Javascript null 时跳转到下一条记录代码修改答案

【问题标题】：Javascript When null skip to the next record code modificationJavascript null 时跳转到下一条记录代码修改
【发布时间】：2026-01-31 03:55:02
【问题描述】：

我有一些运行良好的代码，但我遇到了问题。

基本上，当它到达一个为 NULL 的记录时，它会在数组中添加一个 0...

在这种情况下，第二条记录为 NULL，所以我得到：

[10, 0, 20]

我需要它做的是，如果 thsub 为 NULL，则不向数组中添加任何内容并继续下一条记录。

所以在这种情况下，期望的结果是：

[10, 20]

这是完整的代码：

var data = {
      "cars": [{
          "id": "1",
          "name": "name 1",
          "thsub": [{
            "id": "11",
            "name": "sub 1",
            "stats": {
              "items": 5,
            },
            "ions": null
          }, {
            "id": "22",
            "name": "sub 2",
            "stats": {
              "items": 5,
            },
            "translations": null
          }],
          "image": null
        },
    
        {
          "id": "2",
          "name": "name 2",
          "thsub": null, //this will break the code
          "image": null
        },
        {
          "id": "54",
          "name": "name something",
          "thsub": [{
            "id": "65",
            "name": "sub 1",
            "stats": {
              "items": 10,
            },
            "ions": null
          }, {
            "id": "22",
            "name": "sub 2",
            "stats": {
              "items": 10,
            },
            "translations": null
          }],
          "image": null
        }
      ]
    }
    
    
    
    var thCount = [];
    
    for (var l = 0, m = data.cars.length; l < m; l++) {
      thCount[l] = 0;
      if (data.cars[l].thsub) {
        for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
          if (data.cars[l].thsub[i].stats) {
            thCount[l]+=data.cars[l].thsub[i].stats.items;
          }
        }
      }
    }
    
    console.log(thCount);

我该怎么做？

【问题讨论】：

标签： javascript arrays data-structures iteration reduce

【解决方案1】：

如果设置了thsub，您只能推送一个值。

var data = { cars: [{ id: "1", name: "name 1", thsub: [{ id: "11", name: "sub 1", stats: { items: 5, }, ions: null }, { id: "22", name: "sub 2", stats: { items: 5, }, translations: null }], image: null }, { id: "2", name: "name 2", thsub: null, image: null }, { id: "54", name: "name something", thsub: [{ id: "65", name: "sub 1", stats: { items: 10, }, ions: null }, { id: "22", name: "sub 2", stats: { items: 10, }, translations: null }], image: null }] },
    thCount = [];

for (var l = 0, m = data.cars.length; l < m; l++) {
    if (data.cars[l].thsub) {
        thCount.push(0);
        for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
            if (data.cars[l].thsub[i].stats) {
                thCount[thCount.length - 1] += data.cars[l].thsub[i].stats.items;
            }
        }
    }
}

console.log(thCount);

【讨论】：

【解决方案2】：

solution

你需要添加变量，然后只有在有东西的时候才会推送到数组。

var data = {
  "cars": [{
      "id": "1",
      "name": "name 1",
      "thsub": [{
        "id": "11",
        "name": "sub 1",
        "stats": {
          "items": 5,
        },
        "ions": null
      }, {
        "id": "22",
        "name": "sub 2",
        "stats": {
          "items": 5,
        },
        "translations": null
      }],
      "image": null
    },

    {
      "id": "2",
      "name": "name 2",
      "thsub": null, //this will break the code
      "image": null
    },
    {
      "id": "54",
      "name": "name something",
      "thsub": [{
        "id": "65",
        "name": "sub 1",
        "stats": {
          "items": 10,
        },
        "ions": null
      }, {
        "id": "22",
        "name": "sub 2",
        "stats": {
          "items": 10,
        },
        "translations": null
      }],
      "image": null
    }
  ]
}



var thCount = [];

for (var l = 0, m = data.cars.length; l < m; l++) {
  if (data.cars[l].thsub) {
    var tmp = 0;
    for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
      if (data.cars[l].thsub[i].stats) {
        tmp+=data.cars[l].thsub[i].stats.items;
      }
      thCount.push(tmp);
    }
  }
}

console.log(thCount);

