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Obtaining Part of a Url

PanginoonPanginoon Members
edited November -1 in Sahi - Open Source
I have the following line in my for loop. Everything is work except I cannot for the life of me figure about to get part of a url. here is the line I have to get it:

record_id = _link(_getText(_cell(usertable, i, 1))).slice(35);

This of course is in a function wrapped in browser tags.

here is a sample of the full code:
<browser>
    function CopyRecords() {
           var usertable = _table(18);
           var records = new Array();
           for(var i = 0; i < usertable.rows.length; i++){
               records[i] = new Array();
          if(i = 0) {
              
          } else {
               //Record Name
               record_name = _getText(_cell(usertable, i, 1));
               //Nation ID
                record_id = _link(_getText(_cell(usertable, i, 1))).slice(35);
            
               records[i][0] = record_name;
               records[i][1] = record_id;
          }
           }
        return records;
    }
</browser>

_set($roster, CopyRecords());

for (var $i=0; $i<$roster.length; $i++){
        _debug(_cell($roster[$i][0]));
        _debug(_cell($roster[$i][1]));
}

Comments

  • PanginoonPanginoon Members
    The _link() function cannot use the link on the page as if we use pagination the next previous has additional links that throw that off.
  • narayannarayan Administrators
    Panginoon,

    _link(text) returns the DOM Element of the link. What you need is _link(text).href
    record_id = _link(_getText(_cell(usertable, i, 1))).href.slice(35);
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