I have this code
String coNum = customerOrderLine.coNum.PadLeft(10 - customerOrderLine.coNum.Length);
I know that customerOrderLine.coNum = "123456" So I should end up with coNum being having 4 empty spaces at the front of it but I end up with it being "123456". How do I fix this? I tried PadRight in case that was the mistake and it also failed to work. I have to have the 4 empty spaces at the beginning to pass it into the API I am working on or it will fail.
PadLeft takes a total length as a parameter, so I think you want
String coNum = customerOrderLine.coNum.PadLeft(10);
This is because you have incorrectly specified the totalWidth parameter of the Pad* method.
From docs:
The number of characters in the resulting string, equal to the number
of original characters plus any additional padding characters.[...] If totalWidth is equal to the length of this instance, the method
returns a new string that is identical to this instance.
PadLeft does not specify a default character to pad with; your second argument should be the character to use for the pad, i.e.:
String coNum = customerOrderLine.coNum.PadLeft(10, ' ');
Edit: Also the first argument should be total desired length, not number of pad characters to add, per #Matthew's answer.
Related
I am using the Excel VBA function "Left" to get the left three characters of a string, but the function only returns two. The string has more than three characters, so I'm at a loss for why it would return fewer. The official documentation indicates that the function should return the number of characters requested or the entire string if the string is shorter than the requested character count.
The error appears in the section below, but Trim() appears to be working correctly. I also can't step into the Left() function to see how it's handling the inputs.
part = Trim(part)
pref = Left(part, 3)
EDIT1: I changed the character count in the Left() function to see if it would change accordingly. Left(part, 2) returned 1 character (just P). It appears the function is systemically returning 1 fewer character than requested.
EDIT2: I also changed the If statement to accept one fewer character for events where Left() returned the incorrect quantity.
In the snippet above, a single V should be accepted but the code skipped to the line indicated by the arrow, which shows that the comparison was still false. This is leading me to believe that there is a non-printing character leading the entire string, but I don't know how to check for that.
As mentioned in the comments, there were non-printing characters preceding the first visible character in the strings. I used the following code to remove other potentially problematic characters:
part = Trim(part)
While Left(part, 1) < Chr(32) And Len(part) > 1
part = Right(part, Len(part) - 1)
Wend
This offered a more general way to remove problematic leading characters.
I have a bunch of users in a list called UserList.
And I do not want the output to have the square brackets, so I run this line:
UserList = [1,2,3,4...]
UserListNoBrackets = str(UserList).strip('[]')
But if I run:
len(UserList) #prints22 (which is correct).
However:
len(UserListNoBrackets) #prints 170 (whaaat?!)
Anyway, the output is actually correct (I'm pretty sure). Just wondering why that happens.
Here:
UserListNoBrackets = str(UserList).strip('[]')
UserListNoBrackets is a string. A string is a sequence of characters, and len(str) returns the numbers of characters in the string. A comma is a character, a white space is a character, and the string represention of an integer has has many characters as there are digits in the integer. So obviously, the length of your UserListNoBrackets string is much greater than the length of you UserList list.
You probably need str.join
Ex:
user_list = [1,2,3,4...]
print(",".join(map(str, user_list)))
Note:
Using map method to convert all int elements in list to string.
I wanted to know how to remove first character of a string in octave. I am manipulating the string in a loop and after every loop, I want to remove the first character of the remaining string.
Thanks in advance.
If it's just a one-line string then:
short_string = long_string(2:end)
But if you have a cell array of strings then either do it as above if you have a loop already, otherwise you can use this shorthand to do it in one line:
short_strings = cellfun(#(x)(x(2:end)), long_strings, 'uni', false)
Or else if you have a matrix of strings (i.e. all the same length), then you can vectorize it as:
short_strings = long_strings(:, 2:end)
I have a data file in csv format.
I am trying to split each line using the basic split command line.split(',')
But when I get a string like this "2,,",
instead of returning an array as thus Array(2,"","")
I just get an Array: Array(2).
I am most definitely missing something basic, could someone help point out the correct way to split a comma separated string here?
This is inherited from Java. You can achieve behavior you want by using the split(String regex, int limit) overload:
"2,,".split(",", -1) // = Array(2, "", "")
Note the String instead of Char.
