Bash: Running scripts inside other script throws errors i don't get running them separately

Question

Here's my problem:

I have 4 scripts to generate diverse files. I have a source file i pass as an argument to generate the other output files. The 4 scripts are thought to be run consecutively. They have code like this:

while read line; do                                                             
    if echo "${line: -3}" | grep -q ','                                         
    then                                                                        
        if echo "${line:0:4}" | grep -q 'int'                                   
            then ...

Running them separately give me the right outputs. Calling them from a main script performs wrong, with some error messages like:

./FirstScript.sh: 35: ./FirstScript.sh: Bad substitution

Referring i.e. to the line above:

if echo "${line: -3}" | grep -q ','

My main Script to call the others is:

#!/bin/bash                                                            
sh ./FirstScript.sh $1
sh ./SecondScript.sh
sh ./ThirdScript.sh $1

All Scripts have been set to $chmod 755 *.sh

score 2 · Answer 1 · answered Jul 11 '16 at 13:40

2

What is sh on your system? Your #! line looks like you want the scripts to be interpreted with bash, but your main script is using sh , which may not be as feature-complete as bash.

answered Jul 11 '16 at 13:40

eduffy

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It is by definition not as feature-complete as `bash`. Running `bash` as `sh` will disable many `bash` extensions. – chepner Jul 11 '16 at 13:44
That is one thing I still can't differ very well. Is Bash some kind of 'evolved sh script'? In that case how can I call them as Bash Scripts. An why running them separately as 'sh ./xxxx.sh' works fine, but not calling them from another script – Joster Jul 11 '16 at 13:46
That was certainly the problem. – Joster Jul 11 '16 at 13:59

score 2 · Accepted Answer · edited May 23 '17 at 12:08

You are calling them as sh even though your shebang points to bash. If you want to call them as bash just replace sh with bash , eg

#!/bin/bash                                                            
bash ./FirstScript.sh $1
bash ./SecondScript.sh
bash ./ThirdScript.sh $1

Sh lacks some of the features bash holds, which might be causing errors. See Difference between sh and bash for more information about this.

score 0 · Answer 3 · answered Jul 11 '16 at 15:17

In fact, as I stated #!/bin/bash in the first sentence, I can do:

#!/bin/bash                                                            
./FirstScript.sh $1
./SecondScript.sh
./ThirdScript.sh $1

As it is said to run every executable with /bin/bash.

Bash: Running scripts inside other script throws errors i don't get running them separately

3 Answers3