fix send_xrp

This commit is contained in:
Mayukha Vadari
2021-06-03 13:30:50 -04:00
parent ec5b68d876
commit 86d5bee937

View File

@@ -16,7 +16,6 @@ from xrpl.wallet import generate_faucet_wallet
test_wallet = generate_faucet_wallet(client, debug=True)
# Prepare transaction ----------------------------------------------------------
import xrpl.utils # workaround for https://github.com/XRPLF/xrpl-py/issues/222
my_payment = xrpl.models.transactions.Payment(
account=test_wallet.classic_address,
amount=xrpl.utils.xrp_to_drops(22),
@@ -28,9 +27,11 @@ print("Payment object:", my_payment)
signed_tx = xrpl.transaction.safe_sign_and_autofill_transaction(
my_payment, test_wallet, client)
max_ledger = signed_tx.last_ledger_sequence
tx_id = signed_tx.get_hash()
print("Signed transaction:", signed_tx)
print("Transaction cost:", xrpl.utils.drops_to_xrp(signed_tx.fee), "XRP")
print("Transaction expires after ledger:", max_ledger)
print("Identifying hash:", tx_id)
# Submit transaction -----------------------------------------------------------
validated_index = xrpl.ledger.get_latest_validated_ledger_sequence(client)
@@ -41,10 +42,9 @@ print(f"Can be validated in ledger range: {min_ledger} - {max_ledger}")
# to send the transaction and wait for the results to be validated.
try:
prelim_result = xrpl.transaction.submit_transaction(signed_tx, client)
except xrpl.transaction.XRPLReliableSubmissionException as e:
except xrpl.clients.XRPLRequestFailureException as e:
exit(f"Submit failed: {e}")
print("Preliminary transaction result:", prelim_result)
tx_id = prelim_result.result["tx_json"]["hash"]
# Wait for validation ----------------------------------------------------------
# Note: If you used xrpl.transaction.send_reliable_submission, you can skip this