How to find length of digits in an integer?
If you want the length of an integer as in the number of digits in the integer, you can always convert it to string like str(133) and find its length like len(str(123)).
If you want the length of an integer as in the number of digits in the integer, you can always convert it to string like str(133) and find its length like len(str(123)).
Bottom Line Use either COUNT(field) or COUNT(*), and stick with it consistently, and if your database allows COUNT(tableHere) or COUNT(tableHere.*), use that. In short, don’t use COUNT(1) for anything. It’s a one-trick pony, which rarely does what you want, and in those rare cases is equivalent to count(*) Use count(*) for counting Use * for … Read more
Perhaps table is what you are after? dummyData = rep(c(1,2, 2, 2), 25) table(dummyData) # dummyData # 1 2 # 25 75 ## or another presentation of the same data as.data.frame(table(dummyData)) # dummyData Freq # 1 1 25 # 2 2 75
You have a couple of options how to get the value of COUNT(*) from the SQL. The easiest three are probably this: $sql = “SELECT COUNT(*) FROM news”; $result = mysqli_query($con, $sql); $count = mysqli_fetch_assoc($result)[‘COUNT(*)’]; echo $count; or using column alias: $sql = “SELECT COUNT(*) as cnt FROM news”; $result = mysqli_query($con, $sql); $count = … Read more
This will do what you want (list of towns, with the number of users in each): select town, count(town) from user group by town You can use most aggregate functions when using GROUP BY: (COUNT, MAX, COUNT DISTINCT etc.) Update (following change to question and comments) You can declare a variable for the number of … Read more
It’s not recommended to use reserved words for names in SQL. So I call the count result cnt instead. As your function is scalar, i.e. you expect only one value back, you can use: $count = $mysqli->query(“select count(*) as cnt from cars”)->fetch_object()->cnt;
Convert your variable to a factor, and set the categories you wish to include in the result using levels. Values with a count of zero will then also appear in the result: y <- c(0, 0, 1, 3, 4, 4) table(factor(y, levels = 0:5)) # 0 1 2 3 4 5 # 2 1 0 … Read more
There’s three ways to get this sort of count, each with their own tradeoffs. If you want a true count, you have to execute the SELECT statement like the one you used against each table. This is because PostgreSQL keeps row visibility information in the row itself, not anywhere else, so any accurate count can … Read more
Here’s a solution with the dplyr package – it has n_distinct() as a wrapper for length(unique()). df %>% group_by(color) %>% mutate(unique_types = n_distinct(type))