Javascript正则表达式解析点和空格

Question

In Javascript I have several words separated by either a dot or one ore more whitepaces (or the end of the string).
I'd like to replace certain parts of it to insert custom information at the appropriate places.

Example:

var x = "test1.test     test2 test3.xyz test4";

If there's a dot it should be replaced with ".X_"
If there's one or more space(s) and the word before does not contain a dot, replace with ".X "

So the desired output for the above example would be:

"test1.X_test test2.X test3.X_xyz test4.X"

Can I do this in one regex replace? If so, how?
If I need two or more what would they be?

Thanks a bunch.

Answer 1

To answer this:

If there's a dot it should be replaced with ".X_"

If there's one or more spaces it should be replaced with ".X"

Do this:

x.replace(/\./g, '.X_').replace(/\s+/g, '.X');

Edit: To get your desired output (rather than your rules), you can do this:

var words = x.replace(/\s+/g, ' ').split(' ');
for (var i = 0, l = words.length; i < l; i++) {
    if (words[i].indexOf('.') === -1) {
        words[i] += ".X";
    }
    else {
        words[i] = words[i].replace(/\./g, '.X_');
    }
}
x = words.join(' ');

Basically...

1. Strip all multiple spaces and create an array of "words"
2. Loop through each word.
3. If it doesn't have a period in it, then add ".X" to the end of the word
4. Else, replace the periods with ".X_"
5. Join the "words" back into a string and separate it by spaces.

Edit 2 :

Here's a solution using only javascript's replace function:

x.replace(/\s+/g, ' ')  // replace multiple spaces with one space
 .replace(/\./g, '.X_') // replace dots with .X_
 // find words without dots and add a ".X" to the end
 .replace(/(^|\s)([^\s\.]+)($|\s)/g, "$1$2.X$3");

Answer 2

Try this:

var str = 'test1.test     test2 test3.xyz test4';

str = str.replace(/(\w+)\.(\w+)/g, '$1.X_$2');
str = str.replace(/( |^)(\w+)( |$)/g, '$1$2.X$3');

console.log(str);

In the first replace it replaces the dot in the dotted words with a .X_ , where a dotted word is two words with a dot between them .

In the second replace it adds .X to words that have no dot, where words that have no dot are words that are preceded by a space OR the start of the string and are followed by a space OR the end of the string .