如何在不调用函数的情况下使用 Observable.If

Question

目前，每个 IF 都会调用函数 isEven 和 isOdd。有没有一种方法只有当IF的计算对应于函数的逻辑路径时才调用函数？

示例：JSBin 的引用：http://jsbin.com/wegesaweti/1/edit?html,js,output)

var isEven = function(x) {
  console.log('function isEven called')
  return Rx.Observable.return(x + ' is even');
};    
var isOdd = function(x) {
  console.log('function isOdd called')
  return Rx.Observable.return(x + ' is odd');
};
var source = Rx.Observable.range(1,4)
  .flatMap((x) => Rx.Observable.if(
    function() { return x % 2; },
    isOdd(x),
    isEven(x)
));    
var subscription = source.subscribe(
    function (x) {
        console.log('Next: ' + x);
    });

当前输出：

function isOdd called
function isEven called
Next: 1 is odd
function isOdd called
function isEven called
Next: 2 is even
function isOdd called
function isEven called
Next: 3 is odd
function isOdd called
function isEven called
Next: 4 is even

预期产出

function isOdd called
Next: 1 is odd
function isEven called
Next: 2 is even
function isOdd called
Next: 3 is odd
function isEven called
Next: 4 is even

谢谢！

Answer 1

基于 RXJS docuemntation，thenSource 和 elseSource 需要 Observable（或 Scheduler elseSource）。

您的问题的替代方案：

var source = Rx.Observable.range(1,4).flatMap((x) => 
    (x % 2) == 1 ? isOdd(x) : isEven(x)
);

工作示例：http://jsbin.com/godubemano/1/edit?html,js,output

如何在不调用函数的情况下使用 Observable.If

How to use Observable.If without calling functions

rxjs5