Question

我正在尝试使用splice提取嵌套数组。我得到的是每个元素上带有undefined的所需大小的数组

  var testCurrProds = [ 'Prod1', 'prduct1', 'Prod1'] 
  var testCurrProds1 = [ 'Prod2', 'prduct2', 'Prod2', 'Prod2', "Prod9"] 
  var testCurrProds2 = [ 'Prod3', 'prduct3', 'Prod3', 'Prod3', "Prod9"] 
  var testCurrProds3 = [ 'Prod4', 'prduct4', 'Prod4', 'Prod4', "Prod9", "Prod9", "Prod98"]   


  var testString = [1, 'tName1', 4, 69, 'Haircare', "T1", "false", "false", testCurrProds]
  var testString1 = [2, 'tName2', 1, 69, 'Skincare', "T1", "false", "true", testCurrProds1]
  var testString2 = [3, 'tName3', 3, 69, 'Haircare', "T1", "true", "true", testCurrProds2]
  var testString3 = [4, 'tName4', 5, 69, 'Skincare', "T1", "true", "false", testCurrProds3]
  var testPackages = [testString, testString1, testString2, testString3];

  var currPackage = popArray(testPackages, 1, 'tName1');
  console.log(currPackage);
  
  function popArray(x, findID, findName)
  {
      //return array containing submitted ID 
      for (var k = 0; k < x.length; k++ )
      {
          //console.log(x[k]);  
          if((x[k][0]==findID) & (x[k][1]==findName))
          {
              //var temp = x.splice(k,1)
              //console.log(temp);
              return x.splice(k,1);
          }
      }
  }

Answer 1

这里可以使用数组推导，我认为比迭代方法更容易理解，并且更不容易出错。

function popArray(arrayOfArrays, findID, findName) {
    const index = arrayOfArrays.findIndex(a => a[0] == findID && a[1] == findName);
    return index !== -1 && arrayOfArrays.splice(index, 1)[0];
}

基本上，这只是使用内置的findIndex函数来找到匹配项，然后将其从数组中剪接并返回。

Answer 2

分析代码后，我没有发现逻辑和代码有任何问题，只不过您对'and'逻辑条件使用了单个'＆'符号。

function popArray(x, findID, findName)
{
  //return array containing submitted ID 
  for (var k = 0; k < x.length; k++ )
   {
    //console.log(x[k]);  
    if((x[k][0]==findID) && (x[k][1]==findName))
    {
        var temp = x.splice(k,1)
        console.log(temp);
        //return x.splice(k,1);
    }
  }
}

查看此附件图像。我运行了这段代码并获得了预期的输出。 http://prntscr.com/m0baai

使用拼接提取嵌套数组

2 个答案: