I have created an alias to open an older git version of a file in sublime text.
My alias accepts 2 arguments, $1 is the revision and $2 is the file path,
but I would like to parse the arguments to get the file extension of the file to open it in its correct format.
My existing code is here:
revise=git show $1:$2 > redirected.txt $T C:/PROGRA~1/SUBLIM~2/sublime_text.exe redirected.txt
However I would like to to something like:
revise=git show $1:$2 > redirected.{$2.extesnion} $T C:/PROGRA~1/SUBLIM~2/sublime_text.exe redirected.{$2.extesnion}
environment details:
cmder version: 1.3.15.1010
in cmd.exe session you can read parameters from the command line using the $* placeholder to get everything that comes after the alias, alias example:
vi=vim $*
alias usage:
D:\
λ vi test.txt
Or you can read space-delimited argument using placeholders $1, $2, $n... alias example:
example= echo param one: $1 param two: $2
alias usage:
D:\
λ example hi how are you?
alias result:
param one: hi param two: how
note -are- and -you?- where no echoed
Related
how do i store the output of a command into a variable for use in a github-actions .yaml?
docker images --format='{{.ID}}' | select -first 1
gives me:
fc6e040841a1
i've seen stuff online about select-object..but i honestly have no idea, just trying to push an image to a registry...
the following cmd doesn't work in powershell:
for /f "delims=" %a in ("docker images --format='{{.ID}}' | select -first 1") do #set "%_img%=%a"
the following cmd doesn't work in powershell:
for /f "delims=" %a in ("docker images --format='{{.ID}}' | select -first 1") do #set "%_img%=%a"
That's because this is Command Prompt syntax. Specifically, everything outside of the () would only work under cmd.exe. The PowerShell equivalent for assigning a command result to a variable is:
$variableName = COMMAND
To apply it to your use case:
$imageId = docker images --format='{{.ID}}' | Select-Object -First 1
Note that select is an alias of Select-Object and either can be used interchangeably.
Edit: While not required for setting variables unlike in the Command Prompt, for syntax is still different in PowerShell when batch scripting. You can read up on PowerShell's for, foreach, and ForEach-Object constructs when you want to learn how they are used in PowerShell scripts, and watch for this gotcha when using the foreach "statement" as part of a pipeline.
While not part of the original scope of the question, since OP did ask and I answered in the comments, I will put the bash equivalent here for the sake of completeness and how I transposed this from the PowerShell method I used above:
imageId=$(docker images --format="{{.ID}}" | head -n 1)
This is similar to the PowerShell syntax with a few changes: remove the $ from the variable name on assignment, and Select-Object is replaced by head. You can't pad the = with whitespace, and you have to subshell the command with $().
I am trying to use grep with the pwd command.
So, if i enter pwd, it shows me something like:
/home/hrq/my-project/
But, for purposes of a script i am making, i need to use it with grep, so it only prints what is after hrq/, so i need to hide my home folder always (the /home/hrq/) excerpt, and show only what is onwards (like, in this case, only my-project).
Is it possible?
I tried something like
pwd | grep -ov 'home', since i saw that the "-v" flag would be equivalent to the NOT operator, and combine it with the "-o" only matching flag. But it didn't work.
Given:
$ pwd
/home/foo/tmp
$ echo "$PWD"
/home/foo/tmp
Depending on what it is you really want to do, either of these is probably what you really should be using rather than trying to use grep:
$ basename "$PWD"
tmp
$ echo "${PWD#/home/foo/}"
tmp
Use grep -Po 'hrq/\K.*', for example:
grep -Po 'hrq/\K.*' <<< '/home/hrq/my-project/'
my-project/
Here, grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
\K : Cause the regex engine to "keep" everything it had matched prior to the \K and not include it in the match. Specifically, ignore the preceding part of the regex when printing the match.
SEE ALSO:
grep manual
perlre - Perl regular expressions
I am trying to check whether variable GROUP exist in SAS data set file or not from the UNIX command but unfortunately it's showing that GROUP variable does not exist in the data set,However GROUP variable is present in SAS data set.
In my command for case sensitive and whole word match I am using i and w options of grep command respectively. But still UNIX command is not giving the expected result.I s there any way to fix this issue?
Below is the command which I am using:
sasfile="sasdata"
rwords="GROUP"
cat $sasfile | grep -iqw "$rwords"
Thank you
As mentioned in earlier comment
SAS data sets are stored in disk files using a proprietary format.
There may be encodings and storage methodologies that do not yield the
information you seek in a plain text examination of said disk file.
Running SAS code in a SAS session is the definitive way to glean information about a data set.
What will that code look like ?
Proc CONTENTS
Data step or macro code that uses VARNAME function
... many other ways ...
In UNIX SAS can use stdio.
From "SAS(R) 9.2 Companion for UNIX Environments", STDIO System Option: UNIX
Details
This option tells SAS to take its input from standard input (stdin),
to write its log to standard error (stderr), and to write its output
to standard output (stdout).
This option is designed for running SAS
in batch mode or from a shell script. If you specify this option
interactively, SAS starts a line mode session.
