Bash for loop with function

Question

How to make the command inside this for loop execute?

#!/bin/bash

for i in {8..12}
do
    printf "curl \"http://myurl?hour=20140131%02s\" -o file%02s.txt\n" "$i" "$i"
done

the current version of my script prints out the following lines:

curl "http://myurl?hour=2014013108" -o file08.txt
curl "http://myurl?hour=2014013109" -o file09.txt
curl "http://myurl?hour=2014013110" -o file10.txt
curl "http://myurl?hour=2014013111" -o file11.txt
curl "http://myurl?hour=2014013112" -o file12.txt

I'd like to change the script so that each line is executed and a file is saved at each curl request.

Not sure I understand, which double quotes? all of them? If I remove all but the escaped ones it doesn't work. — Fra
– Fra, Commented Feb 4, 2014 at 17:31

anubhava · Accepted Answer · 2014-02-04 18:29:59Z

3

No need to use eval, you can make it simple:

for i in {8..12}; do
    printf -v n %02d $i
    curl -s "http://myurl/?hour=20140131${n}" -o "file${n}.txt"
done

edited Feb 4, 2014 at 18:29

answered Feb 4, 2014 at 17:51

anubhava

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4 Comments

glenn jackman Over a year ago

The builtin printf allows you to: printf -v n %02d $i. That's faster since it does not have to spawn a subshell, but suffers in readability.

anubhava Over a year ago

Thanks @glennjackman: I often forget that -v var option in printf

anubhava Over a year ago

@chepnerL That worked for me on GNU bash, version 3.2.48(1)-release

chepner Over a year ago

There I go again, forgetting which version of bash adds various non-POSIX extensions :)

chepner · Accepted Answer · 2014-02-04 18:08:07Z

1

If you are using bash 4, there's no need for printf:

for i in {08..12}
do
    curl "http://myurl?hour=20140131$i" -o "file$i.txt"
done

answered Feb 4, 2014 at 18:08

chepner

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Comments

Малъ Скрылевъ · Accepted Answer · 2014-02-04 18:20:21Z

1

Try to evaluate the generated command:

#!/bin/bash

for i in {8..12}
do
    eval curl $(printf "'http://myurl?hour=20140131%.2i' -o file%.2i.txt" "$i" "$i")
done

Also you can use the simplified code:

for i in $(seq -f "%02g" 08 12)
do
    curl "http://myurl?hour=20140131${i}" -o file${i}.txt
done

edited Feb 4, 2014 at 18:20

answered Feb 4, 2014 at 17:40

Малъ Скрылевъ

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1 Comment

chepner Over a year ago

I strongly recommend against this. While it would work, there are safer alternatives that don't involve eval.

michael501 · Accepted Answer · 2014-02-04 19:39:09Z

0

just add another line without printf

#!/bin/bash

for i in {8..12}
do
   printf "curl \"http://myurl?hour=20140131%02s\" -o file%02s.txt\n" 
   x=$(printf "%02d" "$i")
   curl "http://myurl?hour=20140131$x" -o file$x.txt 
done

edited Feb 4, 2014 at 19:39

answered Feb 4, 2014 at 17:44

michael501

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2 Comments

chepner Over a year ago

The point of the printf is to force a leading zero for single digit values of i.

michael501 Over a year ago

@chepner ok, i made the changes , %02d adds zero and not "%02s"

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