如何比較數組並創建與數組a匹配的新數組？

Question

我想匹配兩個數組，如果數組b包含數組ai的字符串值要創建新數組，則使用下面的代碼將所有對象從數組b推入。 我如何只推入匹配的物體？

ctrl.js

 var a = ["standard","industry","4578595"]
 var b = [{file:"server.log",value:"lorem ipsum industry standard 4578595"}, {file:"server.log",value:"lorem ipsum industry standadr"},{file:"server.log",value:"lorem ipsum industry"}];
 var finalyArray = [];
 function compareArrays() {   
     for (i = 0; i < a.length; i++) {
         for (z = 0; z < a.length; z++) {
             if (a[i] === b[z]) {
                 finalArray.push(b[z]);
                 console.log(finalArray); 
             }
         }
     }
}

compareArrays();

Answer 1

無需使用嵌套循環和常規迭代。

數組過濾更加干凈，非常適合此操作：

 const a = ["standard","industry","4578595"] const b = [{file:"server.log",value:"lorem ipsum industry standard 4578595"}, {file:"server.log",value:"lorem ipsum industry standadr"},{file:"server.log",value:"lorem ipsum industry"}]; const matchingCondition = ({ itemB, itemA }) => itemB.value.includes(itemA); // filter all items from B that satisfy a matching condition with at least one item from A const matches = b.filter( itemB => a.some( itemA => matchingCondition({ itemB, itemA }) ) ); console.log(matches); 

Answer 2

    // return a subset array of b, which contains all the strings of array a 

    var a = ["standard","industry","4578595"]
    var b = [
      {file:"server.log",value:"lorem ipsum industry standard 4578595"}, 
      {file:"server.log",value:"lorem ipsum industry standadr"},
      {file:"server.log",value:"lorem ipsum 4578595 industry"}];

    function filterArray(a,b) {
      var aLen = a.length;
      var bLen = b.length;
      var filterArray = []

      // if a is empty it should return empty array
      if (!aLen){
        return filterArray;
      }

      for(var i = 0; i< bLen; i++) {
        var bValuesArray = b[i].value.split(" ");
        var bValuesLength = bValuesArray.length;
        var addObject = 0;

        for(var j=0; j< aLen; j++) {
           for(var k=0; k <bValuesLength; k++ ){
              if(a[j] === bValuesArray[k]){
                addObject++;
                continue;
              }
           }
        }
        // add if all elements of a are present in b array
        if(addObject === aLen) {
          filterArray.push(b[i]);
        }
      }

      return filterArray;
    };

    console.log(filterArray(a,b));