sed：替换方括号之间的字母

Question

我有以下字符串：

signal[i]
signal[bg]
output [10:0]
input [i:1]

我想要的是替换方括号之间的字母（例如通过下划线）并保留表示表声明的其他字符串：

signal[_]
signal[__]
output [10:0]
input [i:1]

感谢

Answer 1

替代 gawk 解决方案：

awk -F'\\[|\\]' '$2!~/^[0-9]+:[0-9]$/{ gsub(/./,"_",$2); $2="["$2"]" }1' OFS= file

输出：

signal[_]
signal[__]
output [10:0]

• -F'\\[|\\]' - 将[和]视为字段分隔符

• $2!~/^[0-9]+:[0-9]$/ - 如果第二个字段不代表表格声明

，则执行操作

• gsub(/./,"_",$2) - 用_

替换每个字符

Answer 2

尝试：

awk '{gsub(/\[[a-zA-Z]+\]/,"[_]")} 1'  Input_file

全局替换（括号）字母直到最长匹配，然后用[_]代替。提及1将打印行（编辑或不编辑）。

编辑：上面会用一个_替换所有字母，所以要获得尽可能多的下划线，以下可能会有所帮助。

awk '{match($0,/\[[a-zA-Z]+\]/);VAL=substr($0,RSTART+1,RLENGTH-2);if(VAL){len=length(VAL);;while(i<len){q=q?q"_":"_";i++}};gsub(/\[[a-zA-Z]+\]/,"["q"]")}1'   Input_file

或

awk '{
        match($0,/\[[a-zA-Z]+\]/);
        VAL=substr($0,RSTART+1,RLENGTH-2);
        if(VAL){
                len=length(VAL);
                while(i<len){
                                q=q?q"_":"_";
                                i++
                            }
               };
        gsub(/\[[a-zA-Z]+\]/,"["q"]")
     }
     1
    '   Input_file

很快会添加解释。

EDIT2：以下是OP和用户的解释目的。

awk '{
        match($0,/\[[a-zA-Z]+\]/);            #### using match awk's built-in utility to match the [alphabets] as per OP's requirement.
        VAL=substr($0,RSTART+1,RLENGTH-2);    #### Creating a variable named VAL which has substr($0,RSTART+1,RLENGTH-2); which will have substring value, whose starting point is RSTART+1 and ending point is RLENGTH-2.
                                                   RSTART and RLENGTH are the variables out of the box which will be having values only when awk finds any match while using match.
        if(VAL){                              #### Checking if value of VAL variable is NOT NULL. Then perform following actions.
                len=length(VAL);              #### creating a variable named len which will have length of variable VAL in it.
                while(i<len){                 #### Starting a while loop which will run till the value of VAL from i(null value).
                                q=q?q"_":"_"; #### creating a variable named q whose value will be concatenated it itself with "_".
                                i++           #### incrementing the value of variable i with 1 each time.
                            }
               };
        gsub(/\[[a-zA-Z]+\]/,"["q"]")         #### Now globally substituting the value of [ alphabets ] with [ value of q(which have all underscores in it) then ].
     }
     1                                        #### Mentioning 1 will print (edited or non-edited) lines here.
    '  Input_file                             #### Mentioning the Input_file here.

Answer 3

这可能适合你（GNU sed）;

sed ':a;s/\(\[_*\)[[:alpha:]]\([[:alpha:]]*\]\)/\1_\2/;ta' file

使用任意数量的_和至少一个字母字符匹配开始和结束方括号，并用下划线替换所述字符并重复。

Answer 4

awk '{sub(/\[i\]/,"[_]")sub(/\[bg\]/,"[__]")}1' file

signal[_]
signal[__]
output [10:0]
input [i:1]

解释如下：由于括号是特殊字符，因此必须将其转义为字面处理，然后才能轻松使用sub。