How to show line number when executing bash script - linux

I have a test script which has a lot of commands and will generate lots of output, I use set -x or set -v and set -e, so the script would stop when error occurs. However, it's still rather difficult for me to locate which line did the execution stop in order to locate the problem.
Is there a method which can output the line number of the script before each line is executed?
Or output the line number before the command exhibition generated by set -x?
Or any method which can deal with my script line location problem would be a great help.
Thanks.

You mention that you're already using -x. The variable PS4 denotes the value is the prompt printed before the command line is echoed when the -x option is set and defaults to : followed by space.
You can change PS4 to emit the LINENO (The line number in the script or shell function currently executing).
For example, if your script reads:
$ cat script
foo=10
echo ${foo}
echo $((2 + 2))
Executing it thus would print line numbers:
$ PS4='Line ${LINENO}: ' bash -x script
Line 1: foo=10
Line 2: echo 10
10
Line 3: echo 4
4
http://wiki.bash-hackers.org/scripting/debuggingtips gives the ultimate PS4 that would output everything you will possibly need for tracing:
export PS4='+(${BASH_SOURCE}:${LINENO}): ${FUNCNAME[0]:+${FUNCNAME[0]}(): }'

In Bash, $LINENO contains the line number where the script currently executing.
If you need to know the line number where the function was called, try $BASH_LINENO. Note that this variable is an array.
For example:
#!/bin/bash
function log() {
echo "LINENO: ${LINENO}"
echo "BASH_LINENO: ${BASH_LINENO[*]}"
}
function foo() {
log "$#"
}
foo "$#"
See here for details of Bash variables.

PS4 with value $LINENO is what you need,
E.g. Following script (myScript.sh):
#!/bin/bash -xv
PS4='${LINENO}: '
echo "Hello"
echo "World"
Output would be:
./myScript.sh
+echo Hello
3 : Hello
+echo World
4 : World

Workaround for shells without LINENO
In a fairly sophisticated script I wouldn't like to see all line numbers; rather I would like to be in control of the output.
Define a function
echo_line_no () {
grep -n "$1" $0 | sed "s/echo_line_no//"
# grep the line(s) containing input $1 with line numbers
# replace the function name with nothing
} # echo_line_no
Use it with quotes like
echo_line_no "this is a simple comment with a line number"
Output is
16 "this is a simple comment with a line number"
if the number of this line in the source file is 16.
This basically answers the question How to show line number when executing bash script for users of ash or other shells without LINENO.
Anything more to add?
Sure. Why do you need this? How do you work with this? What can you do with this? Is this simple approach really sufficient or useful? Why do you want to tinker with this at all?
Want to know more? Read reflections on debugging

Simple (but powerful) solution: Place echo around the code you think that causes the problem and move the echo line by line until the messages does not appear anymore on screen - because the script has stop because of an error before.
Even more powerful solution: Install bashdb the bash debugger and debug the script line by line

If you're using $LINENO within a function, it will cache the first occurrence. Instead use ${BASH_LINENO[0]}

Related

Linux Bash Trying to write checklist progress "bar" of sorts. Replace the same line

