[BUGFIX] Update example code for service instanciation

Author: Francois Suter

ChangeLog

@@ -1,3 +1,7 @@
+2023-05-02 Francois Suter (Idéative)  <typo3@ideative.ch>
+
+	* Update example code for service instanciation
+
 2022-12-27 Francois Suter (Idéative)  <typo3@ideative.ch>
 
 	* Relase version 4.0.0

Documentation/Developer/Index.rst

@@ -16,21 +16,12 @@ easy task. The first step is to get the proper service object:
 
 .. code-block:: php
 
-	$services = \TYPO3\CMS\Core\Utility\ExtensionManagementUtility::findService('connector', 'sql');
-	if ($services === false) {
-		// Issue an error
-	} else {
-		$connector = \TYPO3\CMS\Core\Utility\GeneralUtility::makeInstanceService('connector', 'sql');
-	}
+   $registry = GeneralUtility::makeInstance(\Cobweb\Svconnector\Registry\ConnectorRegistry::class);
+   $connector = $registry->getServiceForType('sql');
 
-On the first line, you get a list of all services that are of type
-“connector” and subtype “sql”. If the result if false, it means no
-appropriate services were found and you probably want to issue an
-error message.
-
-On the contrary you are assured that there's at least one valid
-service and you can get an instance of it by calling
-:code:`\TYPO3\CMS\Core\Utility\GeneralUtility::makeInstanceService()` .
+An additional step could be to check if the service is indeed available,
+by calling :php:`$connector->isAvailable()`, although - in this particular
+case - the SQL connector service is always available.
 
 The next step is simply to call the appropriate method from the API –
 with the right parameters – depending on which format you want to have