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2525

answers:

4

As far as I can tell there's no simple way of retrieving a character offset from a TextRange object in Internet Explorer. The W3C Range object has a node, and the offset into the text within that node. IE seems to just have pixel offsets. There are methods to create, extend and compare ranges, so it would be possible to write an algorithm to calculate the character offset, but I feel I must be missing something.

So, what's the easiest way to calculate the character offset of the start of an Internet Explorer TextRange?

A: 

You can iterate through the body element's TextRange.text property using String.substring() to compare against the TextRange for which you want the character offset.

function charOffset(textRange, parentTextRange)
 { var parentTxt = parentTextRange.text;
   var txt       = textRange.text;
   var parentLen = parentTxt.length;

   for(int i=0; i < parentLen ; ++i) 
    { if (parentTxt.substring(i, txt.length+i) == txt) 
       { var originalPosition = textRange.getBookmark();

         //moves back one and searches backwards for same text
         textRange.moveStart("character",-1);
         var foundOther = textRange.findText(textRange.text,-parentLen,1);

         //if no others were found return offset
         if (!foundOther) return i;

         //returns to original position to try next offset
         else textRange.moveToBookmark(originalPosition);
       }
    }

   return -1;
 }

[Reference for findText()]

Mark Cidade
In this case, you're better off using bodyTxt.indexOf(txt), and this was indeed my first approach. However, if the text in the range is repeated anywhere, it'll throw this number off.To be clear, it's the offset within the node containing the selection I'm interested in, not within the whole body.
Thom
I updated my solution to accept the parent node and to check for repeated text.
Mark Cidade
+8  A: 

I use a method based on this caret position trick:

// Assume r is a range:
var offsetFromBody = Math.abs( r.moveEnd('character', -1000000) );

Since moveEnd returns the number of characters actually moved, offset should now be the offset from the start of the document. This works fine for testing primitive caret movement, but for expanded selections and for getting the exact node that holds the range anchor you'll need something more complex:

// where paramter r is a range:
function getRangeOffsetIE( r ) {
  var end = Math.abs( r.duplicate().moveEnd('character', -1000000) );
  // find the anchor element's offset
  var range = r.duplicate();
  r.collapse( false );
  var parentElm = range.parentElement();
  var children = parentElm.getElementsByTagName('*');
  for (var i = children.length - 1; i >= 0; i--) {
    range.moveToElementText( children[i] );
    if ( range.inRange(r) ) {
      parentElm = children[i];
      break;
    }
  }
  range.moveToElementText( parentElm );
  return end - Math.abs( range.moveStart('character', -1000000) );
}

This should return the correct caret text offset. Of course, if you know the target node already, or are able to provide a context, then you can skip the whole looping search mess.

Borgar
+2  A: 

I'd suggest this: IERange (http://code.google.com/p/ierange/), or just the TextRange-to-DOM Range algorithm from it.

Tim Down
+1  A: 

I used a slightly simpler solution using the offset values of a textRange:

function getIECharOffset() {
  var offset = 0;

  // get the users selection - this handles empty selections
  var userSelection = document.selection.createRange();

  // get a selection from the contents of the parent element
  var parentSelection = userSelection.parentElement().createTextRange();

  // loop - moving the parent selection on a character at a time until the offsets match
  while (!offsetEqual(parentSelection, userSelection)) {
    parentSelection.move('character');
    offset++;
  }

  // return the number of char you have moved through
  return offset;
}

function offsetEqual(arg1, arg2) {
  if (arg1.offsetLeft == arg2.offsetLeft && arg1.offsetTop == arg2.offsetTop) {
    return true;
  }
  return false;
}
Tom Berthon