Today=$(date)
for i in {2..15}
do
week_{$i}=$(date -d "$Today +$i week")
echo ${week_2}
done
I am getting no values in output in BASH.
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One error easily spotted is "Today=$date" which assigns the value of the variable "date" to the variable named "Today". Try Today=$(date);echo $Today for a start.Gyro Gearloose– Gyro Gearloose2016-02-07 17:40:03 +00:00Commented Feb 7, 2016 at 17:40
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I faintly remember that a for-loop opens a sub-shell, and any variable is local to that sub-shell and no way to set those values in the parent shell.Gyro Gearloose– Gyro Gearloose2016-02-07 18:13:56 +00:00Commented Feb 7, 2016 at 18:13
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no-way except to place the values in a file (or a pipe??) and extract them afterwards.Gyro Gearloose– Gyro Gearloose2016-02-07 18:14:52 +00:00Commented Feb 7, 2016 at 18:14
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1@Starkeen it looks like there is a typo in your edit. Putting "Today=$date for i in {2..15} do" in one line gives a syntax error. Should be 2 lines or a ";" befor the "for".Gyro Gearloose– Gyro Gearloose2016-02-07 18:37:47 +00:00Commented Feb 7, 2016 at 18:37
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and another newline or semicolon before the "do"Gyro Gearloose– Gyro Gearloose2016-02-07 18:46:43 +00:00Commented Feb 7, 2016 at 18:46
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1 Answer
One way...
Today=$(date)
for i in {2..15}
do
tmp=$(date -d "$Today +$i week")
eval week_${i}=\$tmp
eval echo \$week_${i}
done
Second way...
Today=$(date)
for i in {2..15}
do
week_[$i]=$(date -d "$Today +$i week")
echo ${week_[$i]}
done
Third way... Today=$(date)
for i in {2..15}
do
eval echo \${week_${i}:=$(date -d "$Today +$i week")} > /dev/null
done
for i in {2..15}
do
eval echo \$week_${i}
done
2 Comments
Gyro Gearloose
Could you please explain a bit? I'm no rookie to bash, but I mostly just used it. Transferring variable assignments from within a loop to the outside, I could never manage, and I had some try about it. It was a while loop, but I can' believe "for" could be much different.
Roman Bronshtein
In this examples variables available after loop. See second loop in "Third way".
