如何在「發佈行動」中插入當前網址Facebook JavaScript功能

Question

我可以將以下操作發佈到Facebook。我現在想要做的是自動將當前頁面URL插入到MYURL中。然後，我可以將此代碼放入外部文件中，而不必手動插入網址並將其包含在每個網頁上。我是JavaScript新手（我一起花了很長時間纔看到下面的代碼）。我已經搜查了這個死亡，並找不到辦法做到這一點。我會非常感謝任何幫助。

<script type="text/javascript"> 

function postRead() 
    { 

FB.api('/me/namespace:action&object=http://MYURL','post', function(response) { 
      if (!response || response.error) { 
     FB.login(function(response) { 
    if (response.authResponse) { 
    console.log('Welcome! Fetching your information.... '); 
    FB.api('/me', function(response) { 
     console.log('Good to see you, ' + response.name + '.'); 
    }); 

FB.api('/me/namespace:action&object=http://MYURL','post', function(response) { 
      if (!response || response.error) { 
       alert('Error occured'); 
       } else { 
       document.getElementById("postlink").innerHTML="Posted!"; 

       } 
     }); 

    } else { 
    console.log('User cancelled login or did not fully authorize.'); 
    } 
}); 
       } else { 
       document.getElementById("postlink").innerHTML="Posted!"; 

       } 
     }); 
    } 


</script> 

Answer 1

我解決了這個問題 - >strip characters from url on javascript 'onclick'-command。不幸的是我不能投票，因爲我沒有名譽，這是一個恥辱。

答案代碼是雙星，這是意味着膽子很抱歉。

function postRead() 
    { 
FB.api('/me/namespace:action&object='**+encodeURIComponent(location.href)**,'post', function(response) { 
      if (!response || response.error) { 
     FB.login(function(response) { 
    if (response.authResponse) { 
    console.log('Welcome! Fetching your information.... '); 
    FB.api('/me', function(response) { 
     console.log('Good to see you, ' + response.name + '.'); 
    }); 

FB.api('/me/namespace:action&object='+encodeURIComponent(location.href),'post', function(response) { 
      if (!response || response.error) { 
       alert('Error occured'); 
       } else { 
       document.getElementById("postlink").innerHTML="Posted!"; 

       } 
     }); 

    } else { 
    console.log('User cancelled login or did not fully authorize.'); 
    } 
}); 
       } else { 
       document.getElementById("postlink").innerHTML="Posted!"; 

       } 
     }); 
    }