pass named argument to program in shell

Question

I want to pass some additional named arguments to a programe in shell script. Say I have following script:

#!/bin/bash
FILE="/root/serv er/conf/config.yml"
EXEARG="-configFile \"$FILE\""
"/root/serv er/bin/server" $EXEARG

I put EXEARG on separate line because I may change it in another process(for example jenkins build). But bash -x myscript gives me this:

+ FILE='/root/serv er/conf/config.yml'
+ EXEARG='-configFile "/root/serv er/conf/config.yml"'
+ '/root/serv er/bin/server' -configFile '"/root/serv' 'er/conf/config.yml"'

I'm expecting:
'/root/serv er/bin/server' -configFile "/root/serv er/conf/config.yml"
How should I do that?

WOW! Didn't know this excellent tool before. Problem solved. Thanks @Cyrus — sify, Jul 07 '19 at 12:19
BTW, this is also the topic of [BashFAQ #50](http://mywiki.wooledge.org/BashFAQ/050). — Charles Duffy, Jul 07 '19 at 13:00

score 0 · Answer 1 · answered Jul 07 '19 at 12:26

0

Using the tool recommended by @Cyrus, I solved it by SC2089

answered Jul 07 '19 at 12:26

sify

645
9
19

pass named argument to program in shell

1 Answers1