How to display items in a list in a label using tkinter

Question

I currently have this code:

for element in tagall:  
    for item in element: 
       items = item
        print(item)

but I can't seem to incorporate it into changing the text attribute on a label in tkinter. When I do:

label.config(text=(item))

the label only displays the last item and not all of them.
Thanks in advance

P.S. I fogot to mention that I'm using an sql server and 'tagall' is a variable of which a column of data has been set to

What about something like: label.config(text='\n'.join(element)) — user590028
– user590028, Commented Feb 19, 2014 at 16:32

user590028 · Accepted Answer · 2014-02-19 16:42:09Z

1

Try this:

label.config(test=repr(tagall))

That should get you going down the right path

answered Feb 19, 2014 at 16:42

user590028

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3 Comments

BigGibb Over a year ago

This does not separate the fields onto their own line - it just displays them

user590028 Over a year ago

I wasn't trying to format the response -- I was trying to help you with your bug "...the label only displays the last item and not all of them...". I assumed once you saw the output from repr() you'd figure out how to format the output appropriately.

BigGibb Over a year ago

newScreenLabel.config(text=("\n".join(str.split(repr(tagall))))) displays each value on a new line

Joan Smith · Accepted Answer · 2014-02-19 16:49:18Z

1

Think about what label.config(text=(item)) does. When you call the function, it overwrites the existing text of label with the new text stored in item. Therefore, it shouldn't surprise you that you only see the last item.

Now, it depends on what you want the display of the list of items to look like. Do you want them each on their own line? Separated by spaces? That's going to determine the string you use to join the list of items.

If you want them separated by new lines, replace your loop with

label.config(text=("\n".join(list_of_items)))

Then you should be good to go.

edited Feb 19, 2014 at 16:49

answered Feb 19, 2014 at 16:42

Joan Smith

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3 Comments

BigGibb Over a year ago

TypeError: sequence item 0: expected str instance, pyodbc.Row found

Joan Smith Over a year ago

Sorry, dropped the extra parens,label.config(text=("\n".join(list_of_items)))

BigGibb Over a year ago

Where does the parens go? EDIT: Added the parens around the list of items but this only separates the last field into its individual letters of individual lines

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