Semaphore thinks lock file already exists
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I'm using link to create a semaphore - the idea is to lock out of writing to a db.
Here I have a script to create a table in a database:
#!/bin/bash
if [ "$#" -lt 3 ]; then
echo "Not enough parameters"
exit 1
elif [ "$#" -gt 3 ]; then
echo "Too many parameters"
exit 1
fi
if [ ! -d "$1" ]; then
echo "That database doesn't exist!"
exit 1
fi
./P.sh $1
if [ -f "$1/$2.txt" ]; then
echo "That table already exists!"
./V.sh $1
exit 1
else
touch "$1/$2.txt"
fi
./V.sh $1
echo "$3" > "$1/$2.txt"
echo "Ok, table created"
exit 0
Here's my P file:
#!/bin/bash
if [ -z "$1" ]; then
echo "Usage $0 mutex-name"
exit 1
elif [ ! -e "$1" ]; then
echo "Target for the lock must exist"
exit 2
else
while ! ln "$1" "$1.lock"; do
sleep 1
done
exit 0
fi
and my V:
#! /bin/bash
if [ -z "$1" ]; then
echo "Usage $1 mutex-name"
exit 1
else
rm "$1.lock"
exit 0
fi
let's say I create a table by running ./create_table people footballers age,height
This should create a file footballers.lock (created by P) and then once the writing has happened the V should remove it. But for some reason the P thinks that the .lock file already exists, and it definitely doesn't.
Can anyone spot what's going wrong?
bash semaphore ln
add a comment |
up vote
0
down vote
favorite
I'm using link to create a semaphore - the idea is to lock out of writing to a db.
Here I have a script to create a table in a database:
#!/bin/bash
if [ "$#" -lt 3 ]; then
echo "Not enough parameters"
exit 1
elif [ "$#" -gt 3 ]; then
echo "Too many parameters"
exit 1
fi
if [ ! -d "$1" ]; then
echo "That database doesn't exist!"
exit 1
fi
./P.sh $1
if [ -f "$1/$2.txt" ]; then
echo "That table already exists!"
./V.sh $1
exit 1
else
touch "$1/$2.txt"
fi
./V.sh $1
echo "$3" > "$1/$2.txt"
echo "Ok, table created"
exit 0
Here's my P file:
#!/bin/bash
if [ -z "$1" ]; then
echo "Usage $0 mutex-name"
exit 1
elif [ ! -e "$1" ]; then
echo "Target for the lock must exist"
exit 2
else
while ! ln "$1" "$1.lock"; do
sleep 1
done
exit 0
fi
and my V:
#! /bin/bash
if [ -z "$1" ]; then
echo "Usage $1 mutex-name"
exit 1
else
rm "$1.lock"
exit 0
fi
let's say I create a table by running ./create_table people footballers age,height
This should create a file footballers.lock (created by P) and then once the writing has happened the V should remove it. But for some reason the P thinks that the .lock file already exists, and it definitely doesn't.
Can anyone spot what's going wrong?
bash semaphore ln
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm using link to create a semaphore - the idea is to lock out of writing to a db.
Here I have a script to create a table in a database:
#!/bin/bash
if [ "$#" -lt 3 ]; then
echo "Not enough parameters"
exit 1
elif [ "$#" -gt 3 ]; then
echo "Too many parameters"
exit 1
fi
if [ ! -d "$1" ]; then
echo "That database doesn't exist!"
exit 1
fi
./P.sh $1
if [ -f "$1/$2.txt" ]; then
echo "That table already exists!"
./V.sh $1
exit 1
else
touch "$1/$2.txt"
fi
./V.sh $1
echo "$3" > "$1/$2.txt"
echo "Ok, table created"
exit 0
Here's my P file:
#!/bin/bash
if [ -z "$1" ]; then
echo "Usage $0 mutex-name"
exit 1
elif [ ! -e "$1" ]; then
echo "Target for the lock must exist"
exit 2
else
while ! ln "$1" "$1.lock"; do
sleep 1
done
exit 0
fi
and my V:
#! /bin/bash
if [ -z "$1" ]; then
echo "Usage $1 mutex-name"
exit 1
else
rm "$1.lock"
exit 0
fi
let's say I create a table by running ./create_table people footballers age,height
This should create a file footballers.lock (created by P) and then once the writing has happened the V should remove it. But for some reason the P thinks that the .lock file already exists, and it definitely doesn't.
Can anyone spot what's going wrong?
bash semaphore ln
I'm using link to create a semaphore - the idea is to lock out of writing to a db.
Here I have a script to create a table in a database:
#!/bin/bash
if [ "$#" -lt 3 ]; then
echo "Not enough parameters"
exit 1
elif [ "$#" -gt 3 ]; then
echo "Too many parameters"
exit 1
fi
if [ ! -d "$1" ]; then
echo "That database doesn't exist!"
exit 1
fi
./P.sh $1
if [ -f "$1/$2.txt" ]; then
echo "That table already exists!"
./V.sh $1
exit 1
else
touch "$1/$2.txt"
fi
./V.sh $1
echo "$3" > "$1/$2.txt"
echo "Ok, table created"
exit 0
Here's my P file:
#!/bin/bash
if [ -z "$1" ]; then
echo "Usage $0 mutex-name"
exit 1
elif [ ! -e "$1" ]; then
echo "Target for the lock must exist"
exit 2
else
while ! ln "$1" "$1.lock"; do
sleep 1
done
exit 0
fi
and my V:
#! /bin/bash
if [ -z "$1" ]; then
echo "Usage $1 mutex-name"
exit 1
else
rm "$1.lock"
exit 0
fi
let's say I create a table by running ./create_table people footballers age,height
This should create a file footballers.lock (created by P) and then once the writing has happened the V should remove it. But for some reason the P thinks that the .lock file already exists, and it definitely doesn't.
Can anyone spot what's going wrong?
bash semaphore ln
bash semaphore ln
asked Nov 22 at 15:47
Raph117
171111
171111
add a comment |
add a comment |
1 Answer
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up vote
0
down vote
Found it - you can't use ln on directories...
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Found it - you can't use ln on directories...
add a comment |
up vote
0
down vote
Found it - you can't use ln on directories...
add a comment |
up vote
0
down vote
up vote
0
down vote
Found it - you can't use ln on directories...
Found it - you can't use ln on directories...
answered Nov 22 at 16:01
Raph117
171111
171111
add a comment |
add a comment |
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