How could i skip a row if a csv field is empty-CodePudding

I have a csv file where I have some names and the date from when students have passed the exam. if they did not pass the exam the date will be empty. how could I filter out the empty field so I only have a list with people that have passed the test?

Here is the code that I have. the row[0] is the names of the student and row[12] is the date they have passed it.

with open("MYCSV_FILE", 'r') as read_obj:
       csv_dict_reader = csv.reader(read_obj, delimiter=';')
for row in csv_dict_reader:
    if row[12]:
        with open('fed_results\Behaalde_studenten.csv', 'a') as f_object:  
            # Pass the CSV  file object to the writer() function
            writer_object = csv.writer(f_object)
            # Result - a writer object
            # Pass the data in the list as an argument into the writerow() function
            writer_object.writerow(row[0])  
            # Close the file object
            f_object.close()

CodePudding user response：

row[12] can be checked for an empty result like this:

if row[12]:
    # do something #

So your code can check for non-empty rows by:

with open("MYCSV_FILE", 'r') as read_obj:
    csv_dict_reader = csv.reader(read_obj, delimiter=';')
    for row in csv_dict_reader:
        if row[12]:
            print(row[0], row[12])

Or you can consider an if/else statement:

with open("MYCSV_FILE", 'r') as read_obj:
    csv_dict_reader = csv.reader(read_obj, delimiter=';')
    for row in csv_dict_reader:
        if row[12]:
            print(row[0], row[12])
        else:
            # what to do if it isn't there #

This all assumes that each row is delimited by a semicolon even when empty values are given, such that:

a;b;c;d;e;f;g;h;i;j;k;l
1;2;3;4;5;6;7;8;9;10;11;