Chrome擴展：鏈接收集

Question

popup.js：

function GetUrls() 
    { 
    var fourmTabs = new Array(); 
    chrome.tabs.query({}, function (tabs) { 
     for (var i = 0; i < tabs.length; i++) { 
      fourmTabs[i] = tabs[i]; 
     } 
     for (var i = 0; i < fourmTabs.length; i++) { 
      if (fourmTabs[i] != null) 
      { 
      document.write("<b>" + fourmTabs[i].title + "</b>" + "<br/><a href='" + fourmTabs[i].url + "'>" + fourmTabs[i].url + "</a><br/><br/>"); 
      } 
      else { 
       document.write("??" + i); 
      } 
     } 
    }); 
    } 
    window.addEventListener("DOMContentLoaded", GetUrls()); 

    function OpenInNewTab(url) 
    { 
    var win=window.open(url, '_blank'); 
     win.focus(); 
    } 
    //document.addEventListener('??', OpenInNewTab(??)); 

現在新標籤一步我..How可以調用OpenNewTab功能與處於GetUrls函數的局部變量的網址？或者有沒有更好的想法？

Answer 1

我不確定你想要什麼。在任何情況下，getSelected被棄用，取而代之的query，所以你應該使用：

window.addEventListener("DOMContentLoaded", function() { 
    chrome.tabs.query({ 
     // All tabs in current window 
     currentWindow: true 
    }, 
    function(tabs) { 
     var olTag = document.createElement("ol"); 
     for (var i = 0, len = tabs.length; i < len; i++) { 
      // Create a list-item for each tab 
      var liTag = document.createElement("li"); 
      liTag.appendChild(document.createTextNode(tabs[i].url)); 
      olTag.appendChild(liTag); 
     } 
     document.body.appendChild(olTag); 
    }); 
}); 

這個代碼列出了所有標籤的網址，你有開放的，就像這樣：

These are links you open: 

    1. http://stackoverflow.com/posts/20557596/edit 
    2. chrome://extensions/