Remove quotes from subshell call in bash script

Always no quotes for $() statement.

We don't need quotes to hold blanks in result:
 # i=$(echo 1 2 3)
 # echo $i
 1 2 3
 #

These quotes can make something wrong in some case:
 # i=$(echo '!')
 #
 # i="$(echo '!')"
 -bash: !: event not found
 #

No real problem for current code, only to use a better code style.

Change-Id: I5909636bdc8de3d44a305d033c8c892af446acf3
Signed-off-by: Zhao Lei <zhaolei@cn.fujitsu.com>
This commit is contained in:
Zhao Lei 2015-09-23 18:48:11 +08:00
parent 6b82210f9d
commit c01517925d

View File

@ -63,7 +63,8 @@ function config_exists() {
function inspect_interface() {
local interface=$1
local mac_addr_type="$(cat /sys/class/net/${interface}/addr_assign_type)"
local mac_addr_type
mac_addr_type=$(cat /sys/class/net/${interface}/addr_assign_type)
echo -n "Inspecting interface: $interface..."
if config_exists $interface; then
@ -72,25 +73,20 @@ function inspect_interface() {
echo "Device has generated MAC, skipping."
else
ip link set dev $interface up &>/dev/null
HAS_LINK="$(get_if_link $interface)"
TRIES=10
while [ "$HAS_LINK" == "0" -a $TRIES -gt 0 ]; do
HAS_LINK="$(get_if_link $interface)"
if [ "$HAS_LINK" == "1" ]; then
break
else
sleep 1
fi
TRIES=$(( TRIES - 1 ))
local has_link
local tries
for ((tries = 0; tries < 10; tries++)); do
has_link=$(get_if_link $interface)
[ "$has_link" == "1" ] && break
sleep 1
done
if [ "$HAS_LINK" == "1" ] ; then
if [ "$has_link" == "1" ]; then
enable_interface "$interface"
else
echo "No link detected, skipping"
fi
fi
}
if [ -n "$INTERFACE" ]; then