如何停止由於超時而導致的輪詢？

Question

所以我投票的東西非常標準

(function poll(){ 
    $.ajax({ ... }) 
}); 

...它工作得很好。但是現在，我希望能夠每隔幾秒鐘繼續輪詢，如果兩分鐘後沒有得到迴應，請停止輪詢並提出錯誤。

我該如何做超時？

Answer 1

這樣的事情如何？初始化，跟蹤並重置ajax承諾中的輪詢。

var pollingTimer  = null, // stores reference to the current timer id 
    firstTimeoutResponse = null; // stores the start of what might be a series of timeout responses 

function poll(){ 
    $.ajax({ 
    // your options here... 
    }).done(function() { 
    // reset the "timeout" timer 
    firstTimeoutResponse = null; 
    }).fail(function(jqXHR, textStatus) { 
    // if the failure wasn't a timeout, short-circuit, 
    // but only after resetting the timeout timestamp 
    if (textStatus !== 'timeout') { 
     firstTimeoutResponse = null; 

     return; 
    } 

    // if it was a timeout failure, and the first one (!), init the timeout count 
    if (firstTimeoutResponse = null) { 
     firstTimeoutResponse = (new Date).getTime(); 
    } 
    }).always(function() { 
    // if 2 min have passed and we haven't gotten a good response, stop polling/chort-circuit 
    if ((new Date).getTime() - firstTimeoutResponse > 120000) { // 120000ms = 2min 
     window.clearTimeout(pollingTimer); 

     return; 
    } 

    // queue the next ajax call 
    pollingTimer = window.setTimeout(poll, 3000); // poll every 3s 
    }); 
} 

// kick things off! 
poll();