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plugins/pay: eliminate worst channel if we go over fee / delay threshold.

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But keep the error in this case, so we don't always report "no route".

Reported-by: @niftynei
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
This commit is contained in:
Rusty Russell
2019-01-15 20:34:08 +10:30
committed by Christian Decker
parent e2777642c0
commit 9d33a3d752
2 changed files with 95 additions and 4 deletions
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6
tests/test_pay.py
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@@ -74,14 +74,20 @@ def test_pay_limits(node_factory):
assert err.value.error['code'] == PAY_ROUTE_TOO_EXPENSIVE
# It should have retried (once without routehint, too)
assert len(l1.rpc.call('paystatus', {'bolt11': inv['bolt11']})['pay'][0]['attempts']) == 3
# 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.
assert len(l1.rpc.call('paystatus', {'bolt11': inv['bolt11']})['pay'][1]['attempts']) == 3
# This works, because fee is less than exemptfee.
l1.rpc.call('pay', {'bolt11': inv['bolt11'], 'msatoshi': 100000, 'maxfeepercent': 0.0001, 'exemptfee': 2000})
assert len(l1.rpc.call('paystatus', {'bolt11': inv['bolt11']})['pay'][2]['attempts']) == 1
def test_pay0(node_factory):