Bad substitution on bash substringing operation

Question

I am trying to get this simple function working:

p4edit(){
    p4 edit ${$1:25}
}

I read the other popular bad substitution question on SO and it did not seem to help me or be related to my problem. What am I doing wrong here? I want to cut off the first 25 characters of the argument provided to my function.

I have noticed a simple echo ${"test":3} fails the same way, but this succeeds:

test="test"
echo ${test:3}

I am just running this in a bash instance.

Provide a link to the referenced question, explain what is going wrong. What does "fails the same way" mean? How do you have 13K rep and not know this? — Mad Physicist, Aug 03 '17 at 14:52
I did explain what is going wrong. You read the question I posted above and you don't know what is going wrong? — temporary_user_name, Aug 03 '17 at 14:53
You were close with `echo ${"test":3}` -- just use `echo ${test:3}` — Jeff Schaller, Aug 03 '17 at 14:57

Doktor OSwaldo · Answer 1 · 2017-08-03T15:01:35.947

2

Why two times a $ ?

p4edit(){
  echo ${1:25}
}

works fine for me. String functions in bash are a bit tricky, since they are not really consistent. But ${} already defines, that you are looking for a variable. So only submit the name to it. There are some stringfunctions with ${#var} but as far as I know, there is never a $ inside a ${}

edited Aug 03 '17 at 15:01

answered Aug 03 '17 at 14:56

Doktor OSwaldo

5,732
20
41

No, there is a case where `$` is inside. I fixed the error by adding the second `$` here: `for str in ${$(someVar)//;/ } ; do` (but a result is not that I expected). – CoolMind Mar 02 '23 at 14:47

score 2 · Accepted Answer · answered Aug 03 '17 at 14:56

2

You have too much money! (too many dollar signs). Use:

p4edit(){
    p4 edit ${1:25}
}

To extract the 25th-and-onwards characters from $1.

answered Aug 03 '17 at 14:56

Jeff Schaller

2,352
5
23
38

Bad substitution on bash substringing operation

2 Answers2