在文本文件中查找特定单词并打印输出

Question

I need help on python code

I have a text file with around 10 lines of paragraph. Below is an example

"406%2C318482214%2C318484497%2C318486317%2C318484676%2C318483611%2C318484802%2C318487059%2C318489974%2C318482672%2C318475417&tag_for_child_directed_treatment=0&_c_csdk_npa_o=false&_c_req_tfua=false&_c_req_npa=true& npa=1 & tfua =0&guci=0.0.0.0.0.0.0.8&rbv=1&u_w=839&u_h=424&msid=com.ea.gp.test&_package_name=com.ea.gp.test&an=30178.android.com.ea.gp.test&net=wi&u_audio=3&u_so=l&preqs_in_session=0&support_transparent_background=true"

How can i write a python code to find the value assigned to the specific word "npa" & "tfua". Lets assume the highlighted once in the above text. It might happen that those words repeat multiple times also.

Answer 1

Suppose your string is myStr

myStr = "406%2C318482214%2C318484497%2C318486317%2C318484676%2C318483611%2C318484802%2C318487059%2C318489974%2C318482672%2C318475417&tag_for_child_directed_treatment=0&_c_csdk_npa_o=false&_c_req_tfua=false&_c_req_npa=true&npa=1&tfua=0&guci=0.0.0.0.0.0.0.8&rbv=1&u_w=839&u_h=424&msid=com.ea.gp.test&_package_name=com.ea.gp.test&an=30178.android.com.ea.gp.test&net=wi&u_audio=3&u_so=l&preqs_in_session=0&support_transparent_background=true"
myLst = [lst.split("=") for lst in myStr.split("&")]
myDict = {lst[0] : lst[1] for lst in myLst}

Now myDict["npa"] gives the desired value. Likewise for myDict["tfua"]

If you are sure you only want those two you could restrict the dict to only those values:

myLst = [lst.split("=") for lst in myStr.split("&")]
myDict = {lst[0] : lst[1] for lst in myLst if lst[0] in ["npa", "tfua"]}

Answer 2

Using the re module

import re
x = "406%2C318482214%2C318484497%2C318486317%2C318484676%2C318483611%2C318484802%2C318487059%2C318489974%2C318482672%2C318475417&tag_for_child_directed_treatment=0&_c_csdk_npa_o=false&_c_req_tfua=false&_c_req_npa=true&npa=1&tfua=0&guci=0.0.0.0.0.0.0.8&rbv=1&u_w=839&u_h=424&msid=com.ea.gp.test&_package_name=com.ea.gp.test&an=30178.android.com.ea.gp.test&net=wi&u_audio=3&u_so=l&preqs_in_session=0&support_transparent_background=true"

npa=re.findall('(?<=npa=)(.*?)(?=&)', x)
tfua=re.findall('(?<=tfua=)(.*?)(?=&)', x)


print(npa)
>>> ['True','1']

print(tfua)
>>> ['False','0']