I want to add a line break in the Output Statement in Flowgarithm. For example, I want:
Line 1
Line 2
NOT:
Line 1 Line 2
I have been working in the program, and it does not accept many special symbols, such as |, <<, /, etc...
I have tried scouring the website (http://www.flowgorithm.org/index.htm) to no avail.
I have found, however, a section on said website, under Documentation>Documentation>Output, there is a section that has details about "newline", "first", and "last".
Thank you!
Yes, you need to use ToChar(13) in between the lines.
Like this:
This gives an output like:
13 is the ASCII code for carriage return. Hope this is what you need.
Related
This may be a more general question so sorry in advance. I am creating a script and thought it would be good to use multi-line strings instead of using multiple printf or echo statements. Say I have the following:
while :
do
printf "line 1
line 2
line 3"
done
The second and third lines would be printed with a space in front because of the indentation in the file.
l1
line 2
line 3
Is there a way to prevent that aside from removing the indentation on the code? Also, is it considered a better practice to just multiple printf/echo statements if you need to output information that spans multiple lines?
Indent with tabs (here whitespace) and use a heredoc (with <<-)
cat <<- EOF
line 1
line 2
line 3
EOF
Multi-line strings will always look a bit bad, or have some other downsides, I'm afraid. The most legible way to embed them in bash code is probably the here-doc, which shows the string (almost) exactly like it will look when output. As an extra knack, you can use extra punctuation to make the here-doc delimiter to stand out from the string itself too, like so:
if true
then
some commands
cat <<"____EndOfTextBlock____"
This text here
spans multiple
lines.
____EndOfTextBlock____
some other commands
even more commands
fi
I can use
:5,12s/foo/bar/g
to search for foo and replace it by bar between lines 5 and 12. How can I do that only in line 5 and 12 (and not in the lines in between)?
Vim has special regular expression atoms that match in certain lines, columns, etc.; you can use them (possibly in addition to the range) to limit the matches:
:5,12s/\(\%5l\|\%12l\)foo/bar/g
See :help /\%l
You can do the substitution on line 5 and repeat it with minimal effort on line 12:
:5s/foo/bar
:12&
As pointed out by Ingo, :& forgets your flags. Since you are using /g, the correct command would be :&&:
:5s/foo/bar/g
:12&&
See :help :& and friends.
You could always add a c to the end. This will ask for confirmation for each and every match.
:5,12s/foo/bar/gc
Interesting question. Seems like there's only range selection and no multiple line selection:
http://vim.wikia.com/wiki/Ranges
However, if you have something special on line 5 and 12, you could use the :g operator. If your file looks like this (numbers only for reference):
1 line one
2 line one
3 line one
4 line one
5 enil one
6 line one
7 line one
8 line one
9 line one
10 line one
11 line one
12 enil one
And you want to replace one by eno on the lines where there's enil instead of line:
:g/enil/s/one/eno/
You could use ed - a line oriented text editor with similar commands to vi and vim. It probably predates vi and vim.
In a script (using a here document which processes input till the EndCommand marker) it would look like:
ed file <<EndCommands
5
s/foo/bar/g
7
s/foo/bar/g
wq
EndCommands
Obviously, the ed commands can be used on the command line also.
I am writing a swirl lesson using swirlify package functions in RStudio.
Below is how lesson.yaml file looks like now
- Class: text
Output: Welcome to Part 1 Playing with Numbers!!!
Output for which looks like
How to insert a new line or line break after Welcome to Part 1 in lesson.yaml file above, so that it displays the output as below when I run the demo_lesson() command again after saving the lesson.yaml file
| Welcome to Part 1
| Playing with Numbers!!!
Using YAML, you can use any of these equivalent approaches:
Quoted string with escape
- Class: text
Output: "Welcome to Part 1\nPlaying with Numbers!!!"
Literal scalar
- Class: text
Output: |-
Welcome to Part 1
Playing with Numbers!!!
