Print line next to pattern in python

Input file: 'file.txt' contains results of a set of students in the following format (i.e. for any student result precedes the student id)

$ cat file.txt
Result:Pass
id:502
Result:Fail
id:909
Result:Pass
id:503
Result:Pass
id:501
Result:Fail
id:802

Required: Print the Ids of the students who have passed the exam.

The python program:

fp = open("passedids.txt","w")
data = open("file.txt").readlines()
for i in range(len(data)):
        if data[i].startswith("Result:Pass"):
                fp.write(data[i+1].split(":")[1])

Executing it:

$ python printnext.py
$ cat passedids.txt
502
503
501

Another python alternative:

fp=open('file.txt','r')
previous_line = ""

for current_line in fp:
    if 'Result:Pass' in previous_line:
        print current_line.split(":")[1],
    previous_line = current_line
fp.close()

Executing it:

$ python printnext1.py
502
503
501

Related post:

- Print line above pattern in python