使用bash替换以两行之间开头的字符串

Question

有没有办法在不使用循环的情况下使用 sed 替换整个文件中字符串的开头？

例如，我的源数据如下：

str_address: 123 main street
str_desc: Apt3
str_desc: 2nd floor
str_city: new york
str_desc: mailing address

现在，该文件将有数千个地址，但我希望“str_desc”出现在“str_address”之后和“@”之前987654324@" 将替换为 "str_address"，但在 str_city 之后出现的任何 "str_desc" > 保持不变。

期望的输出：

str_address: 123 main street
str_address: Apt3
str_address: 2nd floor
str_city: new york
str_desc: mailing address

我可以提取此信息，

cat file | awk '/str_city/{f=0} f; /str_address/{f=1}'

给了

str_desc: Apt3
str_desc: 2nd floor

但我无法将第一个“str_desc”更改为“str_address”。

Answer 1

您的awk 提取代码中几乎有完整的解决方案：

awk '/str_city/{f=0} f; /str_address/{f=1}'

这个想法是：

• 看到str_address时打开标志。
• 看到str_city时关闭标志。
• 如果标志打开，请将 str_desc 替换为 str_address。

基本上是这样（可读形式，顺序很重要）：

awk '
    $1 == "str_address:"           { flag = 1 }
    $1 == "str_desc:" && flag == 1 { $1 = "str_address:" }
    $1 == "str_city:"              { flag = 0 }
                                   { print }
    ' < inputFile >outputFile

这是一个显示它在行动中的成绩单：

pax$ echo '
     str_address: 123 main street
     str_desc: Apt3
     str_desc: 2nd floor
     str_city: new york
     str_desc: mailing address
     ' | awk '
         $1 == "str_address:"           { flag = 1 }
         $1 == "str_desc:" && flag == 1 { $1 = "str_address:" }
         $1 == "str_city:"              { flag = 0 }
                                        { print }'

str_address: 123 main street
str_address: Apt3
str_address: 2nd floor
str_city: new york
str_desc: mailing address

当然还有缩小版：

awk '$1=="str_address:"{f=1}$1=="str_desc:"&&f==1{$1="str_address:"}$1=="str_city:"{f=0}{print}' < inputFile >outputFile

Answer 2

您可以在 sed 中使用地址范围：

$ sed '/str_address/,/str_city/s/str_desc/str_address/' infile
str_address: 123 main street
str_address: Apt3
str_address: 2nd floor
str_city: new york
str_desc: mailing address

这使/str_address/,/str_city/ 范围之外的所有str_desc 保持不变，并用str_address 替换其他s/str_desc/str_address/ 部分。