HHL algorithm, How can I get result from register $|b\rangle$?

From the paper A survey on HHL algorithm: From theory to application in quantum machine learning , I use qasm code from here. I try to follow the example in page 7. with Ax = b and  From the algorithm 2 in page 7 from the paper, They applied measurement on qubit q.
If q = $$|1\rangle$$, the answer will be stored in q[.3] or register b. But in the example here, I got state $$|0\rangle$$ from ancilla bit if I use the same rotation angle as the example.

When I measure register b, I always got state $$|1\rangle$$ even if the result on ancilla bit is $$|1\rangle$$ (I change ancilla bit rotation angel from the example to 3.6591 and 1.8296).

Does the problem is the rotation angle on ancilla bit?
How can I get the correct result on register b? Can you give me a good example to do this?

In your code the measure clause means one shot, so the state collapses to one basis state as here: By removing that last line the Statevector option shows the state before measurement, i.e. as a superposition of basis states: Now you can see that for the basis states with q=1 (that is 0001 and 1001) we get the right components of the solution vector $$x$$ (you can check that the numbers coincide with those reported in Table 2, row 1 of the paper you mentioned).