如果不在第二个数组中，则从一个数组中删除项目

Question

我声明我在写这篇文章之前尝试了很长时间post。

对于 InDesign 脚本，我使用两个列表项数组。现在我试图删除一个数组中不在第二个数组中的项目，但我被卡住了。

考虑到我使用以下 javascript 代码（效果很好）删除两个数组之间的相等项：

function check_dupli(arr_A, arr_B) {
    for(var i = arr_B.length - 1; i >= 0; i--) {
        for(var j = 0; j < arr_A.length; j++) {
            if(arr_B[i] === arr_A[j]) {
                arr_B.splice(i, 1);
            }
        }
    }
    arr_B.sort();
}

arr_A = ["a","b","d","f","g"]
arr_B = ["a","c","f","h"]

check_dupli(arr_A, arr_B) --> arr_B = ["c","h"]
check_dupli(arr_B, arr_A) --> arr_B = ["b","d","g"]

我想修改它以忽略不在两个数组中的项目，并获得我想要的，但是出了点问题，因为我也得到了不需要的数据:

function get_dupli(arr_A, arr_B, arr_C) {
    for(var e = arr_B.length - 1; e >= 0; e--) {
        for(var k = 0; k < arr_A.length; k++) {
            if(arr_B[e] === arr_A[k]) {
                arr_C.push(arr_B[e]);
            }
        }
    }
    arr_C.sort();
}

arr_A = ["a","b","d","f","g"]
arr_B = ["a","g","k"]
arr_C = ["h"]

get_dupli(arr_A, arr_B, arr_C) --> arr_C = ["a","g","h","k"] instead of --> ["a","g","h"]
get_dupli(arr_B, arr_A, arr_C) --> arr_C = ["a","b","d","f","g","h"] instead of --> ["a","g","h"]

我哪里错了？ pure javascript中还有其他方法可以解决问题吗？

在此先感谢您的帮助。

Answer 1

这样做：

//the array which will loose some items
var ar1 = ["a", "b", "c"];
//the array which is the template
var ar2 = ["d", "a", "b"];

var tmpar = [];

for(var i = 0; i < ar1.length; i++){
  if(ar2.indexOf(ar1[i]) !== -1){
    tmpar.push(ar1[i]);
  }
}

ar1 = tmpar;

alert(ar1);

我们创建一个临时数组来存储有效值。

我们确保第一个数组中值的索引不是“-1”。如果它是“-1”，则未找到索引，因此该值无效！我们存储所有不是“-1”的东西（所以我们存储每个有效值）。

Answer 2

你可以利用Array.prototype.filter和Array.prototype.concat来简化它：

arr_A = ["a","b","d","f","g"]
arr_B = ["a","g","k"]
arr_C = ["h"]

function getCommonItems(arrayA, arrayB, result) {
 result = result || [];
 result = result.concat(arrayA.filter(function(item) {
            return arrayB.indexOf(item) >= 0;    
          }));
 return result.sort();
}

alert(getCommonItems(arr_A, arr_B, arr_C).join(", "));
alert(getCommonItems(arr_B, arr_A, arr_C).join(", "));

对于第一种情况：

arr_A = ["a","b","d","f","g"]
arr_B = ["a","c","f","h"]

function getDifference(arrayA, arrayB, result) {
 return arrayB.filter(function(item) {
            return arrayA.indexOf(item) === -1;    
  }).sort();
}

alert(getDifference(arr_A, arr_B).join(", "));
alert(getDifference(arr_B, arr_A).join(", "));

Answer 3

Array.prototype.contains =  function ( object )
{
 var i = 0, n = this.length;
 
 for ( i = 0 ; i < n ; i++ )
 {
  if ( this[i] === object )
  {
   return true;
  }
 }
 
 return false;
}

Array.prototype.removeItem = function(value, global) {
 var idx;
 var  n = this.length;
 while ( n-- ) {
  if ( value instanceof RegExp && value.test ( this[n]) 
  || this[n] === value ) {
   this.splice (n, 1 );
   if ( !global ) return this;
  }
 }
 return this;
};

arr_A = ["a","b","d","f","g"];
arr_B = ["a","c","f","h"];

var item
while ( item = arr_A.pop() ) {
 arr_B.contains ( item ) && arr_B.removeItem ( item  );
}

arr_B;

Answer 4

arr_A = ["a","b","d","f","g"];
arr_B = ["a","c","f","h"];
var newArr = [];
var item
while ( item = arr_B.shift() ) {
 arr_A.contains ( item ) && newArr[ newArr.length ] = item ;
}
newArr;// ["a", "f"];

Answer 5

Opsss ....我相信我已经给出了答案并关闭了这个post ......抱歉!!!

尽管我做了所有检查，但我的脚本失败是由一个愚蠢的错误引起的……传递给函数的数组 arr_A 是原始数组的修改副本。

感谢大家的关心和帮助。再次抱歉...

如果不在第二个数组中，则从一个数组中删除项目

Remove items from one array if not in the second array

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arrays

adobe-indesign