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+--- Thread: search text file (/thread-14350.html)
search text file - jacklee26 - Nov-26-2018
I have one question with my code if I couldn't find match string it will keep print not find(if your text has 10 line, and could find your match, it will print 10 times). I wish if the string does not match just print one line not found.
Example:
enter your string: sss
not found
not found
not found
not found
i wish to just print one line not find, how can i changed my code
import sys
substring = raw_input("enter your string : ")
for line in open('POD.txt'):
#if ("") in line:
if substring in line:
print (line)
else:
print ("not found")
RE: search text file - ichabod801 - Nov-26-2018
Have a count variable, and increase it by one each time the line is found. After the loop, print not found if the count is zero.
RE: search text file - jacklee26 - Nov-26-2018
i wish to occur like this:
enter your string: sss
not found
do i have to put a if else statement inside the if loop
RE: search text file - ichabod801 - Nov-26-2018
You need the if in the loop to check for matches, and add an increment to a count variable to it. You need to take the else out of the loop (unindent it), and convert it to an
if count == 0:
RE: search text file - jacklee26 - Nov-27-2018
ok thanks if figure it out thanks.
import sys
substring = raw_input("enter your string : ")
count =0
for line in open('POD.txt'):
#if ("") in line:
if substring in line:
count +=1
print (line)
if count==0:
print ("string not found")
But this only support python2, how to let python3 also support.
if i run using python3 it will occur
File "test_file.py", line 10
print (line)
^
TabError: inconsistent use of tabs and spaces in indentation
RE: search text file - ichabod801 - Nov-27-2018
You need to make sure your indentation is consistent: either all spaces or all tabs. Try changing all tabs to four spaces or vice versa. Note that line 9 (
print(line)) should be indented one more level to be even with count += 1.