如果某个字符出现在另一个字符之前，则替换该字符的所有实例

Question 1

和sed：

sed -e :1 -e 's/^\([^=]*\)\./\1_/;t1'

即替换.行首以外的字符序列，后跟.相同的序列和_，并重复该过程，直到不再匹配。

和awk：

awk -F = -v OFS== '{gsub(/\./, "_", $1); print}'

现在，如果右侧包含=shell 特有的字符（\“$&();'#~<>...`、空格、制表符、其他空格...），您可能需要引用它:

sed "s/'/'\\\\''/g;:1"'
     s/^\([^=]*\)\./\1_/;t1'"
     s/=/='/;s/\$/'/"

或者：

awk -F = -v q="'" -v OFS== '
   {gsub(q, q "\\" q q)
    gsub(/\./, "_", $1)
    $2 = q $2
    print $0 q}'

Answer

和sed：

sed -e :1 -e 's/^\([^=]*\)\./\1_/;t1'

即替换.行首以外的字符序列，后跟.相同的序列和_，并重复该过程，直到不再匹配。

和awk：

awk -F = -v OFS== '{gsub(/\./, "_", $1); print}'

现在，如果右侧包含=shell 特有的字符（\“$&();'#~<>...`、空格、制表符、其他空格...），您可能需要引用它:

sed "s/'/'\\\\''/g;:1"'
     s/^\([^=]*\)\./\1_/;t1'"
     s/=/='/;s/\$/'/"

或者：

awk -F = -v q="'" -v OFS== '
   {gsub(q, q "\\" q q)
    gsub(/\./, "_", $1)
    $2 = q $2
    print $0 q}'

Question 2

使用bash：

while IFS='=' read -r i j; do echo "${i//./_}=$j" ; done

我们使用参数扩展模式将变量名称中的所有s${i//./_}替换为 s 。._

例子：

$ cat file.txt 
my.variable.var1=a-long-ling.with.lot_of_different:characters:and_numbers
my.variable.another.var2=another-long-ling.with.lot_of_different:characters:and_numbers
my.variable.number3=yet_another_long-ling.with.lot_of_different:characters:and_numbers

$ while IFS='=' read -r i j; do echo "${i//./_}=$j" ; done <file.txt 
my_variable_var1=a-long-ling.with.lot_of_different:characters:and_numbers
my_variable_another_var2=another-long-ling.with.lot_of_different:characters:and_numbers
my_variable_number3=yet_another_long-ling.with.lot_of_different:characters:and_numbers

Answer

使用bash：

while IFS='=' read -r i j; do echo "${i//./_}=$j" ; done

我们使用参数扩展模式将变量名称中的所有s${i//./_}替换为 s 。._

例子：

$ cat file.txt 
my.variable.var1=a-long-ling.with.lot_of_different:characters:and_numbers
my.variable.another.var2=another-long-ling.with.lot_of_different:characters:and_numbers
my.variable.number3=yet_another_long-ling.with.lot_of_different:characters:and_numbers

$ while IFS='=' read -r i j; do echo "${i//./_}=$j" ; done <file.txt 
my_variable_var1=a-long-ling.with.lot_of_different:characters:and_numbers
my_variable_another_var2=another-long-ling.with.lot_of_different:characters:and_numbers
my_variable_number3=yet_another_long-ling.with.lot_of_different:characters:and_numbers

Question 3

这是另一个sed：

sed 'h;s/\./_/g;G;s/=.*=/=/'

这个只进行两次替换，无论=so 前面有多少个点，输入如下：

my.var.an.other.var.with.many.dots=line.with.many:chars:and_numbers.and.stuff

结果是

my_var_an_other_var_with_many_dots=line.with.many:chars:and_numbers.and.stuff

当每行只有一个=字符时（如示例输入中所示），此方法可以正常工作。一种更通用的方法，即使每行有多个字符，也
始终仅替换第一个字符=（并且仅当该行至少包含一个时）：==

sed '/=/{                   # if line matches =
h                           # copy pattern space over the hold space
s/\./_/g                    # replace all . with =
G                           # append hold space content to pattern space
s/=.*\n[^=]*=/=/            # replace from the first = up to the first = after
}'                          # the newline character with a single =

或者

sed -e '/=/{h;s/\./_/g;G;s/=.*\n[^=]*=/=/' -e '}'

所以输入如下：

my.var.with.many.dots.but.no.equal.sign.and.stuff
my.var.with.many.dots=line.with.many:chars:numbers_and.stuff
other.var.with.many.dots=and.with.more.than=one.equal.sign=and.stuff

它输出：

my.var.with.many.dots.but.no.equal.sign.and.stuff
my_var_with_many_dots=line.with.many:chars:numbers_and.stuff
other_var_with_many_dots=and.with.more.than=one.equal.sign=and.stuff

Answer