I am using a the vim-screen plugin that enable me to write scripts , start an interpreter in the same window and send lines the the interpreter. Problem is that the interpreter do not accept statements written on several lines.
exemple:
This will work f:{[x] y:y+1; Z:y+1; :Z; };
But this won't
f:{[x] y:y+1;
Z:y+1;
:Z;
};
How can I write a vim function that I could call to reshape the lines in order to be sent to the interpreter?
EDIT:
I had no success in making this function, I would like to create a function that would, from a input like this (that would be visually selected)
F:{[a;b;r]
//ccc1
aaa1;
aaa2;
//ccc2
aaa3;
};
output something like this F:{[a;b;r] aaa1; aaa2; aaa3; };
So I created a bounty
If you want to actually modify the buffer, J / :join do that. If you just want to join the lines that are sent to the interpreter (but keep them split in the buffer), you can retrieve the selected lines with getline(), and then join() them. Here's an example command:
:command! -range Invoke echomsg join(getline(<line1>,<line2>), '')
Edit
Based on that, you can "massage" the List of lines returned by getline(). E.g. to ignore the commented lines:
:command! -range Invoke echomsg join(filter(getline(<line1>,<line2>), 'v:val !~# "^\\s*//"'), '')
Additionally strip leading whitespace (this becomes unwieldy in a single line; better use a function now):
:command! -range Invoke echomsg join(map(filter(getline(<line1>,<line2>), 'v:val !~# "^\\s*//"'), 'substitute(v:val, "^\\s\\+", " ", "g")'), '')
Standard continuation character in vimscript scripts is backslash in the beginning of the next line. So, this
f:{[x] y:y+1;
\ Z:y+1;
\ :Z;
\ };
should work.
Related
I'm currently playing around on PowerShell, and I'm wondering how I can call multiple functions after eachother, but giving them each their own line.
I currently have this:
$leet= $text.replace("a", "4").replace("e", "1").replace("e", "3");
But I want it more like this:
$leet= $text
.replace("a", "4")
.replace("e", "1")
.replace("e", "3");
But PowerShell doesn't really like the newlines, and it doesn't work either when I add the ` to the end of each line followed by another one.
So, am I missing something, or is this not possible in PowerShell?
As Jeroen Mostert notes you can get line continuation for free by specifying the operator (.) and then placing the whitespace between it and the right-hand operand:
$leet = $text.
Replace("a", "4").
Replace("e", "1").
Replace("e", "3")
(note that the last call, Replace("e", "3"), does nothing - all the e's have already been replaced by the preceding call)
I'm writing a reusable function, so I need the argument to be as flexible as possible.
Consider a simple example:
function testf(){
print ./*.$1
}
This works. For example, with testf mp3 it lists all the files ending with .mp3 in an array, making possible the use of for loops. But this way it only allows me to work with the extension name.
Therefore, I tried:
function testf(){
print ./$1
}
However, it doesn't work. Using testf *.mp3, unlike using print *.mp3 in the terminal, it will only pass the first matching string instead of the whole array.
Any suggestion?
ists all the files ending with .mp3 in an array ... there is no array involved in your question.
But to your problem: First, you want to pass to your function a wildcard pattern, but this is not what you are actually doing. testf *.mp3 expands the pattern before the function is invoked (this process is called filename generation), and your testf gets just a list of files as parameters. You can pass a pattern, but you have to ask the shell not to expand it:
testf '*.mp3'
In this case, your $1 indeed will contain the string *.mp3. However, your print ./$1 will still not work. The reason is that filename generation occurs before parameter expansion (which is the process where $1 is replaced by the string it contains). Again, you have to ask the shell to do it the other way round:
print ./${~1}
The shell performs several types of expansions before launching the command. When you enter
testf *.mp3
the shell will expand the glob first, and pass each filename as a separate argument to the function
Your function could look like this:
function testf(){
printf './%s\n' "$#"
}
If you iterate over .VARIABLES and print each, any true variable can be printed correctly with the following rule:
print_variables: $(foreach V,$(.VARIABLES),print-$(V)) .phony_explicit
print-%: .phony_explicit; #echo "$* = \"$($*)\""
.PHONY: .phony_explicit ...
A 0- or 1-line function will still work, but any more will result in Syntax error: Unterminated quote string. Just one multiline function will break the entire print_variables rule. As a workaround, I have added ;\ to each line in my function definitions, but that won't fix existing multiline functions (either via includes from this makefile or via other makefiles including this one.) What can I do? Is there a container of just function variables, or a way to test if a variable is a function definition?
