bash substring and positional parameters

Question

I have to compile some mysql c api code and tired of writing all this in command line:

gcc main.c -o main `mysql_config --cflags --libs`

I wrote a shell script in bash and pass positional parameter in bash:

gcc $1 -o ${"$1":0:2} 'mysql_config --cflags --libs' but this doesn't work. I get error message: ./compile: line 2: ${"$1":0:-2}: bad substitution. Can someone please tell me what I am doing wrong?

The only way I got this to work is by assigning a new variable:

filename=$1;
gcc $filename -o ${filename:0:-2} `mysql_config --cflags --libs`

Is this the only way to do it or is there a way to fix what I am doing wrong in the first case?

Time to start learning about make... Or any other build-management tool... — twalberg
– twalberg, Commented Mar 5, 2014 at 19:27
@twalberg I thought make would only be appropriate for larger programs. This is just a small test program. thanks. — user1527227
– user1527227, Commented Mar 5, 2014 at 19:33
While it's true that it provides the most benefit for large complicated builds, it works perfectly fine for tiny single-source builds too... — twalberg
– twalberg, Commented Mar 5, 2014 at 19:35
@twalberg would you might showing me how it works in this case? — user1527227
– user1527227, Commented Mar 5, 2014 at 23:48
Posted an answer below to maintain formatting. Note the spacing at the beginning of the gcc line must be a single tab character. — twalberg
– twalberg, Commented Mar 6, 2014 at 0:48

Manny D · Accepted Answer · 2014-03-05 18:21:36Z

3

You almost had it:

${1:0:2}

You don't need another reference to $1 inside the brackets since everything in it will be interpreted as the name of the variable, as in the case of ${filename:0:-2}.

answered Mar 5, 2014 at 18:21

Manny D

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1 Comment

chepner Over a year ago

+1 Some explanation for the curious: 1 is the name of the parameter. However, the positional parameters aren't variables (i.e., parameters with valid identifiers as names), so you can't assign to it using variable assignment; 1=somestring doesn't work. To set the value of a positional parameter, you need to use the set command.

twalberg · Accepted Answer · 2014-03-06 00:48:07Z

1

In response to the comments under the question, here's an example makefile for this situation:

MSQL_FLAGS := $(shell mysql_config --cflags)
MSQL_LIBS  := $(shell mysql_config --libs)

main : main.c
        gcc $(MSQL_FLAGS) -o $@ $< $(MSQL_LIBS)

answered Mar 6, 2014 at 0:48

twalberg

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