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# 1  
Old 10-17-2007
string

I have 2007-10-03 and I want to write it as 03/10/2007
how can I do this?Smilie
# 2  
Old 10-17-2007
[code]
string="2007-10-03"
IFS="-" && echo "$string" | read year month day
string="$day/$month/$year"
echo $string
[/code
# 3  
Old 10-17-2007
Lightbulb

Quote:
Originally Posted by jim mcnamara
[code]
string="2007-10-03"
IFS="-" && echo "$string" | read year month day
string="$day/$month/$year"
echo $string
[/code
thanks but it only writes //

and I'm afraid I couldnt understand the part:

IFS="-" && echo "$string" | read year month day

I have a question can I make an array that
its first element is "2"
second element is "0"
third element is "0"
fourth element is "7"
.....
# 4  
Old 10-17-2007
Quote:
Originally Posted by yoq_bise
I have 2007-10-03 and I want to write it as 03/10/2007
how can I do this?Smilie
Code:
# echo "2007-10-03" | awk 'BEGIN{FS="-";OFS="/"}{print $3,$2,$1}'
03/10/2007

# 5  
Old 10-17-2007
thanks Smilie
# 6  
Old 10-18-2007
Some other ways of doing this

You could do this with cut this way:

Code:
#!/bin/bash

OLDSTRING="2007-10-03"
YEAR=$(echo "${OLDSTRING}" | cut -d"-" -f1)
MONTH=$(echo "${OLDSTRING}" | cut -d"-" -f2)
DAY=$(echo "${OLDSTRING}" | cut -d"-" -f3)
NEWSTRING="${DAY}/${MONTH}/${YEAR}"

Just another way. Could someone give the awk example?

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