0.002 responseHeadersStart
0.082 responseHeadersEnd
0.082 responseBodyStart
0.082 responseBodyEnd
0.083 connectionReleased
0.083 callEnd
```

Notice how no connect events are fired for the second call. It reused the connection from the first request for dramatically better performance.

### EventListener.Factory

### 3. Generate Self-contained Access Tokens

By default, a UUID is issued as an id_token in WSO2 Identity Server, which is of the first type above. But, it also can be configured to issue a self-contained id_token (JWT), which is of the second type above.

- Open the `<IS_HOME>/repository/conf/identity/identity.xml` file and uncomment the following entry under `<OAuth>` element.

```

		return
	}

	durationStr := r.Form.Get(peerRESTDuration)
	duration, err := time.ParseDuration(durationStr)
	if err != nil || duration.Seconds() == 0 {
		duration = time.Second * 10
	}
	result := netperf(r.Context(), duration.Round(time.Second))
	peersLogIf(r.Context(), gob.NewEncoder(w).Encode(result))
}

func (s *peerRESTServer) HealBucketHandler(mss *grid.MSS) (np grid.NoPayload, nerr *grid.RemoteErr) {

        throw (T) t;
      }
    }
    new SneakyThrower<Error>().throwIt(t);
  }

  /**
   * Returns a {@link Stream} containing the elements of the first stream, followed by the elements
   * of the second stream, and so on.
   *
   * <p>This is equivalent to {@code Stream.of(streams).flatMap(stream -> stream)}, but the returned
   * stream may perform better.
   *
   * @see Stream#concat(Stream, Stream)
   */

		Data:    "something else",
	}

	result := DB.Clauses(clause.OnConflict{
		OnConstraint: "some_id_other_id_unique",
		UpdateAll:    true,
	}).Create(&thing2)
	if result.Error != nil {
		t.Errorf("creating second thing: %v", result.Error)
	}

	var things []Thing
	if err := DB.Find(&things).Error; err != nil {
		t.Errorf("Failed, got error: %v", err)
	}

	if len(things) > 1 {

     * approach would be for each caller CAS it from null to a Set populated with its exception. But
     * there's another race: If the first thread fails with an exception and a second thread
     * immediately fails with the same exception:
     *
     * Thread1: calls setException(), which returns true, context switch before it can CAS
     * seenExceptions to its exception
     *

    //    were added to the queue. It's easily possible for one thread to take the head of the
    //    queue, immediately followed by another thread taking the next element in the queue. That
    //    second thread can then dispatch to the subscriber it took before the first thread does.
    //
    // All this makes me really wonder if there's any value in queueing here at all. A dispatcher

            // this will run at most twice. The first time, the firstRun flag is turned off, and if the retry flag
            // is set on the first run, it will be turned off and not re-set on the second try. This is because the
            // only way the retry flag can be set is if ( firstRun == true ).
            while (firstRun || retry) {
                ChecksumObserver md5ChecksumObserver = null;

For example you could have a file `main.py` with:

```Python hl_lines="3"
import os

name = os.getenv("MY_NAME", "World")
print(f"Hello {name} from Python")
```

/// tip

The second argument to <a href="https://docs.python.org/3.8/library/os.html#os.getenv" class="external-link" target="_blank">`os.getenv()`</a> is the default value to return.

      }
    }
  }

  private fun ensureAllTaskQueuesIdle() {
    val entryTime = System.nanoTime()

    for (queue in TaskRunner.INSTANCE.activeQueues()) {
      // We wait at most 1 second, so we don't ever turn multiple lost threads into
      // a test timeout failure.
      val waitTime = (entryTime + 1_000_000_000L - System.nanoTime())
      if (!queue.idleLatch().await(waitTime, TimeUnit.NANOSECONDS)) {