结合2个bookmaklets

Question

我有2只独立的小书签

javascript: (function() { var url = document.location; var links = document.getElementsByTagName('link'); var myvar = 'none'; for (var i = 0, l; l = links[i]; i++) { if (l.getAttribute('rel') == 'next') { myvar = l.getAttribute('href'); break; } } alert(myvar); })(); 

和

javascript: (function() { var url = document.location; var links = document.getElementsByTagName('link'); var myvar = 'none'; for (var i = 0, l; l = links[i]; i++) { if (l.getAttribute('rel') == 'prev') { myvar = l.getAttribute('href'); break; } } alert(myvar); })(); 

，我想他们在结合一个所以无论是在单个书签显示出来。我希望这会做，但很明显，我塞进了

javascript: (function() { var url = document.location; var links = document.getElementsByTagName('link'); var myvar = 'none'; var myvar1 = 'none'; for (var i = 0, l; l = links[i]; i++) { if (l.getAttribute('rel') == 'next') { myvar = l.getAttribute('href'); break; } } { if (l.getAttribute('rel') == 'prev') { myvar1 = l.getAttribute('href'); break; } } alert(myvar + ' ' + myvar1); })(); 

Answer 1

试试这个：

javascript: (function() { var url = document.location; var links = document.getElementsByTagName('link'); var prev = 'none', next = 'none'; for (var i = 0, l; l = links[i]; i++) { if (l.getAttribute('rel') == 'prev') { prev = l.getAttribute('href'); } else if(l.getAttribute('rel') == 'next') { next = l.getAttribute('href'); } if(prev != 'none' && next != 'none') { break; }} alert(prev + '\n' + next); })(); 

此外，这里是一个缩小的版本：

javascript:(function(){for(var e=document.getElementsByTagName("link"),b="none",c="none",d=0,a;(a=e[d])&&!("prev"==a.getAttribute("rel")?b=a.getAttribute("href"):"next"==a.getAttribute("rel")&&(c=a.getAttribute("href")),"none"!=b&&"none"!=c);d++);alert(b+"\n"+c)})();