JavaScript不按邏輯工作

Question

我一直在努力尋找最後一整晚找到錯誤，但找不到。我的JavaScript代碼不能正常工作。

var currCab = 0; 
var cabBooked = 0; 

function sendMessage(cabId) 
{ 
    if(cabBooked == 1) { 
     return; 
    } 

    // get data from html page 

    $.post("sendMessage.php",data, 
      function(data,status){     
       $("#status").html(data); 
      }); 
} 

function checkBookingStatus() 
{ 
    var data = "bookingId="+$("#bookingId").val(); 
    $.post("checkBookingStatus.php",data, 
      function(data,status){ 
       // callback here 
       var ans = data.indexOf("true"); 
       if(ans!= -1) { 
        console.log("returning true from checkBookingStatus"); 
        $("#simpleStatus").html("Cab booked. You can close tab now"); 
        cabBooked = 1; 
        return true; 
       } else { 
        console.log("returning false from checkBookingStatus"); 
        return false; 
       } 
      }); 
} 


function tryCabAlloc() 
{ 
    if(!checkBookingStatus()) { 
     if(cabBooked == 1) { 
      return; 
     } 
     var checkid = document.getElementById("cabId"+currCab); 
     if(checkid==null) { 
      $("#simpleStatus").html("Cab could not be booked."); 
      $("#status").html("No more clients to communicate with."); 
     } else { 
      var cid = $("#cabId"+currCab).val(); 
      sendMessage(cid); 
      updateTrying(cid); 
      currCab = currCab + 1; 
     } 
    } 
} 

$(document).ready(function() { 
    $("#simpleStatus").html("Please wait. The communication is being started..."); 
    $("#status").html("GCM push initiating ..."); 

    tryCabAlloc(); 

    setInterval(function(){ 
     if(cabBooked==0) { 
      tryCabAlloc(); 
     } else { 
      throw ''; 
     } 
    },95000); 

}); 

的代碼應該停止時，checkBookingStatus返回true，cabBooked設置爲1，但不停止所有工作。我不知道爲什麼。

Answer 1

if(!checkBookingStatus()) {將繼續如果它是假的，而不是如果它是真實的你的評論。