3

我想知道如何使用 javascript 过滤值数组如何使用“provider-send”和“provider-receive”分隔数组

 var filterradio = id.filter(function(e){       
       return e.id.split("-")[0] == "provider-send"
    })  

var id=["provider-send-credit-transfer", "provider-send-debit-fund","provider-receive-credit-transfer","provider-receive-debit-fund"]


Expected Output:
result_sn: ["provider-send-credit-transfer", "provider-send-debit-fund"]
result_rcn:["provider-receive-credit-transfer","provider-receive-debit-fund"]
4

5 回答 5

1

如果总是“provider-receive-...”和“provider-send...”，那么您可以执行以下操作将它们分开

for (i = 0; i < id.length; i++) {
    if (id[i].split("provider-send-").length > 1) {
        result_sn.push(id[i]);
    } else if (id[i].split("provider-receive-").length > 1) {
        result_rcn.push(id[i])
    }
}
于 2019-04-09T01:57:15.353 回答
0

2作为第二个参数传递给split()，然后再次传递join()给-。第二个参数split()指定结果数组中的最大元素数。 

var id=["provider-send-credit-transfer", "provider-send-debit-fund","provider-receive-credit-transfer","provider-receive-debit-fund"]

var filterradio = id.filter(function(id){    
    return id.split("-",2).join('-') === "provider-send"
})  
         
 console.log(filterradio)

于 2019-04-09T01:50:28.427 回答
0

我建议使用reduce而不是filteras reduce 用于将数组大小减小为单个返回元素。在这里，我将数组缩减为一个具有两个键result_sn和result_rcn.

var id = ["provider-send-credit-transfer", "provider-send-debit-fund", "provider-receive-credit-transfer", "provider-receive-debit-fund"]


const result = id.reduce((obj, str) => {
  if (str.match(/^provider-send/g))
    obj['result_sn'].push(str);
  else
    obj['result_rcn'].push(str);
  return obj;
}, {
  'result_sn': [],
  'result_rcn': []
});

console.log(result)

于 2019-04-09T02:00:05.047 回答
0

试试这个 ：

var id = [ "provider-send-credit-transfer", "provider-send-debit-fund","provider-receive-credit-transfer","provider-receive-debit-fund" ] ;

var result_sn = [] , result_rcn = [] ;

for( value of id ) {
  var twoWords = value.split('-',2).join('-') ;
  if ( twoWords === "provider-send" ) 
    result_sn.push( value ) ;
  else if ( twoWords === "provider-receive" )
    result_rcn.push( value ) ;
}

console.log( result_sn ) ;
console.log( result_rcn ) ;

于 2019-04-09T02:24:26.473 回答
0

使用.filter()将要求您为要匹配的每个分配给变量的模式编写 1 个过滤器。

推荐使用.reduce()，并且更容易扩展以支持更多模式。

一开始它可能看起来很吓人，但实际上你是在使用accumulator临时变量，它被存储并带到每次迭代中。并且数组的每次迭代都会为您提供当前的迭代值currentValue。

我添加了something-else作为添加新模式的示例。

var id = [
  "provider-send-credit-transfer",
  "provider-send-debit-fund",
  "provider-receive-credit-transfer",
  "provider-receive-debit-fund",
  "something-else-credit-transfer",
  "something-else-debit-fund"
];

const {
  'provider-send': result_sn,
  'provider-receive': result_rcn,
  'something-else': result_ste
} = id.reduce(function(accumulator, currentValue) {
  let prefix = currentValue.match(/^\w+-\w+/)[0];
  return {...accumulator, [prefix]: (accumulator[prefix] || []).concat(currentValue)}
}, {});

console.log(result_sn);
console.log(result_rcn);
console.log(result_ste);

于 2019-04-09T03:00:14.107 回答