Sorry if I am not giving you enough info, this is my first time posting here.
I am trying to make this in a bash script.
Downloading...............
"run bash commands and when they are done, replace the "Downloading..." text with the text bellow in the same line aka space."
Downloading............... DONE!
go to next line and show
Installing................
"run bash commands again and when they are done, replace the "Installing..." text with the text bellow in the same line aka space."
Installing................ DONE!
I hope you get what I mean. Thanks in advance.
I've tried:
#/bin/bash
tput sc # save cursor
printf "Something that I made up for this string"
sleep 1
tput rc;tput el # rc = restore cursor, el = erase to end of line
printf "Another message for testing"
sleep 1
tput rc;tput el
printf "Yet another one"
sleep 1
tput rc;tput el
But it doesn't make new lines, it just uses one line to show all text.
I'm assuming you pulled the tput code from somewhere, and I'm guessing that 'somewhere' also explained that tput is being used to overwrite the same line multiple times (as your script actually does).
From your description it doesn't sound like you need to overwrite any lines so using tput is the wrong solution.
If I understand your description correctly you should be able to do everything you want with some (relatively) simple printf commands, eg:
printf "Downloading .... " # no '\n' in the output so cursor remains at end of current line
# run your bash commands here
printf "DONE!\n" # append to end of current line and then add a new line (\n)
printf "Installing .... " # no '\n' in the output so cursor remains at end of current line
# run more bash commands here
printf "DONE!\n" # append to end of the current line and then add a new line (\n)
Keep in mind that if any of your 'bash commands' generate any output then the cursor will be moved (probably to a new line) thus messing up your output. Net result is that you'll need to make sure your 'bash commands' do not generate any output to stdout/stderr (alternatively, make sure all output - stdout/stderr - is redirected to files).
If your requirement is to have the 'bash commands' send output to the terminal then you may need to go back to using tput ... but that's going to depend on exactly how you want the output to appear.
NOTE: If this (above) does not meet your requirement then please update the question with more details.

Bash when i try to apped to a string its overwrite

i run the following function in (git-)bash under windows:
function config_get_container_values() {
local project_name=$1
local container_name=$2
#local container_name="gitea"
echo "###"
buildcmd="jq -r \".containers[]."
echo "$buildcmd"
buildcmd="${buildcmd}${container_name}"
echo "$buildcmd"
buildcmd="${buildcmd}foobar"
echo "$buildcmd"
echo "###"
}
The output of this is the following. Whyever, after using the variable to extend the string, he starts to overwrite $buildcmd. I tried this also with everything in one line as well with the append command (=+). Everytime the same result.
###
jq -r ".containers[].
jq -r ".containers[].gitea
foobar".containers[].gitea
###
The really strange thing is: When i enable the line local container_name="gitea" everything works as expected. The output is:
###
jq -r ".containers[].
jq -r ".containers[].gitea
jq -r ".containers[].giteafoobar
###
When i put this all into a news file, its also works as expected. So i think something goes wrong in the thousands of line before calling this function. Any idea, what could be cause of this behavior?
Regards
Dave
This is not how you should build up the command, DOS line endings aside. Use --arg to pass the name into the filter as a variable. For example,
config_get_container_values() {
local project_name=$1
local container_name=$2
jq -r --arg n "$container_name " '.containers[][$n+"foobar"]'
}
config_get_container foo gitea < some.json
If the function is invoked with
config_get_container_values proj gitea
it produces the "expected" output. If it is invoked with
config_get_container_values proj $'gitea\r'
it produces output that looks like the first output example. $'gitea\r' expands to a string that consists of 'gitea' followed by a Carriage return (CR) character.
One possible cause of the problem is that the container name (gitea) was read from a file that had Windows/DOS line endings (CR-LF). Problems like that are common. See the first question ("Check whether your script or data has DOS style end-of-line characters") in the "Before asking about problematic code" section of the Stack Overflow 'bash' Info page.

Assign full text file path to a variable and use variable as file path in sh file

I am trying to create a shell script for logs and trying to append data into a text file. I have write this sample "test.sh" code for testing:
#!/bin/sh -e
touch /home/sample.txt
SPTH = '/home/sample'.txt
echo "MY LOG FILE" >> "$SPTH"
echo "DUMP started at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
echo "DUMP finished at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
but in above code all lines are working correct except one line of code i.e.
echo "MY LOG FILE" >> "$SPTH"
It is giving error:
test.sh: line 6: : No such file or directory
I want to replace this full path of file "/home/sample.txt" to variable "$SPATH".
I am executing my shell script using
sh test.sh
What I am doing wrong.
Variable assignments in bash shell does not allow you to have spaces within. It will be actually interpreted as command with = and the subsequent keywords as arguments to the first word, which is wrong.
Change your code to
SPTH="/home/sample.txt"
That is the reason why SPTH was not assigned to the actual path you intended it to have. And you have no reason to have single-quote here and excluding the extension part. Using it fully within double-quotes is absolutely fine.
The syntax for the command line is that the first token is a command, tokens are separated by whitespace. So:
SPTH = '/home/sample'.txt
Has the command as SPTH, the second token is =, and so on. You might think this is daft, but most shells behave like this for historical reasons.
So you need to remove the whitespace:
SPTH='/home/sample'.txt

