How to make a button open a specific .csv file using Python Tkinter?

you can use a lambda function in order to pass an argument to your load function. Unfortunately pastebin ist down right now so i cannot have a look at your code. The idea is something like this:

for filename in filenames: ...Button(...,command=lambda i=filename:loadFile(i),...

so in your loadFile function you have the filename as first parameter

I also tried this the way heinzderaugust suggested at first:

 button[i] = ttk.Button(loadProject, text=data_files[i], command = lambda i=data_files[i]:load(i)).grid(column=1, row=i, sticky=W)  

I was also able to solve this by using a class as the intermediary part of the callback:(This class was inserted inside viewProject )

class Command:            def __init__(self, callback, *args, **kwargs):                    self.callback = callback                    self.args = args                    self.kwargs = kwargs            def __call__(self):                    return(self.callback(*self.args, **self.kwargs))

Then I changed your button definition to:

button[i] = ttk.Button(loadProject, text=data_files[i], command = Command(load,data_files[i])).grid(column=1, row=i, sticky=W)     

And finally I changed the load function to accept a argument:

def load(myFile):                            spamReader = csv.reader(open(myFile))            for row in spamReader:                    a = list(row)            ttk.Label(loadProject, text=(a[0])).grid(column=4, row=0, sticky=W)            ttk.Label(loadProject, text=("\n"+a[1])).grid(column=4, row=1, sticky=W)            ttk.Label(loadProject, text=("\n"+a[2])).grid(column=4, row=2, sticky=W)            ttk.Label(loadProject, text=(a[3])).grid(column=4, row=3, sticky=W)            ttk.Label(loadProject, text=(a[4])).grid(column=4, row=4, sticky=W)            ttk.Label(loadProject, text=("\n"+a[5])).grid(column=4, row=5, sticky=W)            ttk.Label(loadProject, text=("\n"+a[6])).grid(column=4, row=6, sticky=W)            ttk.Label(loadProject, text=("\n\n"+a[7])).grid(column=4, row=7, sticky=W)

I am unsure which would be best practice, but both do work.

(Certainly the lambda approach does sort of seem more 'pythonic')

Source: https://code.activestate.com/recipes/66521-avoiding-lambda-in-writing-callback-functions/