write output files with default name

Question

[sata72]

I have prepared a shell script file and for the output I want to have a default name, but something is wrong. Can anybody revise this command?

calldir=/profile/variant/input/
base=$(echo $sam | sed "s/.sam.*/_sorted/g")
sam=/profile/variant/input/s5000W_b2.sam
--output $calldir/$(basename $base)_series_name.sam.gz

But in the output the file name didn't change and for all files its:

 _series_name.sam.gz 

I expected this name based on input, I want if the input change the name of output change:

s5000W_b2_series_name.sam.gz 

[graeme.thorn]

Try putting the base= command after you define the sam= variable, as $sam is empty when base is defined.

Also, the sed command won't change antthing as there is no pattern ".sam.*" to match in the same variable

[seidel]

You haven't fully explained what you'd like to achieve. You should carefully define your inputs and your outputs. On your example, you modify the value of the $sam variable, but then immediately redeclare it with a new value - which doesn't make sense. I have no idea what the second line is attempting to do. However here is a simple example that takes your $sam and $calldir variables and spits out the value you seem to expect (I think), a few lines in a script called test.sh:

#!/bin/bash
calldir=$1
sam=$2

base=$(basename $sam .sam)
output=$calldir"/"$base"_series_name.sam.gz"

execute it:

./test.sh /profile/variant/input /profile/variant/input/s5000W_b2.sam

output:

/profile/variant/input/s5000W_b2_series_name.sam.gz

[Ram]

What is the 4th line of your command? Why does it start with --output?

[sata72]

the 4th line specify the name of output

[sata72]

i know something is wrong in the line 4th, "$(basename $base) " this section should get the name of input file "s5000W_b2" and create the s5000W_b2_series_call.vcf.gz

[Ram]

Why does it start with a -- though? No commands start with a hyphen.