[英]For bash scripting
#!/bin/bash
read –p “Enter the number you seek ” NUM
for VALUE in $@; do
if [ $VALUE –eq $NUM ]; then COUNT=$((COUNT+1)); fi
done
echo $NUM appeared $COUNT times
是否可以修改上述腳本,以便:從用戶輸入兩個輸入,而不僅僅是 NUM,並計算等於或介於兩者之間的參數數量。 例如,如果用戶輸入 12 和 20,我們將有 5 個匹配項(12、18、19、12、18)。 假設第一個輸入小於第二個(即你不需要擔心用戶輸入20然后輸入12)
這就是我想出的:
#1/bin/bash
read - p "Enter two numbers ' NUM1 NUM2
for VALUE in $@; do
if [$VALUE <= $NUM1 ] && [ $VALUE => $NUM2 ]; then COUNT=$((COUNT+1))
fi
done
echo we have $COUNT numbers between $NUM1 and $NUM2, including $NUM1 and $NUM2
#1
是#!
.[
后需要一個空格。-ge
和-le
,因為>
和<
是 shell 腳本中的 I/O 重定向運算符。"
它與結束"
,而不是'
。#!/bin/bash
COUNT=0
read - p "Enter two numbers " NUM1 NUM2
for VALUE in "$@"; do
if [ "$VALUE" -ge "$NUM1" ] && [ "$VALUE" -le "$NUM2" ]; then COUNT=$((COUNT+1))
fi
done
echo "we have $COUNT numbers between $NUM1 and $NUM2, including $NUM1 and $NUM2"
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