I'm trying to create a ZSH auto completion script for Symfony's Console, I almost finished it but I block in the last part. I need to create an array thanks to a function returns, but I don't know how to process.
Here's the code I wrote:
_find_console () {
echo "php $(find . -maxdepth 2 -mindepth 1 -name 'console' -type f | head -n 1)"
}
_console_get_command_list () {
`_find_console` --no-ansi | \
sed "1,/Available commands/d" | \
awk '/ [a-z]+/ { print $0 }' | \
sed -E 's/^[ ]+//g' | \
sed -E 's/[:]+/\\:/g' | \
sed -E 's/[ ]{2,}/\:/g'
}
_console () {
local -a commands
commands=(`_console_get_command_list`)
_describe 'commands' commands
}
compdef _console php console
compdef _console console
If I execute the command line, it formats it in the proper way. But used with ZSH the output is like this:
acme:hello -- Hello
assets:install -- Installs
cache:warmup -- Warms
and the formatted function returns:
acme\:hello:Hello World example command
assetic\:dump:Dumps all assets to the filesystem
cache\:warmup:Warms up an empty cache
The words that are not displayed, are used as a command completion. Here's an example: Asciinema example.
If I take the returns and use it directly into the array, it works well.
Related
I'm trying to pipe a series of manipulations into an xargs call that I can use to swap the first value with the second using the sed command (sed is optional if there's a better way).
Basically I'm grabbing method signature in camel case and appending a prefix while trying to retain camel case.
So it should take...
originalMethodSignature
and replace it with...
givenOriginalMethodSignature
Because I'm using a series of pipes to find and modify the text, I was hoping to use multiple params with xargs, but it seems that most of the questions involving that use sh -c which would be fine but in order for the sed command to be interactive on a Mac terminal I need to use single quotes inside the shell calls' single quotes.
Something like this, where the double quotes preserve the functionality of the single quotes in the sed command...
echo "somePrecondition SomePrecondition" | xargs -L1 sh -c 'find ~/Documents/BDD/Definitions/ -type f -name "Given$1.swift" -exec sed -i "''" "'"s/ $0/ given$1/g"'" {} +'
assuming there's a file called "~/Documents/BDD/Definitions/GivenSomePrecondition.swift" with below code...
protocol GivenSomePrecondition { }
extension GivenSomePrecondition {
func somePrecondition() {
print("empty")
}
}
The first awk is going through a list of swift protocols that start with the Given keyword (e.g. GivenSomePrecondition), then they strip it down to "somePrecondition SomePrecondition" before hitting the final pipe. My intent is that the final xargs call can replace $0 with given$1 interactively (overwriting the file).
The original command in context...
awk '{ if ($1 ~ /^Given/) print $0;}' ~/Documents/Sell/SellUITests/BDDLite/Definitions/HasStepDefinitions.swift \
| tr -d "\t" \
| tr -d " " \
| tr -d "," \
| sort -u \
| xargs -I string sh -c 'str=$(echo string); echo ${str#"Given"}' \
| awk '{ print tolower(substr($1,1,1)) substr($1, 2)" "$1 }' \
| xargs -L1 sh -c '
find ~/Documents/Sell/SellUITests/BDDLite/Definitions/ \
-type f \
-name "Given$1.swift" \
-exec sed -i '' "'"s/ $0/ given$1/g"'" {} +'
You don't need xargs or sh -c, and taking them out reduces the amount of work involved.
echo "somePrecondition SomePrecondition" |
while read -r source replace; do
find ~/Documents/BDD/Definitions/ -type f -name "Given${replace}.swift" -print0 |
while IFS= read -r -d '' filename; do
sed -i '' -e "s/ ${source}/ given${replace}/g" "$filename"
done
done
However, to answer your questions as opposed to sidestepping it, you can write functions that use any kind of quotes you want, and export them into your subshell, either with export -f yourFunction in a parent process or by putting "$(declare -f yourFunction)" inside the string passed after bash -c (assuming that bash is the same shell used in the parent process defining those functions).
#!/usr/bin/env bash
replaceOne() {
local source replace
source=$1; shift || return
replace=$1; shift || return
sed -i '' -e "s/ $1/ given$2/g" "$#"
}
# substitute replaceOne into a new copy of bash, no matter what kind of quotes it has
bash -c "$(declare -f replaceOne)"'; replaceOne "$#"'
I found a nice zsh command online that uses fzf to fuzzy-search shell history:
fh() {
eval $( ([ -n "$ZSH_NAME" ] && fc -l 1 || history) | fzf +s --tac | sed 's/ *[0-9]* *//')
}
However, instead of eval-ing the command, I'd like to place it in my prompt and edit it further before running it, something like:
$ fh grep
(search happens)
$ grep -RI foo . <--- cursor
I tried replacing eval with echo, which is better but doesn't give me the command to edit. Is there a way to do this in bash/zsh?
Here is a related thread with a suggestion to use xvkbd, but I was hoping there is something simpler.
In zsh, the vared builtin for the zsh line editor (ZLE) will let you edit an environment variable.
After editing, the updated variable can be used to execute the command:
fh() {
fzfresult=$( ([ -n "$ZSH_NAME" ] && fc -l 1 || history) | fzf +s --tac | sed 's/ *[0-9]* *//')
vared -p 'fzfout> ' -ac fzfresult
${fzfresult[#]}
}
In the vared command, -p sets the prompt. The -ac option changes the fzfresult variable to an array so we can execute it in the next step.
