views:

82

answers:

4

I have a code:

print "bug " + data[str.find(data,'%')+2:-1]
temp = data[str.find(data,'%')+2:-1]
time.sleep(1)
print "bug tuple " + tuple(temp.split(', '))

And after this my application displays:

bug 1, 2, 3 Traceback (most recent call last): File "C:\Python26\Lib\site-packages\pythonwin\pywin\framework\scriptutils.py", line 312, in RunScript exec codeObject in main._dict_ File "C:\Documents and Settings\k.pawlowski\Desktop\atsserver.py", line 165, in print "bug tuple " + tuple(temp.split(', ')) TypeError: cannot concatenate 'str' and 'tuple' objects

I don't know what I make wrong. print tuple('1, 2, 3'.split(', ')) works properly.

+4  A: 
print tuple(something)

may work because print will do an implicit str() on the argument, but and expression like

"" + ()

does not work. The fact that you can print them individually doesn't make a difference, you can't concatenate a string and a tuple, you have to convert either one of them. I.e.

print "foo" + str(tuple("bar"))

However, depending on str() for conversion probably won't give the desired results. Join them neatly using a separator using ",".join for example

Ivo van der Wijk
+2  A: 

Change it to

print "bug tuple ", tuple(temp.split(', '))
Moe
this is it ;) thanks. I am stupid after many hours of coding.. ;)
Carolus89
get some sleep...
Paul McGuire
+3  A: 

Why do you think it should work?

try:

print "bug tuple " + str(tuple(temp.split(', ')))
Maciej Kucharz
A: 

Why tuple by splitting, you have string for one ready except the paranthesis, why not:

print "bug tuple (%s)" % '1, 2, 3'
Tony Veijalainen