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连续检查值 - Bash 脚本

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连续检查值 - Bash 脚本

我想编写一个脚本，从标准输入获取其参数（每天的学习时间）并检查学生是否定期学习。定期学习是指学生连续7天每天至少学习4小时，并且这7天中至少2天每天至少学习6小时（这不必是连续的）。我试过这个：

four_h=0
six_h=0
for i in $@ ; do
        if [ $i -ge 6 ]; then let six_h=six_h+1 
        elif [ $i -ge 4 ]; then let four_h=four_h+1 
    else :
    fi
done

if [ $four_h -ge 7 ] && [ $six_h -ge 2 ];
    then echo "you are regularly studying"
else echo "you are not regularly studying"
fi

但它不起作用。我想我无法迭代，但不明白为什么。另外我不知道如何检查学生是否“连续”学习超过4小时。

答案1

听起来您必须确保满足两个条件：

  1. 任意 7 个连续值中的所有数字都必须为 4 或更大。
  2. 任意 7 个连续数字中的两个数字必须至少为 6。

因此，您似乎必须保留一个数字数组，并且需要从该数组恰好包含 7 个数字时开始检查这两个条件。

之后，您可以开始替换以循环方式读取数组中的数字，并针对每个新数字再次检查条件。

下面的bash脚本就是这样做的。正如这个出现作为一个家庭作业问题，除了代码中的注释所说之外，我不会再多说什么。考虑一下，如果在不知道代码的作用的情况下使用它作为家庭作业的答案，你会对自己造成伤害。如果代码的质量与您或您的同事所看到的不同，您也可能会因作弊而被指责。

#!/bin/bash

has_studied () {
        # Tests the integers (given as arguments) and returns true if
        # 1. All numbers are 4 or larger, and
        # 2. There are at least two numbers that are 6 or greater

        local enough large

        enough=0 large=0
        for number do
                [ "$number" -ge 4 ] && enough=$(( enough + 1 ))
                [ "$number" -ge 6 ] && large=$(( large + 1 ))
        done

        # Just some debug output:
        printf 'Number of sufficient study days: %d\n' "$enough" >&2
        printf 'Number of long study days (out of those): %d\n' "$large" >&2

        [ "$enough" -eq 7 ] && [ "$large" -ge 2 ]
}


# Array holding the read integers
ints=()

echo 'Please enter hours of study, one integer per line' >&2
echo 'End by pressing Ctrl+D' >&2

# Loop over standard input, expecting to read an integer at as time.
# No input validation is done to check whether we are actually reading
# integers.

while read integer; do
        # Add the integer to the array in a cyclic manner.
        ints[i%7]=$integer
        i=$(( i + 1 ))

        # If the array now holds 7 integers, call our function.
        # If the function returns true, continue, otherwise terminate
        # with an error.
        if [ "${#ints[@]}" -eq 7 ]; then
                if ! has_studied "${ints[@]}"; then
                        echo 'Has not studied!'
                        exit 1
                fi
        fi
done

# If the script hasn't terminated inside the loop, then all input has
# been read and the student appears to have studied enough.

echo 'Appears to have studied'

答案2

awk读取的版本stdin

awk 'BEGIN{print "Enter study hours and finish input with ctrl+d:"}
        {if ($0 >=0) {count++; print $0" hours on day "count}
        if ($0 >= 4 && $0 < 6) hrs[4]++;
        if ($0 >= 6) hrs[6]++}
    END{print "\nStudy scheme over "count" days:";
        for (d in hrs) print d" hours studied on "hrs[d]" days";
        if (hrs[4]+hrs[6] < 7 || hrs[6] < 2) not=" not";
        print "You are"not" studying regularly"
    }' -

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