I don't have fzf installed, so this isn't completely tested, but the result should look like this:
% fh grep
fzfout> grep -RI foo . <-- edit, hit enter, get output:
file1: text with foo
file4: more text with foobar
I ended up using this:
fh() {
print -s $( ([ -n "$ZSH_NAME" ] && fc -l 1 || history) | fzf +s --tac | sed 's/ *[0-9]* *//')
}
ie silently print, followed by up-arrow, which brings up the command for editing. Would love to hear if there is a simpler way.
Trying to run a command as a variable but I am getting strange results
Expected result "1" :
grep -i nosuid /etc/fstab | grep -iq nfs
echo $?
1
Unexpected result as a variable command:
cmd="grep -i nosuid /etc/fstab | grep -iq nfs"
$cmd
echo $?
0
It seems it returns 0 as the command was correct not actual outcome. How to do this better ?
You can only execute exactly one command stored in a variable. The pipe is passed as an argument to the first grep.
Example
$ printArgs() { printf %s\\n "$#"; }
# Two commands. The 1st command has parameters "a" and "b".
# The 2nd command prints stdin from the first command.
$ printArgs a b | cat
a
b
$ cmd='printArgs a b | cat'
# Only one command with parameters "a", "b", "|", and "cat".
$ $cmd
a
b
|
cat
How to do this better?
Don't execute the command using variables.
Use a function.
$ cmd() { grep -i nosuid /etc/fstab | grep -iq nfs; }
$ cmd
$ echo $?
1
Solution to the actual problem
I see three options to your actual problem:
Use a DEBUG trap and the BASH_COMMAND variable inside the trap.
Enable bash's history feature for your script and use the hist command.
Use a function which takes a command string and executes it using eval.
Regarding your comment on the last approach: You only need one function. Something like
execAndLog() {
description="$1"
shift
if eval "$*"; then
info="PASSED: $description: $*"
passed+=("${FUNCNAME[1]}")
else
info="FAILED: $description: $*"
failed+=("${FUNCNAME[1]}")
done
}
You can use this function as follows
execAndLog 'Scanned system' 'grep -i nfs /etc/fstab | grep -iq noexec'
The first argument is the description for the log, the remaining arguments are the command to be executed.
using bash -x or set -x will allow you to see what bash executes:
> cmd="grep -i nosuid /etc/fstab | grep -iq nfs"
> set -x
> $cmd
+ grep -i nosuid /etc/fstab '|' grep -iq nfs
as you can see your pipe | is passed as an argument to the first grep command.
For some reason I cannot pass the 2nd parameter to a function which is on a another file, exactly here:
$lsValidLocal | xargs -n 1 -I {} bash -c 'Push "{}" "**$inFolder**"
The Push function on functions.sh does not read the 2nd parameter $inFolder.
I tried several different ways, the only working way till now is exporting the variable to make it globally accessible (not a good solution though)
script.sh
#!/bin/bash
#other machine
export otherachine="IP_address_otherachine"
#folders
inFolder="$HOME/folderIn"
outFolder="$HOME/folderOut"
#loading functions.sh
. /home/ec2-user/functions.sh
export lsValidLocal="lsValid $inFolder"
echo $inFolder
#execution
$lsValidLocal | xargs -n 1 -I {} bash -c 'Push "{}" "$inFolder"'
functions.sh
function Push() {
local FILE=$1
local DEST=$2
scp $FILE $otherachine:$DEST &&
rm $FILE ${FILE}_0 &&
ssh $otherachine "touch ${FILE}_0"
}
function lsValid() { #from directory
local DIR=$1
ls $DIR/*_0 | sed 's/.\{2\}$//'
}
export -f Push
export -f Pull
export -f lsValid
The problem with the code you have written is that $inFolder is inside single quotes (') which will prevent it being expanded.
$lsValidLocal | xargs -n 1 -I {} bash -c 'Push "{}" "**$inFolder**"'
This will be executed as three separate layers of processes
bash <your scrpit>
|
\xargs ...
|
\bash -c Push ...
Your code is not transferring the value across from the outer shell to inner shell... But you are expanding the variable inFolder using the inner shell. As you correctly point out it can be done with an exported environment variable.
The alternative is to have the outer shell expand it before passing to xargs.
$lsValidLocal | xargs -n 1 -I {} bash -c "Push '{}' '**$inFolder**'"
Notice I have reversed ' and " to allow $inFolder to be expanded before xargs is called.
I am trying to create a script that opens automatically any files containing a particular pattern.
This is what I achieved so far:
xargs -d " " vim < "$(grep --color -r test * | cut -d ':' -f 1 | uniq | sed ':a;N;$!ba;s/\n/ /g')"
The problem is that vim does not recognize the command as separate file of list, but as a whole filename instead:
zsh: file name too long: ..............
Is there an easy way to achieve it? What am I missing?
The usual way to call xargs is just to pass the arguments with newlines via a pipe:
grep -Rl test * | xargs vim
Note that I'm also passing the -l argument to grep to list the files that contain my pattern.
Use this:
vim -- `grep -rIl test *`
-I skip matching in binary files
-l print file name at first match
Try to omit xargs, becouse this leads to incorrect behaviour of vim:
Vim: Warning: Input is not from a terminal
What I usually do is append the following line to a list of files:
> ~/.files.txt && vim $(cat ~/.files.txt | tr "\n" " ")
For example :
grep --color -r test * > ~/.files.txt && vim $(cat ~/.files.txt | tr "\n" " ")
I have the following in my .bashrc to bind VV (twice V in uppercase) to insert that automatically :
insertinreadline() {
READLINE_LINE=${READLINE_LINE:0:$READLINE_POINT}$1${READLINE_LINE:$READLINE_POINT}
READLINE_POINT=`expr $READLINE_POINT + ${#1}`
}
bind -x '"VV": insertinreadline " > ~/.files.txt && vim \$(cat ~/.files.txt | tr \"\\n\" \" \")"'