A simple minimal example would be easier to understand; this has nothing to do with .VARIABLES, pattern rules, etc. (and I'm not sure what the point of the .phony_explicit prereq is..)
define F
foo
bar
endef
print: ; echo "F = $(F)"
will show the problem:
echo "F = foo
/bin/sh: 1: Syntax error: Unterminated quoted string
This is because when make sees a variable that contains newlines in a recipe, it assumes that the newlines mean you want the lines of the variable to become lines in the recipe.
First in general you should use single-quotes around strings you send to the shell, unless you need the shell to expand them; it won't help in this situation but in general it's much safer.
There's no way to undo that, really. You have a number of options.
The first is to not use echo but instead use the make function info:
print-F: ; $(info F = "$(F)")
yields:
F = "foo
bar"
Another option is to use subst to replace the newlines with some other value. The new value cannot itself contain explicit newlines, but you can ask the shell to print a newline for you:
# Create a variable containing a single newline
# Note this must contain TWO newlines!
define NL
endef
print-F: printf 'F = "$(subst %,%%,$(subst $(NL),\n,$(F))"\n'
Yields:
printf 'F = "foo\nbar"\n'
F = "foo
bar"
One final option is to convert your makefile to use the .ONESHELL feature, but I assume that's a step too far just to get this debugging output available :).
I am navigating a Java-based CLI menu on a remote machine with expect inside a bash script and I am trying to extract something from the output without leaving the expect session.
Expect command in my script is:
expect -c "
spawn ssh user#host
expect \"#\"
send \"java cli menu command here\r\"
expect \"java cli prompt\"
send \"java menu command\"
"
###I want to extract a specific string from the above output###
Expect output is:
Id Name
-------------------
abcd 12 John Smith
I want to extract abcd 12 from the above output into another expect variable for further use within the expect script. So that's the 3rd line, first field by using a double-space delimiter. The awk equivalent would be: awk -F ' ' 'NR==3 {$1}'
The big issue is that the environment through which I am navigating with Expect is, as I stated above, a Java CLI based menu so I can't just use awk or anything else that would be available from a bash shell.
Getting out from the Java menu, processing the output and then getting in again is not an option as the login process lasts for 15 seconds so I need to remain inside and extract what I need from the output using expect internal commands only.
You can use regexp in expect itself directly with the use of -re flag. Thanks to Donal on pointing out the single quote and double quote issues. I have given solution using both ways.
I have created a file with the content as follows,
Id Name
-------------------
abcd 12 John Smith
This is nothing but your java program's console output. I have tested this in my system with this. i.e. I just simulated your program's output with cat. You just replace the cat code with your program commands. Simple. :)
Double Quotes :
#!/bin/bash
expect -c "
spawn ssh user#domain
expect \"password\"
send \"mypassword\r\"
expect {\\\$} { puts matched_literal_dollar_sign}
send \"cat input_file\r\"; # Replace this code with your java program commands
expect -re {-\r\n(.*?)\s\s}
set output \$expect_out(1,string)
#puts \$expect_out(1,string)
puts \"Result : \$output\"
"
Single Quotes :
#!/bin/bash
expect -c '
spawn ssh user#domain
expect "password"
send "mypasswordhere\r"
expect "\\\$" { puts matched_literal_dollar_sign}
send "cat input_file\r"; # Replace this code with your java program commands
expect -re {-\r\n(.*?)\s\s}
set output $expect_out(1,string)
#puts $expect_out(1,string)
puts "Result : $output"
'
As you can see, I have used {-\r\n(.*?)\s\s}. Here the braces prevent any variable substitutions. In your output, we have a 2nd line with full of hyphens. Then a newline. Then your 3rd line content. Let's decode the regex used.
-\r\n is to match one literal hyphen and a new line together. This will match the last hyphen in the 2nd line and the newline which in turn make it to 3rd line now. So, .*? will match the required output (i.e. abcd 12) till it encounters double space which is matched by \s\s.
You might be wondering why I need parenthesis which is used to get the sub-match patterns.
In general, expect will save the expect's whole match string in expect_out(0,string) and buffer all the matched/unmatched input to expect_out(buffer). Each sub match will be saved in subsequent numbering of string such as expect_out(1,string), expect_out(2,string) and so on.
As Donal pointed out, it is better to use single quote's approach since it looks less messy. :)
It is not required to escape the \r with the backslash in case of double quotes.
Update :
I have changed the regexp from -\r\n(\w+\s+\w+)\s\s to -\r\n(.*?)\s\s.
With this way - your requirement - such as match any number of letters and single spaces until you encounter first occurrence of double spaces in the output
Now, let's come to your question. You have mentioned that you have tried -\r\n(\w+)\s\s. But, there is a problem here with \w+. Remember \w+ will not match space character. Your output has some spaces in it till double spaces.
