Extract block of text using regex python

Question

Here is my text file:

YN FILE=“stop.x" DATE=wwepweijfp Life=spfewff FUIL=“keener” CREATED_BY=“xxxxx” 
DY FD=4344334 HSH=OIHF 
DC A_GN=EFEFOK 
DF EFIIEFNFNEE
DF NIFENPEFEF
DF AO EIJHEPFN A0 pjwdwdwd
DF C0 STATICVECTOR 
ZZ ---------------< Fields >--------------------- 
DF A0 VERN FMGL ROJF AO
GS MOB40 99 
GA 4 Y 3 F 00H00M F 
G2 “eiberbnier       " EE 
C0 D56916E65D4500574124B15AD3226746A7F70AD7A3936A550BDB87FC077E9F4C7D36C4E9BFFA7545C4AE8C4173
A0 VDS1000170 34486EF018B1BF89 955486C529989CBC 89BCE463 919BC6E9 C927D0 09BA5356CD7CFFBD 
A0 VDS1000171 1951420D5909073A FF7B6A08E2BDC4AC E9593876 50ABACB9 894A6E 57AA4A23DE6414E0

I would like to extract this block :

The first line of the text always starts with a YN FILE. I want to extract everything until it comes across a new line beginning with A0.

I have tried such :

matchObjj = re.search(r'YN.*\nC0.*\nA0', data, re.DOTALL)

But it also extracts the other A0 beginning in newline.

Serge Ballesta · Accepted Answer · 2019-02-27 08:42:34Z

1

Beware, in a regex, * is gready. That means that if there are other lines starting with 'A0', it will stop at the last and not at the first. So you should use an ungreedy *?. And a group would be enough to get rid of the initial 'A0'. It could end with:

matchObjj = re.search(r'(YN.*?\n)A0', data, re.DOTALL)
block = matchObjj.group(1)

answered Feb 27, 2019 at 8:42

Serge Ballesta

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1 Comment

Knitey Over a year ago

That makes so much sense. Thank you !

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