Question

我有一个要替换的字符串。格式为：

  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 3.33330000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,

请注意，该字符串同时包含双引号和换行符。

我想用另一个字符串替换它：

  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 2.22110000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,

我尝试过：

export OLD_STRING1="  VAL1          = \"D_AC\" ,"
export OLD_STRING2="  VAL2          = \"DRC\" ,"
export OLD_STRING3="  VALX2         = 3.33330000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, "

export NEW_STRING1="  VAL1          = \"D_AC\" ,"
export NEW_STRING2="  VAL2          = \"DRC\" ,"
export NEW_STRING3="  VALX2         = 2.22110000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, "

然后这样称呼grep和sed：

find . -type f -name '*filename*' -exec sed -i '' s/$OLD_STRING1$OLD_STRING2$OLD_STRING3/$NEW_STRING1$NEW_STRING2$NEW_STRING3/ {} +

sed生成错误：

sed: 1: "s/BX2          = "D_AC ...": unterminated substitute pattern

我该如何解决？

Answer 1

由于sed无法理解文字字符串（请参阅Is it possible to escape regex metacharacters reliably with sed），因此尝试对sed进行健壮和可移植的工作将是一场噩梦。

在要更改file的位置输入此内容，以使old中包含的文本成为new中包含的文本：

$ cat old
  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 3.33330000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
$
$ cat new
  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 2.22110000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
$
$ cat file
foo
  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 3.33330000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
bar

和此脚本：

$ cat tst.awk
FILENAME==ARGV[1] { old=$0; next }
FILENAME==ARGV[2] { new=$0; next }
s=index($0,old) {
    $0 = substr($0,1,s-1) new substr($0,s+length(old))
}
{ print }

就这么简单：

$ awk -v RS= -f tst.awk old new file
foo
  VAL1          = "D_AC" ,
  VAL2          = "DRC" ,
  VALX2         = 2.22110000e+04, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
bar

grep和替换字符串分布在几行

1 个答案: