Regular expression, taking the return data (including html) the value of the specified parameter

 

 

Above, a plurality of inner return html "shareid" parameters, we need to remove the back of a arbitrary "shareid", "response" returns data html above, and to output text

= re.findall tr_content2 ( ' "shareid":.? (*), ' , the Response, re.S) # regular traverse html, get shareid

Print tr_content2 

print(tr_content2)
print(tr_content2[0])

 

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Origin www.cnblogs.com/becks/p/12499375.html