我試圖弄清楚如何解決此代碼...echo -n "Enter you complaint : ";read complaint;if [[ -z $complaint ]] || [[ $complaint != 'FIRE' ]] || [[ $complaint != 'INTOXICATION' ]] || [[ $complaint != "INJURY" ]] || [[ $complaint != "BREAKAGE" ]] ;then echo "Not a good option"; exit 1;ficase $complaint in FIRE) = $complaint=$(echo "FIRE");; INTOXICATION) = $complaint=$(echo "INTOXICATION");; INJURY) = $complaint=$(echo "INJURY");; BREAKAGE) = $complaint=$(echo "BREAKAGE");;esac是否可能知道：為什么即使我鍵入有效的內(nèi)容也總是輸入？
1 回答
MM們
TA貢獻1886條經(jīng)驗 獲得超2個贊
中的條件if有誤，應(yīng)為：
if [[ -z $complaint ]] || [[ ( $complaint != 'FIRE' && $complaint != 'INTOXICATION' && $complaint != "BRIS" && $complaint != "BOBO" ) ]] ;then
也就是說，您想進入ifif$complaint是null（-z $complaint）或與任何有效選項都不匹配的情況，因此它必須同時與所有選項都不同，因此應(yīng)&&改為||。
或者，您也可以在case...esac塊中使用默認條件：
case $complaint in
FIRE) = $complaint=$(echo "FIRE");;
INTOXICATION) = $complaint=$(echo "INTOXICATION");;
INJURY) = $complaint=$(echo "INJURY");;
BREAKAGE) = $complaint=$(echo "BREAKAGE");;
*) echo "Not a good option"; exit 1;;
esac
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