The code below shows in invlid syntax in the first elif statement. I have checked and rechecked my code several times, but cant figure out how to solve the error.
fileHandle = open ( 'gra1.txt' )
count=0
count1=0
fileList = fileHandle.readlines()
for fileLine in fileList:
line=fileLine.split()
if line[0] == '0':
print "graph G%d {\n", (count)
count +=1
elif line[0] == '1':
print "} \n"
elif line[0]=='':
continue
else:
count1 += 1
if count1==1: a=line[0]
elif count1==2: relation=line[0]
elif count1==3: b=line[0]
else:
print a, relation, b
count1=0
fileHandle.close()
Your elif
is not indented properly...it should be indented the same way if
is indented. Seeing the else
block, it seems that you have by mistake indented the first if
. Remember that elif/else
should be preceded by an if
always.
EDIT: corresponding to the edited question details: Why is the second else
there? It isn't preceded by an if
. I feel you need to get your conditions organized properly before writing the code.
One way to correct the code is to change this to an elif
block:
else:
count1 += 1
if count1==1: a=line[0]
elif count1==2: relation=line[0]
elif count1==3: b=line[0]
You might want your indentation in Python to get better. Consider reading up a bit on that :-)
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