(| starts a literal scalar and - tells YAML to drop the final line break.)
Multiline scalar
- Class: text
Output:
Welcome to Part 1
Playing with Numbers!!!
(since one line break gets folded into a space, you need two line breaks.)
Since I do not know whether swirlify nicely handles line breaks in the string, I guess you could also do
- Class: text
Output: Welcome to Part 1
- Class: text
Output: Playing with Numbers!!!
Thanks to flyx for answering the question, here is how it works!!
I. Quotes string with escape (Works with two \n\n)
lesson.yaml file
II. Literals
First line in Output: |- Hit ENter once
Indent once by pressing one Tab for first line, Hit Enter twice to have break between header line and paragraph like below, then it works..
lesson.yaml file
III. Mulitline Scalar (Works with three times Enter between two lines)
Press Enter once After Output: in lesson.yaml
Indent once by pressingTab` key once, Write your first line, hit Enter thrice and write the second line. Then it works.
lesson.yaml file
OUTPUT FOR ALL THE ABOVE ANSWERS
I am working on a .do file created by someone else. This person used a semicolon delimiter in the entire file. I am trying to go through this file and see what is going on. I like to do this by selecting a portion of the code and hitting the "Execute Selection (do)" button. However, the delimiter seems to be messing up this. Are there any workarounds for me?
Suppose your do-file looks like this:
#delimit ;
set obs
10 ;
gen x = _n ;
gen y = x^2 ;
gen z = x
^3;
Anytime you highlight a selection and press "Execute selection (do)", Stata creates a temporary, self-contained do-file, with default delimit at cr and runs that:
"When a do-file begins execution, the delimiter is automatically set to
carriage return, even if it was called from another do-file that set the
delimiter to semicolon."
It does not sequentially run those commands from the console. Therefore, if you select the first 2 commands in the do-file above, the temporary do-file includes a call to #delimit whereas if you selected the last 2 commands, the temporary do-file would not have this call and would throw a syntax error for two line commands.
One solution could be to copy-paste selections to a fresh do-file that just had the #delimit command at the beginning, and then run that.
You could also write a script to rid your do-file of semicolons. If a line does not end in a semicolon, then append the next line to the end of the current line, and check this line again. Depending on how complex the syntax is in your do-file, this would be more or less difficult.
Another option is comment out the lines you have already ran by enclosing them with /* */ and to use exit; where you want to stop. You do have to be a little careful with local macros.
I am trying to make a shell script work in Windows. Sorry but I'm not very experienced in Windows (or even that much in shell to be honest). The script works well except for this one line:
print "9\n0\n1\n5\n0\n0\n\n" | /usr/ts23/mm_util
The mm_util is an interactive utility that takes numbers as input. It chooses selection 9 first, then 0, then 1, etc. I've changed the path to use the utility, which has an identical interface in Windows but the output is just the first screen. The "9" input isn't entered, and because of this the output (that is parsed) is incorrect. How can I change this so that the "9" is entered on the first screen?
Here is a method that does not require a file. It works on the command line:
(for %N in (9 0 1 5 0 0 "") do #echo(%~N)|c:\Users\ts23\mm_util
The "" is to get an empty line in the output, as you had in your original question. Your answer does not have the blank line.
The %~N notation strips enclosing quotes from the value.
The echo( is non-intuitive syntax that can reliably print a blank line, in case %~N expands to nothing.
Don't forget to double the percents if you put the code in a batch script.
Try to put that nine-linebreak-zero-stuff in a text file, and then execute print textfile.txt | /usr/ts23/mm_util
And bear in mind that Windows uses the pre-UNIX convention that the linebreak is CR LF, not just LF.
The way I got the output I wanted was by using this:
C:\Users\ts23\mm_util < test.txt
And then just put the following inside test.txt
9
0
1
5
0
0
The output I got was what I needed, hopefully this will help someone trying to do something like this in the future.