匹配后删除字符串以及之后的 word/string

Question

我有一个包含以下模式行的文件。

date=2020-02-22 time=13:32:41 type=text subtype=text ip=1.2.3.4 country="China" service="foo"  id=47291 msg="foo: bar.baz," value=50
date=2020-03-17 time=11:49:54 type=text subtype=anothertext ip=1.2.3.5 country="Russian Federation" service="bar"  id=47324 msg="foo: bar.baz," value=30
date=2020-03-30 time=16:29:24 type=text subtype=someothertext ip=1.2.3.6 country="Korea, Republic of" service="grault, garply"  id=47448 msg="foo: bar.baz," value=60

我想删除类型、子类型和服务以及这些字段的值（= 之后的值）。

期望的输出：

date=2020-02-22 time=13:32:41 ip=1.2.3.4 country="China" id=47291 msg="foo: bar.baz," value=50
date=2020-03-17 time=11:49:54 ip=1.2.3.5 country="Russian Federation" id=47324 msg="foo: bar.baz," value=30
date=2020-03-30 time=16:29:24 ip=1.2.3.6 country="Korea, Republic of" id=47448 msg="foo: bar.baz," value=60

我一直在尝试使用 cut、awk、sed，但仍未接近解决方案。我已经在网上搜索了几个小时，但那也是徒劳的。有人可以帮忙吗？

Answer 1

你可以试试这样：

awk -F " " '{ =""; =""; ="";  print}' file

您基本上是将列设置为空字符串。

Answer 2

您可以使用这个 sed:

sed -E 's/(^|[[:blank:]]+)(subtype|type|service)=[^[:blank:]]+//g' file

date=2020-02-22 time=13:32:41 ip=1.2.3.4 country="China"  id=47291 msg="foo: bar.baz," value=50
date=2020-03-17 time=11:49:54 ip=1.2.3.5 country="Russian Federation"  id=47324 msg="foo: bar.baz," value=30
date=2020-03-30 time=16:29:24 ip=1.2.3.6 country="Korea, Republic of" garply"  id=47448 msg="foo: bar.baz," value=60

Answer 3

您以后可能想重用或构建的内容：

$ cat tst.awk
BEGIN {
    split(s,tmp)
    for (i in tmp) {
        skip[tmp[i]]
    }
    FPAT = "[^ ]+(=\"[^\"]+\")?"
}
{
    c=0
    for (i=1; i<=NF; i++) {
        tag = gensub(/=.*/,"",1,$i)
        if ( !(tag in skip) ) {
            printf "%s%s", (c++ ? OFS : ""), $i
        }
    }
    print ""
}

$ awk -v s='type subtype service' -f tst.awk file
date=2020-02-22 time=13:32:41 ip=1.2.3.4 country="China" id=47291 msg="foo: bar.baz," value=50
date=2020-03-17 time=11:49:54 ip=1.2.3.5 country="Russian Federation" id=47324 msg="foo: bar.baz," value=30
date=2020-03-30 time=16:29:24 ip=1.2.3.6 country="Korea, Republic of" id=47448 msg="foo: bar.baz," value=60

以上使用 GNU awk 进行 FPAT 和 gensub()。

匹配后删除字符串以及之后的 word/string

Remove string after match along with the word/string after that

bash

awk

sed

cut