How to extract h160 address of bitcoin blockchain

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I have with me bitcoin blocks of over 100GB. I want to extract the data from it and read them. I have tried to extract the data using Berkeley DB but it didnt work. I am trying to use python for this but now understanding how to do it.  How can i extract the data using python?
Jul 16, 2018 in Blockchain by slayer
• 29,370 points
2,536 views

1 answer to this question.

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You can extract the data using the software that is available in the following link:
https://github.com/znort987/blockparser

When you download it, you will find the following example in the README file:
. Compute and print the balance for all keys ever used since the beginning of time:

    ./parser all >all.txt

When you run it, you should see something like this:

root@81d54ebe5b25:~/blockparser# ls -alh all.txt
-rw-r--r-- 1 root root 900M Aug 25 09:33 all.txt
root@81d54ebe5b25:~/blockparser# head all.txt
---------------------------------------------------------------------------
          State of the ledger at block 194124 (minted : Thu Aug 16 03:36:13 2012)
---------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------------------------------------------------------
              Balance                      Hash160                             Base58                  nbIn        lastTimeIn          nbOut        lastTimeOut
---------------------------------------------------------------------------------------------------------------------------------------------------------------------
         507335.01317523 8bf24a18a58ab500d30c73bf21dbf4703d31ad2c 1DjwTabcA1scU4ztjuA69LaKe1LmhyxYbV    152 Tue Aug 14 18:11:09 2012      17 Tue Jul 17 02:32:38 2012
         105555.03133700 582431b9e63d2394c8b224d1bc45d07ae95d2379 195Pgoq2mPV2oV4k3mTDVrgZnGMMfcxbfy     48 Fri Jun 22 16:26:43 2012       0 Thu Jan  1 00:00:00 1970
          79957.03133700 a0b0d60e5991578ed37cbda2b17d8b2ce23ab295 1FuuPsGb2Thr72Lohsoy1McTsQJfwoEckV      4 Sun Jul 15 20:36:59 2012       0 Thu Jan  1 00:00:00 1970
          53000.03133700 3d9e561f21d312f9b8b46e74169263e2452d5591 16Trt1jJsDHCcBAMxyJTgKzXAfKqwZFkAA     16 Sun Jul 15 20:36:59 2012       9 Sun May 13 12:13:16 2012
answered Jul 16, 2018 by digger
• 26,740 points

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