How can Input variables into bash commands to be run inline? Debug bash ctrlZ? [duplicate]

Question

UPDATE: I learned input variables can be accessed with commands like xargs getopts() and using $@ $% and $# as well as !* for the output of last command

To clarify the question when I run a command like

rep="replace" && echo foobar | sed -e 's/.*/${rep}/'

the text is not replaced with variable contents I tried wrapping in brackets quotes and percent signs. I also tried seds insert command.

Any there any methods to debug or step thru Bash? I noticed pressing ctrl Z rather than ctrl C lets me see return values.

For those who like to say "duplicate" rather than answer; These posts didnt help me:

Replace string variable with string variable using Sed

"sed" special characters handling

Other notes: //WARNING// in this command I set an alias that runs itself is that bad practice? I think I was trying codegolf code and ending up making directories with [[ in my $home directory on live disc so careful on your install.

alias a="espeak $1" && b=$($a) //the b part runs it and can be omited

for example this works fine after the alias of a seq 11 -1 0 |a

alias countDown="seq $1 -1 0|a" doesn't work with countDown 10 as an example run.

This method omits sed command: @destenson post at Escape a string for a sed replace pattern

inputvar="foobar" && txt2replace="oo" && txt2replacewith="rench" && outputvar="${inputvar//"$txt2replace"/"$txt2replacewith"}" && echo $outputvar && echo $outputvar|a

helped

does the string $rep contain escape characters how do I debug and see them? $rep|hd? shows me an extra '.' char but so does echo "replace"|hd — openInvent
– openInvent, Commented Nov 9, 2018 at 2:46
Using an alias is bad practice whether or not it is recursive. — William Pursell
– William Pursell, Commented Nov 9, 2018 at 2:53
Pressing ctrl-Z does something very different than ctrl-C, so the behavior is different. One suspends the process (by sending SIGTSTP), the other kills it (by sending SIGINT). You seem surprised that they do different things. — William Pursell
– William Pursell, Commented Nov 9, 2018 at 2:55
One question per question, please. I added a bunch of duplicates but you might want to create a new question which asks a single question if you need more guidance on the debugging topic in particular. — tripleee
– tripleee, Commented Nov 9, 2018 at 5:21

perreal · Accepted Answer · 2018-11-09 02:52:00Z

0

Put the variable out of the single quotes:

rep="replace" && echo foobar | sed -e 's/.*/'${rep}'/'

Double quotes are fine:

rep="replace" && echo foobar | sed -e 's/.*/'"${rep}"'/'

answered Nov 9, 2018 at 2:52

perreal

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4 Comments

openInvent Over a year ago

That worked but is confusing. so what happens with $rep or %$rep% and ${rep} how can I debug bash?

openInvent Over a year ago

$(rep="echo thxM8" && echo foobar | sed -e 's/.*/'"${rep}"'/')|espeak

perreal Over a year ago

You can debug it like this: rep="replace" && echo "echo foobar | sed -e 's/.*/'"${rep}"/"

openInvent Over a year ago

I meant is there a way to debug bash like by alias [="test && $(echo ${1})"

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