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