Unix: What does cat by itself do?

I saw the line data=$(cat) in a bash script (just declaring an empty variable) and am mystified as to what that could possibly do.
I read the man pages, but it doesn't have an example or explanation of this. Does this capture stdin or something? Any documentation on this?
EDIT: Specifically how the heck does doing data=$(cat) allow for it to run this hook script?
#!/bin/bash
# Runs all executable pre-commit-* hooks and exits after,
# if any of them was not successful.
#
# Based on
# http://osdir.com/ml/git/2009-01/msg00308.html
data=$(cat)
exitcodes=()
hookname=`basename $0`
# Run each hook, passing through STDIN and storing the exit code.
# We don't want to bail at the first failure, as the user might
# then bypass the hooks without knowing about additional issues.
for hook in $GIT_DIR/hooks/$hookname-*; do
test -x "$hook" || continue
echo "$data" | "$hook"
exitcodes+=($?)
done
https://github.com/henrik/dotfiles/blob/master/git_template/hooks/pre-commit
cat will catenate its input to its output.
In the context of the variable capture you posted, the effect is to assign the statement's (or containing script's) standard input to the variable.
The command substitution $(command) will return the command's output; the assignment will assign the substituted string to the variable; and in the absence of a file name argument, cat will read and print standard input.
The Git hook script you found this in captures the commit data from standard input so that it can be repeatedly piped to each hook script separately. You only get one copy of standard input, so if you need it multiple times, you need to capture it somehow. (I would use a temporary file, and quote all file name variables properly; but keeping the data in a variable is certainly okay, especially if you only expect fairly small amounts of input.)
Doing:
t#t:~# temp=$(cat)
hello how
are you?
t#t:~# echo $temp
hello how are you?
(A single Controld on the line by itself following "are you?" terminates the input.)
As manual says
cat - concatenate files and print on the standard output
Also
cat Copy standard input to standard output.
here, cat will concatenate your STDIN into a single string and assign it to variable temp.
Say your bash script script.sh is:
#!/bin/bash
data=$(cat)
Then, the following commands will store the string STR in the variable data:
echo STR | bash script.sh
bash script.sh < <(echo STR)
bash script.sh <<< STR

Variable next to ./name.sh in bash

Simple question, I know squat about bash scripts.
I've got a script test.sh and it sends a mail with some parameters of our DB while we run some stuff. I want to add the options 1, 2, 3 next to the ./test.sh so that the mail contains the current step of the process.
Example:
./test.sh 1 #>> Sends the mail with "Pre-aplication" in its subject.
PS: I know where to change the subject of the mail, but don't know how to read the variable from beside the .sh and then choose the words.
Your first command line input is simply stored in the $1 variable within the script. So you can use $1 without any explicit assignment in test.sh to read the number defined in command line. Find an example here. Note that to get the value, you should use double-quote in your script: "$1"
What should be using arguments.
Assuming you have a function set-up as such
function_name () {
echo "Parameter #1 is $1"
}
You can pass in an argument from the command line like so
sh example.sh example
Basically you can pass in any number of arguments and access each one like so..
$1 ('first argument')... $2 ('second argument')... $n('Nth argument)
You can go through this Documentation on Advanced bash-scripting guide to know more
http://www.tldp.org/LDP/abs/html/internalvariables.html#ARGLIST

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