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I am trying to calculate proof that a specific Transaction is mined with a specific difficulty. So far I got the Block via JSON_RPC and all its Transactions.

I am Using the trie Library for Python 3.6

For the sake of testing everything I'm using a Block with no Transactions.

txTrie = HexaryTrie(db={}) #make new trie
print('txRoot: \t', block['transactionsRoot'])
print('txTrieRootS: \t 0x' + txTrie.root_hash.hex())

For this Code both Print-Statements return the same value.


Now I run the same code with a Block that contains one transaciton

txTrie = HexaryTrie(db={}) #make new trie
txTrie[transactionIndex] = transacionHash #add the transaction to the Trie
print('txRoot: \t', block['transactionsRoot'])
print('txTrieRootS: \t 0x' + txTrie.root_hash.hex())

And now the two print statements don't return the same value.

Can anyone help me how to get Proof that a Transaction is in a specific Block at a specific index? I just can't figure this out.

EDIT: Here is the full code:

import requests
import json
import lib.mpt as mpt
import lib.rlp as rlp
from trie import HexaryTrie

url = 'https://mainnet.infura.io/ffn6QLIJrYtke3b07YLp'

headers = {
    "Content-Type": "application/json"
}
data = {}
data['jsonrpc'] = '2.0'
data['id'] = 1
data['method'] = 'eth_getBlockByNumber'
data['params'] = [hex(100004), True]
data = json.dumps(data)

response = requests.post(url, headers=headers, data=str(data) )
block = json.loads(response.text)
result = block['result']
transactions = result['transactions']
print('txRoot: \t', result['transactionsRoot'])

txTrie = HexaryTrie(db={})
print('txTrieRootS: \t 0x' + txTrie.root_hash.hex())
for x in transactions:
    txTrie[b'0'] = bytes.fromhex(x['hash'][2:])
    print(txTrie[str.encode(x['transactionIndex'][2:])])
print('txTrieRootA: \t 0x' + txTrie.root_hash.hex())
  • 1
    It will be easier to help if you show the full process of getting block, transactionIndex, and transactionHash. Including which network you are connecting to and which block number you're using, so that everyone can reproduce along with you. – carver Jul 31 '18 at 0:50

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