如何避免重复以下替换代码？

Question

This is the input:

[ 'markdown',
  [ 'para', '"a paragraph"' ],
  [ 'hr' ],
  [ 'para', '\'another paragraph\'' ],
  [ 'bulletlist', [ 'listitem', '"a list item"' ] ] ]

The following code loops through each element of the array. If the elements is another array the code goes one depth further and applies replace , if not, it applies replace immediately (if I didn't have that if statement replace on an array would cause error.

  for (i = 1; i < tree.length; i++) {
    var node = tree[i]
    var x = node.length - 1
    var y = node[x].length - 1

    if (Array.isArray(node[x])) {
      node[x] = node[x][y].replace(/"(?=\b)/g, '“')
                          .replace(/"(?!\b)/g, "”")
    } else {
      node[x] = node[x].replace(/"(?=\b)/g, '“')
                       .replace(/"(?!\b)/g, "”")
    }
  }

What bothers me is the duplication with replace . How can I modify the code so I just do .replace(/"(?=\\b)/g, '“').replace(/"(?!\\b)/g, "”") once?

Answer 1

Extract logic in a function:

   function replaceMe(element) {
        return element.replace(/"(?=\b)/g, '“')
                      .replace(/"(?!\b)/g, "”");
    }


...
  for (i = 1; i < tree.length; i++) {
    var node = tree[i]
    var x = node.length - 1
    var y = node[x].length - 1

    if (Array.isArray(node[x])) {
      node[x] = replaceMe(node[x][y]);
    } else {
      node[x] = replaceMe(node[x]);
    }
  }
...

Answer 2

use iteration, valid for N-dimensional array

function doWork(tree){
  for (i = 1; i < tree.length; i++) {
    var node = tree[i]
    var x = node.length - 1
    var y = node[x].length - 1

    if (Array.isArray(node[x])) {
      doWork(node[x]);
    } else {
      node[x] = node[x].replace(/"(?=\b)/g, '“')
                       .replace(/"(?!\b)/g, "”")
    }
  }
}

// use it
doWork(tree);

Answer 3

Sometimes two simple regex it's faster than one complex.

But you can try a different approach like this:

var example_string = '"some_string"',
    regex = /"(.*?)"/g;

var result = example_string.replace(regex, "“$1”");

console.log(result); // “some_string”

```