The use of regexp will matter based on your requirements on the input string which is going to get matched. You can customize the regular expressions based on your needs.
Update version 2 :
What is the significance of .*?. If you ask separately, I am going to repeat what you commented. In regular expressions, * is a greedy operator and ? is our life saver. Let us consider the string as
Stackoverflow is already overflowing with number of users.
Now, see the effect of the regular expression .*flow as below.
* matches any number of characters. More precisely, it matches the longest string possible while still allowing the pattern itself to match. So, due to this, .* in the pattern matched the characters Stackoverflow is already over and flow in pattern matched the text flow in the string.
Now, in order to prevent the .* to match only up to the first occurrence of the string flow, we are adding the ? to it. It will help the pattern to behave as non-greedy manner.
Now, again coming back to your question. If we have used .*\s\s, then it will match the whole line since it is trying to match as much as possible. This is common behavior of regular expressions.
Update version 3:
Have your code in the following way.
x=$(expect -c "
spawn ssh user#host
expect \"password\"
send \"password\r\"
expect {\\\$} { puts matched_literal_dollar_sign}
send \"cat input\r\"
expect -re {-\r\n(.*?)\s\s}
if {![info exists expect_out(1,string)]} {
puts \"Match did not happen :(\"
exit 1
}
set output \$expect_out(1,string)
#puts \$expect_out(1,string)
puts \"Result : \$output\"
")
y=$?
# $x now contains the output from the 'expect' command, and $y contains the
# exit status
echo $x
echo $y;
If the flow happened properly, then exit code will have value as 0. Else, it will have 1. With this way, you can check the return value in bash script.
Have a look at here to know about the info exists command.
I'm trying to pull variables from an API in json format and then put them back together with one variable changed and fire them back as a put.
Only issue is that every value has quote marks in it and must go back to the API separated by commas only.
example of what it should see with redacted information, variables inside the **'s:
curl -skv -u redacted:redacted -H Content-Type: application/json -X PUT -d'{properties:{basic:{request_rules:[**"/(req) testrule","/test-body","/(req) test - Admin","test-Caching"**]}}}' https://x.x.x.x:9070/api/tm/1.0/config/active/vservers/xxx-xx
Obviously if I fire them as a plain array I get spaces instead of commas. However I tried outputting it as a plain string
longstr=$(echo ${valuez[#]})
output=$(echo $longstr |sed -e 's/" /",/g')
And due to the way bash is interpreted it seems to either interpret the quotes wrong or something else. I guess it might well be the single ticks encapsulating after the PUT -d as well but I'm not sure how I can throw a variable into something that has single ticks.
If I put the raw data in manually it works so it's either the way the variable is being sent or the single ticks. I don't get an error and when I echo the line out it looks perfect.
Any ideas?
valuez=( "/(req) testrule" "/test-body" "/(req) test - Admin" "test-Caching" )
# Temporarily set IFS to some character which is known not to appear in the array.
oifs=$IFS
IFS=$'\014'
# Flatten the array with the * expansion giving a string containing the array's elements separated by the first character of $IFS.
d_arg="${valuez[*]}"
IFS=$oifs
# If necessary, quote or escape embedded quotation marks. (Implementation-specific, using doubled double quotes as an example.)
d_arg="${d_arg//\"/\"\"}"
# Substitute the known-to-be-absent character for the desired quote+separator+quote.
d_arg="${d_arg//$'\014'/\",\"}"
# Prepend and append quotes.
d_arg="\"$d_arg\""
# insert the prepared arg into the final string.
d_arg="{properties:{basic:{request_rules:[${d_arg}]}}}"
curl ... -d"$d_arg" ...
if you have gnu awk with version 4 and above, which support FPAT
output=$(echo $longstr |awk '$1=$1' FPAT="(\"[^\"]+\")" OFS=",")
Explanation
FPAT #
This is a regular expression (as a string) that tells gawk to create the fields based on text that matches the regular expression. Assigning a value to FPAT overrides the use of FS and FIELDWIDTHS for field splitting. See Splitting By Content, for more information.
If gawk is in compatibility mode (see Options), then FPAT has no special meaning, and field-splitting operations occur based exclusively on the value of FS.
valuez=( "/(req) testrule" "/test-body" "/(req) test - Admin" "test-Caching" )
csv="" sep=""
for v in "${valuez[#]}"; do csv+="$sep\"$v\""; sep=,; done
echo "$csv"
"/(req) testrule","/test-body","/(req) test - Admin","test-Caching"
If it's something you need to do repeatedly, but it into a function:
toCSV () {
local csv sep val
for val; do
csv+="$sep\"$val\""
sep=,
done
echo "$csv"
}
csv=$(toCSV "${valuez[#]}")