计算文件中出现的次数或更高的值 [英] Counting the number of occurences or higher value in a file

问题描述

我想知道是否有一种方法可以使用AWK或SED查找某些值，这些值高于您在文件内为脚本提供的参数.这是

I was wondering if there was a way to use AWK or SED to find certain values that are higher than the argument you gave to the script inside a file. This being

./script.sh CITY VALUE

在这种情况下，值是2美元或第二个参数，它将作为我们搜索的基础，因此，如果我们有一个文件，

The value, in this case, is the $2 or second argument which is going to the basis of our search, so if we have a file,

dummy2.txt

格式:姓名； ID；专业；电子邮件；评分；访问次数；余额

Format: Name;ID;Profession;email;rating;number of visits;balance

John Trevolta;12334;dentist;gmail;0;0;431
Isabella;4567;dentist;gmail;0;0;400

如果输入，

./script.sh CITY 400

它应该能够从这个确切的dummy2.txt中读取每一列(在本例中为第七列($ 7))，并输出它发现的值大于我们给出的值的次数，因此所需的输出为:

It should be able to read every column ( the seventh in this case ($7)) from this exact dummy2.txt and output the number of times it finds a value higher than the one we gave it so the desired output would be:

1

但是我得到的当前输出如下:

But the current Output I'm given is the following:

1 John Trevolta;12334;dentist;gmail;0;0;431

因此，这是我目前拥有的代码:

So with this, this is my current code that I have for it:

#!/bin/bash
if [ "$#" -eq 0 ]
    then
        echo "Please insert arguments"
        exit 1
elif [ "$#" -lt 2 ] || [ -z "$1"  ]
    then
        echo "Error 404: No arguments found"
        exit 1
else
    grep -c "$1" dummy.txt
    awk -v  x=$2 '$7>x{ i++ }END{ print "Found:" i }' dummy2.txt
    fi

我已经在评论中添加了一些建议，但不幸的是，它仅打印出来，

I have already added some suggestions found in the comments but unfortunately, it only prints,

Found:

它不会开始计算i，甚至不会说它的价值.我不知道是否需要初始化它，是否需要在其上放置"/"/$以使其可读.

It doesn't start counting the i or even say it's value. I don't know if I need to initialize it or not and if I need to put ''/""/$ on it for it to be read.

推荐答案

由于(缺少)输入，因此假定以下输入(numbers.txt):

Because of (the lack of ) input, assume the following input (numbers.txt):

een 1
twee 2
drie 3
vier 4
vijf 5
zes 6
zeven 7

然后:

awk -v arg1 = 2'$ 2> arg1 {i ++} END {打印找到"我大于"arg1}'numbers.txt

将打印:找到大于2的5行

和:

awk -v arg1 = 5'$ 2> arg1 {i ++} END {打印找到"我大于"arg1}'numbers.txt

将打印:找到大于5的2行

(因为...)

在创建这样的脚本时(将其命名为 test.sh ):

When creating a script like this (lets name it test.sh):

#!/bin/bash

rm numbers.txt
echo een 1 >>numbers.txt
echo twee 2 >>numbers.txt
echo drie 3 >>numbers.txt
echo vier 4 >>numbers.txt
echo vijf 5 >>numbers.txt
echo zes 6 >>numbers.txt
echo zeven 7 >>numbers.txt
echo acht 8 >>numbers.txt

awk -v arg1=3 '$2>arg1{ i++ }END{ print "Found " i " lines greater than " arg1 }' numbers.txt

awk -v arg1=5 '$2>arg1{ i++ }END{ print "Found " i " lines greater than " arg1 }' numbers.txt

awk -v arg1=$1 '$2>arg1{ i++ }END{ print "Found " i " lines greater than " arg1 }' numbers.txt

输出将是:

$ ./test.sh 7
Found 5 lines greater than 3
Found 3 lines greater than 5
Found 1 lines greater than 7

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