Javascript 如何使用 RxJs distinctUntilChanged？
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How to use RxJs distinctUntilChanged?
提问by Thijs Koerselman
I'm getting started with RxJs (using the v5 beta), but somehow I can't figure out how to work with distinctUntilChanged. The output from the code below if I run it in babel-node is
我开始使用 RxJs（使用 v5 测试版），但不知何故我无法弄清楚如何使用distinctUntilChanged. 如果我在 babel-node 中运行它，下面代码的输出是
[ 'a', 1 ]
{ key: 'a', state: 1 }
Next: { value: 42 }
Completed
That is not what I would expect. Why is only one entry passing distinctUntilChanged? I would expect the output to be
这不是我所期望的。为什么只有一个条目通过distinctUntilChanged？我希望输出是
[ 'a', 1 ]
[ 'a', 0 ]
[ 'a', 1 ]
{ key: 'a', state: 1 }
{ key: 'a', state: 2 }
{ key: 'a', state: 0 }
{ key: 'a', state: 1 }
Next: { value: 42 }
Next: { value: 24 }
Completed
Here's the code
这是代码
import {Observable} from 'rxjs'
Observable.of(['a', 1], ['a', 1], ['a', 0], ['a', 1])
.distinctUntilChanged(x => x[1])
.subscribe(x => console.log(x))
Observable.of({key: 'a', state: 1}, {key: 'a', state: 2}, {key: 'a', state: 0}, {key: 'a', state: 1})
.distinctUntilChanged(x => x.state)
.subscribe(x => console.log(x))
Observable.of({value: 42}, {value: 42}, {value: 24}, {value: 24})
.distinctUntilChanged(x => x.value)
.subscribe(
function (x) {
console.log('Next: ', x)
},
function (err) {
console.log('Error: ' + err)
},
function () {
console.log('Completed')
}
)
The linksin the v5 docs for these functions appear to be dead
这些函数的 v5 文档中的链接似乎已失效
------ edit -----
- - - 编辑 - - -
Some additional debugging:
一些额外的调试：
Observable.of(['a', 1], ['a', 1], ['a', 0], ['a', 1])
.do(x => console.log('before', x))
.distinctUntilChanged(x => x[1])
.do(x => console.log('after', x))
.subscribe(x => console.log(x))
output:
输出：
before [ 'a', 1 ]
after [ 'a', 1 ]
[ 'a', 1 ]
before [ 'a', 1 ]
before [ 'a', 0 ]
before [ 'a', 1 ]
回答by Thijs Koerselman
I got an answer here. Basically the function signature changed from (key selector, comparator) to (comparator, key selector).
我在这里得到了答案。基本上函数签名从（键选择器，比较器）更改为（比较器，键选择器）。
This is how the example is done in v5:
这是在 v5 中完成示例的方式：
Observable.of(['a', 1], ['a', 1], ['a', 0], ['a', 1])
.distinctUntilChanged(null, x => x[1])
.subscribe(x => console.log(x))
回答by user3743222
Here is a samplewith your code is Rxjs V4. You will see that it works correctly.
这是一个示例，您的代码是 Rxjs V4。你会看到它工作正常。
Observable.of(['a', 1], ['a', 1], ['a', 0], ['a', 1])
.distinctUntilChanged(x => x[1])
.subscribe(x => console.log(x))
...
So it seems to be something with the new beta version. Here are the specs for distinctUntilChanged. The operator itself seems to be working as in version 4.
所以它似乎与新的测试版有关。以下是distinctUntilChanged的规格。操作员本身似乎像在版本 4 中一样工作。
To test things out, I recommend you trace the output of each function by inserting a .do(function(x){console.log(x)})in between operators. I can only think of the ofoperator maybe passing on only the last element of the array.
为了进行测试，我建议您通过.do(function(x){console.log(x)})在运算符之间插入 a来跟踪每个函数的输出。我只能想到of运算符可能只传递数组的最后一个元素。
