问题

I want to make some refinement to some code from a previous question:

// the new base url
var base = ' https://www.example.co.uk/gp/wine/order?ie=UTF8&asin=';
var links  = document.getElementsByTagName('a');

for(var i = 0;i < links.length;i++){
    // check each link for the 'asin' value
    var result = /asin=([\d\w]+)/.exec(links[i].getAttribute('href'));
    if(result){
        // make a new url using the 'base' and the 'asin' value
        links[i].setAttribute('href', base+result[1]);
    }
}

Now, instead of it acting on all links, can I get it to only look at links that are from images?

Here is an HTML snippet to show what I mean:

<a href="/shop/product?ie=UTF8&amp;asin=Z00FDLN878&amp;tab=UK_Default" target="_blank"><img width="125" height="125" border="0" src="http://ecx.images-amazon.com/images/I/01W9a7gwosL.jpg" alt="43453"></a>

That's an image link - I do want it to act on that.

Impossible?

My gut instinct is that this isn't actually possible in code - because document.getElementsByTagName('a') can't see the difference between a text link and an image link.

回答1:

Use querySelectorAll to pre-select only the right kinds of nodes. EG:

// the new base url
var base        = 'https://www.example.co.uk/gp/wine/order?ie=UTF8&asin=';
var linkImgs    = document.querySelectorAll ("a > img");

for (var J = linkImgs.length - 1;  J >= 0;  --J) {
    var imgLink = linkImgs[J].parentNode;

    //--- Check each link for the 'asin' value
    var result  = /asin=([\d\w]+)/.exec (imgLink.getAttribute ('href') );
    if( result) {
        // make a new url using the 'base' and the 'asin' value
        imgLink.setAttribute ('href', base+result[1]);
    }
}

回答2:

You could use regex to check for the link inside the HTML of the link:

for(var i = 0;i < links.length;i++) {

    // check each link for the 'asin' value
    var result = /asin=([\d\w]+)/.exec(links[i].getAttribute('href'));

    // check each link for an img tag
    var hasimage = /<img [^>]+>/.test(links[i].innerHTML);

    if(result && hasimage){
        // make a new url using the 'base' and the 'asin' value
        links[i].setAttribute('href', base+result[1]);
    }

}

Also, using regular expressions to search for HTML probably isn't the best bet, but if you control what's being generated, then this is probably the quickest way without a 3rd party html parser.

回答3:

You can filter the links based on whether or not they contain an image.

var links  = document.getElementsByTagName('a');

links = [].filter.call(links, function(item) {
   // test to see if child node is an image
   return item.childNodes[0].nodeName === 'IMG'; 
});

for(var i = 0;i < links.length;i++){
    // do what you gotta do
}

回答4:

You can just test for an IMG child and only process the link if there is one there.

Example on JSFiddle

// the new base url
var base = ' https://www.example.co.uk/gp/wine/order?ie=UTF8&asin=';
var links  = document.getElementsByTagName('a');

for(var i = 0;i < links.length;i++){
    var linkElement = links[i];
    //get the first child of the a element
    var firstChild = linkElement.children[0];
    //if there is a child and it's an IMG then process this link
    if (typeof(firstChild) !== "undefined" && firstChild.tagName=="IMG") { 
      // check each link for the 'asin' value
      var result = /asin=([\d\w]+)/.exec(links[i].getAttribute('href'));
      if(result){
          // make a new url using the 'base' and the 'asin' value
          links[i].setAttribute('href', base+result[1]);
      }}
}

回答5:

// the new base url
var base = ' https://www.example.co.uk/gp/wine/order?ie=UTF8&asin=';
var links  = document.getElementsByTagName('img');
var hrefs = links.parent;

for(var i = 0;i < hrefs.length;i++){
    // check each link for the 'asin' value
    var result = /asin=([\d\w]+)/.exec(hrefs[i].getAttribute('href'));
    if(result){
        // make a new url using the 'base' and the 'asin' value
        hrefs[i].setAttribute('href', base+result[1]);
    }
}

回答6:

There is a links collection, and you can can just check if the link has an image child node:

var link, links = document.links;
var re = /asin=([\d\w]+)/;
for (var i=0, iLen=links.length; i<iLen; i++) {
  link = links[i]

  if (link.getElementsByTagName('img').length && re.test(link.href)) {
    link.href = base + result[1];
  }
}

回答7:

My initial response would be to look into query Select All and then assign a class name to grab on all of the a tags that would be affected by whatever your trying to do. When I get to my laptop I'll edit this with an example.

来源：https://stackoverflow.com/questions/20255051/how-can-i-refine-this-javascript-code-to-refine-it-so-it-only-work-on-links-from

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