As explained by the Java Docs, the limit parameter is used as follows:
The limit parameter controls the number of times the pattern is
applied and therefore affects the length of the resulting array. If
the limit n is greater than zero then the pattern will be applied at
most n - 1 times, the array's length will be no greater than n, and
the array's last entry will contain all input beyond the last matched
delimiter. If n is non-positive then the pattern will be applied as
many times as possible and the array can have any length. If n is zero
then the pattern will be applied as many times as possible, the array
can have any length, and trailing empty strings will be discarded.
Source
Using split(separator: Char) will call the overload above, using a limit of zero.
I've a question about Fortran 77 and I've not been able to find a solution.
I'm trying to store an array of strings defined as the following:
character matname(255)*255
Which is an array of 255 strings of length 255.
Later I read the list of names from a file and I set the content of the array like this:
matname(matcount) = mname
EDIT: Actually mname value is hardcoded as mname = 'AIR' of type character*255, it is a parameter of a function matadd() which executes the previous line. But this is only for testing, in the future it will be read from a file.
Later on I want to print it with:
write(*,*) matname(matidx)
But it seems to print all the 255 characters, it prints the string I assigned and a lot of garbage.
So that is my question, how can I know the length of the string stored?
Should I have another array with all the lengths?
And how can I know the length of the string read?
Thanks.
You can use this function to get the length (without blank tail)
integer function strlen(st)
integer i
character st*(*)
i = len(st)
do while (st(i:i) .eq. ' ')
i = i - 1
enddo
strlen = i
return
end
Got from here: http://www.ibiblio.org/pub/languages/fortran/ch2-13.html
PS: When you say: matname(matidx) it gets the whole string(256) chars... so that is your string plus blanks or garbage
The function Timotei posted will give you the length of the string as long as the part of the string you are interested in only contains spaces, which, if you are assigning the values in the program should be true as FORTRAN is supposed to initialize the variables to be empty and for characters that means a space.
However, if you are reading in from a file you might pick up other control characters at the end of the lines (particularly carriage return and/or line feed characters, \r and/or \n depending on your OS). You should also toss those out in the function to get the correct string length. Otherwise you could get some funny print statements as those characters are printed as well.
Here is my version of the function that checks for alternate white space characters at the end besides spaces.
function strlen(st)
integer i,strlen
character st*(*)
i = len(st)
do while ((st(i:i).eq.' ').or.(st(i:i).eq.'\r').or.
+ (st(i:i).eq.'\n').or.(st(i:i).eq.'\t'))
i = i - 1
enddo
strlen = i
return
end
If there are other characters in the "garbage" section this still won't work completely.
Assuming that it does work for your data, however, you can then change your write statement to look like this:
write(*,*) matname(matidx)(1:strlen(matname(matidx)))
and it will print out just the actual string.
As to whether or not you should use another array to hold the lengths of the string, that is up to you. the strlen() function is O(n) whereas looking up the length in a table is O(1). If you find yourself computing the lengths of these static strings often, it may improve performance to compute the length once when they are read in, store them in an array and look them up if you need them. However, if you don't notice the slowdown, I wouldn't worry about it.
Depending on the compiler that you are using, you may be able to use the trim() intrinsic function to remove any leading/trailing spaces from a string, then process it as you normally would, i.e.
character(len=25) :: my_string
my_string = 'AIR'
write (*,*) ':', trim(my_string), ':'
should print :AIR:.
Edit:
Better yet, it looks like there is a len_trim() function that returns the length of a string after it has been trimmed.
intel and Compaq Visual Fortran have the intrinsic function LEN_TRIM(STRING) which returns the length without trailing blanks or spaces.
If you want to suppress leading blanks or spaces, use "Adjust Left" i.e. ADJUSTF(STRING)
In these FORTRANs I also note a useful feature: If you pass a string in to a function or subroutine as an argument, and inside the subroutine it is declared as CHARACTER*(*), then
using the LEN(STRING) function in the subroutine retruns the actual string length passed in, and not the length of the string as declared in the calling program.
Example:
CHARACTER*1000 STRING
.
.
CALL SUBNAM(STRING(1:72)
SUBROUTINE SYBNAM(STRING)
CHARACTER*(*) STRING
LEN(STRING) will be 72, not 1000