how to create an array

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# 1  
Old 01-18-2012
how to create an array

Hi, is there a faster way to create an array containing list of numbers? something like the ones below.
array=(01 02 03 04 05 06 07 08 09 10 11 ... 365)

thanks much.
# 2  
Old 01-18-2012
arr=$(seq 1 365)

or, if you need zeroes prepended:
arr=$(seq -f "%03.0f" 1 365)

The '1' is not really needed in this case, but illustrates that you can start the sequence at an arbitrary integer.
# 3  
Old 01-18-2012
Hi mirni, thanks for the reply. had a question though, i need to use the contents of the array to name an output file using this syntax. will it possible to do such.
arr=(01 02 03 04); n=0
for i in $list; do
    command > "1999${arr[n++]}.grd"

desired output: 199901, 199902 199903 199904

Last edited by ida1215; 01-18-2012 at 03:35 PM..
# 4  
Old 01-18-2012
arr=(01 02 03 04); n=0 
for i in $list; do
     command > "1999${arr[$((n++))]}.grd" 

See man bash and search for "arithmetic expansion".
# 5  
Old 01-18-2012
an array seems useless in this case
for (( x=1; x<=365; x++)); do printf -v maVar '%.2d' $x; command >1999$maVar; done

btw, arithmetic expansion is done inside array elements without the need of double parenthesis, ${array[n++]} should work.

Last edited by daPeach; 01-18-2012 at 11:30 PM..
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