMÉE à partir de PHP 7.0.0.
Description
mssql_guid_string
( string
$binary
[, bool $short_format = FALSE
] ) : stringConvertit un GUID binaire de 16 octets en une chaîne de caractères.
Liste de paramètres
-
binary -
Un GUID binaire, de 16 octets.
-
short_format -
TRUEs'il faut utiliser le format court.
Valeurs de retour
Retourne la chaîne convertie en cas de succès.
Exemples
Exemple #1 Exemple avec mssql_guid_string()
<?php
$binary = '19555081977808608437941339997619274330352755554827939936';
var_dump(mssql_guid_string($binary));
var_dump(mssql_guid_string($binary, true));
?>
L'exemple ci-dessus va afficher :
string(36) "35353931-3035-3138-3937-373830383630" string(32) "31393535353038313937373830383630"
add a note
User Contributed Notes 3 notes
darylblake at gmail dot com ¶
8 years ago
Corvus Corax ¶
16 years ago
How can a binary GUID be identified as a binary GUID?
- mssql_field_type() returns 'blob' for GUID fields...
- mssql_guid_string() returns strings for all sort of arbitrary input data, regardless of length and content (though not all are necessarily valid GUID string represenations, since length varies)
- is_string() on GUID variables is true
therefore a mssql_is_guid() function would be a nice thing.
Unfortunately, there is very little that makes a GUID a GUID. Aside from the length of 16 bytes there are only 2 indicators (and even those only for microsoft system call compatibly generated GUIDs):
the upper 4 bits of character 7 encode the version of the GUID, and are either 0001 (v1) , 0011 v3), or 0100 (v4)
(ASCII characters 10 - 1f, 30 ('0') to 3f ('?') or 40 ('@') to 4f ('O'))
depending on the encoding version of the GUID.
GUIDs generated by mssql server 2005s "NEWID()" statement are version 4, indicating that the GUID is completely randomly generated
the upper 1 to 3 bits of character 8 which encode the GUID variant. In the case of GUIDs generated by MsSQL server 2005 the variant always seems to be 'Standard' which is indicated by bit sequence 10.
((non)ASCII characters 81 to BF)
so as long as no custom hacked or 3rd party inserted GUIDs are used this should be working:
<?php
/**
* determine wether binary data represents a valid GUID
*/
function mssql_is_guid($guid)
{
if (!is_string($guid)) return false;
if (strlen($guid)!=16) return false;
$version=ord(substr($guid,7,1))>>4;
// version 1 : Time-based version Uses timestamp, clock sequence, and MAC network card address
// version 2 : Reserverd
// version 3 : Name-based version Constructs values from a name for all sections
// version 4 : Random version Use random numbers for all sections
if ($version<1 || $version>4) return false;
$typefield=ord(substr($guid,8,1))>>4;
$type=-1;
if (($typefield & bindec(1000))==bindec(0000)) $type=0; // type 0 indicated by 0??? Reserved for NCS (Network Computing System) backward compatibility
if (($typefield & bindec(1100))==bindec(1000)) $type=2; // type 2 indicated by 10?? Standard format
if (($typefield & bindec(1110))==bindec(1100)) $type=6; // type 6 indicated by 110? Reserved for Microsoft Corporation backward compatibility
if (($typefield & bindec(1110))==bindec(1110)) $type=7; // type 7 indicated by 111? Reserved for future definition
// assuming Standard type for SQL GUIDs
if ($type!=2) return false;
return true;
}
$valid_guid='xxxxxxx'.'A'.chr(0xA2).'xxxxxxx';
if (mssql_is_guid($valid_guid)) $valid_guid=mssql_guid_string($valid_guid);
echo ($valid_guid);
?>
source:
http://msdn2.microsoft.com/en-us/library/aa446557.aspx
http://en.wikipedia.org/wiki/GUID