The STDIO option
overrides the DMS, DMSEXP, and EXPLORER system options. The STDIO
option does not affect the assignment of the Stdio, Stdin, and Stderr
filerefs. See Filerefs Assigned by SAS in UNIX Environments for more
information.
For example, in the following SAS command, the file
myinput is used as the source program, and files myoutput and mylog
are used for the procedure output and log respectively.
sas -stdio < myinput > myoutput 2> mylog
If you are using the C shell, you should
use parentheses:
(sas -stdio < myinput > myoutput ) >& output_log
With -stdio you want a short SAS program that can indicate if a variable is present in a data set, or perhaps emit a list of variables in a data set for further shell processing. A Proc CONTENTS step is short and sweet.
So looking for your proverbial needle in a haystack
sasfile=<path to data set file>/<dataset>.sas7bdat
needle=GROUP
echo "Proc CONTENTS data=""$sasfile""" | sas -stdio | grep $needle
The default CONTENTS output might contain yield some false matches. So you could also try
echo "Proc CONTENTS noprint data=""$sasfile"" out=list;data _null_;set list;file print;put name;"
| sas -stdio
| grep -i "GROUP"
You could try:
sasfile="sasdata"
rwords="GROUP"
grep -iw "$rwords" "$sasfile"
The only difference between these and your original commands is that I omitted cat and grep's quiet flag -q.
Sample input in sasdata:
fasd group
fdsfds fdsfdsa
fdsfd as GROUP afdsfdsa
Output:
fasd group
fdsfd as GROUP afdsfdsa
The -q flag of grep will suppress standard output but echo $? can retrieve the return value of grep. Using the same input file as before:
grep -iqw "$rwords" "$sasfile" # No stout
echo $? # Prints 0, means grep succeeded
grep -iqw "word" "$sasfile" # No stout
echo $? # Prints 1, means grep failed
I am trying to simulate a path in Xcode which has speed, latitude and longitude information.
There is a site which produces the same: http://www.bikehike.co.uk/mapview.php
I found one awk script which can convert this file to Xcode acceptable format: https://gist.github.com/scotbond/8a61cf1f4a43973e570b
Tried running this command in the terminal: awk -F script.awk bikehike_course >output.gpx
Where script.awk has the script, bikehike_course has the GPX file and output.gpx is the output file name
UPDATE
Tried: awk -f script.awk bikehike_course > output.gpx
Error: awk: syntax error at source line 1 source file adjust_gpx_to_apple_format.awk
context is
awk >>> ' <<<
awk: bailing out at source line 24
I think the syntax of the GPS file is broken.
The script on adjust_gpx_to_apple_format.awk on github is a call of awk with the awk script provided as parameter (in shell syntax).
Thus, the name adjust_gpx_to_apple_format.awk is somehow misleading.
Either the awk ' at the beginning and ' $1' at the end has to be removed. In this case,
awk -f adjust_gpx_to_apple_format.awk
should work (as the script looks like a correct awk script otherwise).
If left as is, the script might be called directly in the shell:
> ./adjust_gpx_to_apple_format.awk input.txt >output.txt
In the latter case, I would suggest two additions:
Insert a "hut" in the first line e.g. #!/bin/bash which makes it more obviously.
Rename the script to adjust_gpx_to_apple_format.sh.
Note:
Remember, that the file suffix does not have the strict meaning in Unix like shells as they have for example in MSDOS. Actually, the suffix could be anything (including nothing). It's more valuable for the user than the shell and should be chosen respectively.
I have a list of file names with environment variables in them. I'd like to read the file line by line and then set a variable to the read in line however have the envirnment variable translated to the appropriate environment variable. Here is my script so far:
#!/bin/ksh
. /test/currentEnv.sh
while read line
do
echo $line
done < $1
if my source file is:
foo1$ENVVAR1.csv
foo2$ENVVAR2.csv
foo3$ENVVAR3.csv
and my Environment variables in currentEnv.sh are
$ENVVAR1=hello; export ENVVAR1
$ENVVAR2=world; export ENVVAR2
$ENVVAR3=test; export ENVVAR3
I'd like the results of the script to be
foo1hello.csv
foo2world.csv
foo3test.csv
currently it just dumps out the original file:
foo1$ENVVAR1.csv
foo2$ENVVAR2.csv
foo3$ENVVAR3.csv
Edit
I was able to get the majority of my files resolved using:
#!/bin/ksh
. /test/currentEnv.sh
while read line
do
eval echo $line
done < $1
however some of my variables are in the middle of string like:
foo3$ENVVAR3_bar.csv
this seems to look for an env variable $ENVVAR3_bar and doesn't find it I need this to output:
foo3test_bar.csv
You declare a variable without the dollar sign:
$var=value # no
var=value # yes
Since underscore is a valid character for a variable name, ksh is trying to expand the variable named ENVVAR3_bar: you need to use braces to separate the variable name from the surrounding text:
foo3$ENVVAR3_bar.csv # no
foo3${ENVVAR3}_bar.csv # yes