如何用shell脚本替换html文件中的多行文本？

Question

I have an html file in which I want to replace a style using a shell script using the function sed . I want to replace:

    pre {
      font-size: inherit;
      line-height: inherit;
    }

with:

    pre {
      font-size: 18px;
      /*line-height: inherit;*/
    }

I tried the following:

    sed -i 's/pre {\n   font-size: inherit;\n   line-height: 
    inherit;\n}/pre {\n font-size: 18px;\n  \/\*line-height: 
    inherit;\*\/\n}/g' /path/file.html

which runs without any error but does not replace the text. Note that I'm including tabs to fine-tune the text pattern. I'm working on ubuntu 18.

Any suggestion will be very grateful.

Answer 1

You can use matched pattern referencing in sed like:

sed -i 's/line-height: inherit;/\/*&*\//g' /path/file.html

• Here, we use line-height: inherit; to match the search pattern. You could also extend this as a regex instead

• & is used to reference the matched text and we put the \\/* and *\\/ as the enclosing characters

Answer 2

sed -Ei '
  /^\s*pre\s*\{\s*$/,/^\s*}\s*$/ {
     /^\s*line-height:\s*inherit\s*;\s*$/ s#^(.*)$#/*&*/#;
  }
' /path/file.html

The first pair of patterns establishes a range between which the code block inside the curlies will be executed - from pre { to } .

Inside the block, the pattern establishes which lines will be edited by the substitution, so that it only comments the line-height directives.

When matched, it comments the whole line (preserving the whitespace).

PS -
don't use -iE, that fails.
-Ei works. ;)
PH

Answer 3

Based on @PaulHodges answer this works perfectly:

    sed -Ei '/^\s*pre\s*\{\s*$/,/^\s*}\s*$/ {/^\s*font-size:\s*inherit\s*;\s*$/ s#^(.*)$#    
    font-size: 18px;#; /^\s*line-height:\s*inherit\s*;\s*$/ s#^(.*)$#/*&*/#;}' /path/file.html

Thank you all for your answers!