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bash question
- From: Andrew DeFaria <ADeFaria at Salira dot com>
- To: cygwin at cygwin dot com
- Date: Thu, 16 May 2002 16:38:53 -0700
- Subject: bash question
- Newsgroups: gmane.os.cygwin
- Reply-to: cygwin at cygwin dot com
I may have asked this before but I'd really like to understand this and
get a fix for it. Given the following script:
#!/bin/bash
declare -i i=0
for x in 1 2 3; do
let i=i+1
echo "item $x"
done
echo "Processed $i items"
cat > /tmp/file <<END
item 1
item 2
item 3
END
declare -i i=0
cat /tmp/file | while read item; do
let i=i+1
echo "Read $item"
done
echo "Processed $i items"
rm -f /tmp/file
I get the following output:
$ test.sh
item 1
item 2
item 3
Processed 3 items
Read item 1
Read item 2
Read item 3
Processed 0 items
The quesiton is why is the integer "i" reporting 0 items processed for
the second loop? I know that it has to do with the fact that a pipe is
involved. However, how can I code this so that "i" survives?
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