Grep variable in for loop

Question

I want to grep a specific line for each loop in a for loop. I've already looked on the internet to see an answer to my problem, I tried them but it doesn't seem to work for me... And I don't find what I'm doing wrong.

Here is the code :

for n in 2 4 6 8 10 12 14 ; do
    for U in 1 10 100 ; do
        for L in 2 4 6 8 ; do
i=0
cat results/output_iteration/occ_"$L"_"$n"_"$U"_it"$i".dat
for k in $(seq 1 1 $L) ; do
        ${'var'.$k}=`grep "   $k    " results/output_iteration/occ_"$L"_"$n"_"$U"_it"$i".dat | tail -n 1`
done

which gives me :

%
%
%  site    density        double occupancy                
    1     0.49791021     0.03866179
    2     0.49891438     0.06077808
    3     0.50426102     0.05718336
    4     0.49891438     0.06077808
./run_deviation_functionL.sh: line 109: ${'var'.$k}=`grep "   $k    " results/output_iteration/occ_"$L"_"$n"_"$U"_it"$i".dat | tail -n 1`: bad substitution

Then, I would like to take only the density number, with something like:

 ${'density'.$k}=`echo "${'var'.$k:10:10}"  | bc -l`

Anyone knows the reason why it fails?

choroba · Accepted Answer · 2015-10-16 11:35:51Z

2

Use declare to create variable names from variables:

declare density$k="`...`"

Use the variable indirection to retrieve them:

var=var$k
echo ${!var:10:10}

edited Oct 16, 2015 at 11:35

answered Oct 16, 2015 at 10:49

choroba

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3 Comments

B_runo Over a year ago

So if I understood well, I have to do declare var$k=grep " $k " file.dat | tail -n 1 ` ? But when I do this, I have this kind of error : ./run.sh: line 109: declare: 1': not a valid identifier ./run.sh: line 109: declare: 0.49791021': not a valid identifier ./run.sh: line 109: declare: `0.03866179': not a valid identifier

B_runo Over a year ago

And when I do than var=var$k, echo $var ; it gives me "var1", "var2"... And not the line I wanted to grep

choroba Over a year ago

@B_runo: Double quote the ... if spaces are involved: declare var$k="`...`".

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