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pay: Fix logic of the intereface find_worst_channel

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trueptolemy
2019-09-14 16:48:01 +08:00
committed by Christian Decker
parent bdbfbf7e55
commit c737fa6b91
2 changed files with 26 additions and 13 deletions
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12
tests/test_pay.py
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@@ -94,25 +94,29 @@ def test_pay_limits(node_factory):
assert err.value.error['code'] == PAY_ROUTE_TOO_EXPENSIVE
# It should have retried (once without routehint, too)
# It should have retried two more times (one without routehint and one with routehint)
status = l1.rpc.call('paystatus', {'bolt11': inv['bolt11']})['pay'][0]['attempts']
# Excludes channel, then ignores routehint which includes that, then
# it excludes other channel.
assert len(status) == 2
assert len(status) == 3
assert status[0]['strategy'] == "Initial attempt"
# Exclude the channel l1->l2
assert status[1]['strategy'].startswith("Excluded expensive channel ")
# With the routehint
assert status[2]['strategy'].startswith("Trying route hint")
# Delay too high.
with pytest.raises(RpcError, match=r'Route wanted delay of .* blocks') as err:
l1.rpc.call('pay', {'bolt11': inv['bolt11'], 'msatoshi': 100000, 'maxdelay': 0})
assert err.value.error['code'] == PAY_ROUTE_TOO_EXPENSIVE
# Should also have retried.
# Should also have retried two more times.
status = l1.rpc.call('paystatus', {'bolt11': inv['bolt11']})['pay'][1]['attempts']
assert len(status) == 2
assert len(status) == 3
assert status[0]['strategy'] == "Initial attempt"
assert status[1]['strategy'].startswith("Excluded delaying channel ")
assert status[2]['strategy'].startswith("Trying route hint")
# This works, because fee is less than exemptfee.
l1.rpc.call('pay', {'bolt11': inv['bolt11'], 'msatoshi': 100000, 'maxfeepercent': 0.0001, 'exemptfee': 2000})