【讨论】：

【解决方案3】：

您可以在 for 循环中创建新变量，如果它不为 0，则使用 push()。

var data = {"cars":[{"id":"1","name":"name 1","thsub":[{"id":"11","name":"sub 1","stats":{"items":5},"ions":null},{"id":"22","name":"sub 2","stats":{"items":5},"translations":null}],"image":null},{"id":"2","name":"name 2","thsub":null,"image":null},{"id":"54","name":"name something","thsub":[{"id":"65","name":"sub 1","stats":{"items":10},"ions":null},{"id":"22","name":"sub 2","stats":{"items":10},"translations":null}],"image":null}]}
var thCount = [];

for (var l = 0, m = data.cars.length; l < m; l++) {
  var count = 0
  if (data.cars[l].thsub) {
    for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
      if (data.cars[l].thsub[i].stats) {
        count += data.cars[l].thsub[i].stats.items;
      }
    }
  }
  if (count != 0) thCount.push(count)
}

console.log(thCount);

【讨论】：

【解决方案4】：

thCount[l] = 0; 行将导致向 thCount 添加一个元素，无论您的条件如何if (data.cars[l].thsub)

您可以使用另一个变量并仅在我们想向我们的数组中添加一些东西时增加它：

	var data = {
  "cars": [{
      "id": "1",
      "name": "name 1",
      "thsub": [{
        "id": "11",
        "name": "sub 1",
        "stats": {
          "items": 5,
        },
        "ions": null
      }, {
        "id": "22",
        "name": "sub 2",
        "stats": {
          "items": 5,
        },
        "translations": null
      }],
      "image": null
    },

    {
      "id": "2",
      "name": "name 2",
      "thsub": null, //this will break the code
      "image": null
    },
    {
      "id": "54",
      "name": "name something",
      "thsub": [{
        "id": "65",
        "name": "sub 1",
        "stats": {
          "items": 10,
        },
        "ions": null
      }, {
        "id": "22",
        "name": "sub 2",
        "stats": {
          "items": 10,
        },
        "translations": null
      }],
      "image": null
    }
  ]
}



var thCount = [];

for (var l = 0, k = -1, m = data.cars.length; l < m; l++) {
  if (data.cars[l].thsub) {
    thCount[++k] = 0;
    for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
      if (data.cars[l].thsub[i].stats) {
        thCount[k] += data.cars[l].thsub[i].stats.items;
      }
    }
  }
}

console.log(thCount); //alert(thCount);

【讨论】：

【解决方案5】：

已经有一些关于如何修复循环的答案，所以基本问题在于thCount[l] = 0 行。您还可以使用高阶函数来遍历对象，这在我看来会产生更好的可读性代码：

var data = {
  "cars": [{
    "id": "1",
    "name": "name 1",
    "thsub": [{
      "id": "11",
      "name": "sub 1",
      "stats": {
        "items": 5,
      },
      "ions": null
    }, {
      "id": "22",
      "name": "sub 2",
      "stats": {
        "items": 5,
      },
      "translations": null
    }],
    "image": null
  },

           {
             "id": "2",
             "name": "name 2",
             "thsub": null, //this will break the code
             "image": null
           },
           {
             "id": "54",
             "name": "name something",
             "thsub": [{
               "id": "65",
               "name": "sub 1",
               "stats": {
                 "items": 10,
               },
               "ions": null
             }, {
               "id": "22",
               "name": "sub 2",
               "stats": {
                 "items": 10,
               },
               "translations": null
             }],
             "image": null
           }
          ]
}





var thCount = data.cars.filter(function(car){
  return car.thsub;
}).map(function(car){
  return car.thsub.reduce(function(a,b){
    return a.stats.items + b.stats.items;
  });
});
console.log(thCount)

首先，您将filter 剔除所有没有thsub 的对象。然后你 map 将这些条目设置为新值。您可以使用reduce-函数获取这些值，该函数汇总了stats 中的所有items-值。

【讨论】：

【解决方案6】：

也可以考虑嵌套reduce 方法，例如像下面这样...

var data = {
  "cars": [{
    "id": "1",
    "name": "name 1",
    "thsub": [{
      "id": "11",
      "name": "sub 1",
      "stats": {
        "items": 5
      },
      "translations": null
    }, {
      "id": "22",
      "name": "sub 2",
      "stats": {
        "items": 5
      },
      "translations": null
    }],
    "image": null
  }, {
    "id": "2",
    "name": "name 2",
    "thsub": null,
    "image": null
  }, {
    "id": "54",
    "name": "name something",
    "thsub": [{
      "id": "65",
      "name": "sub 1",
      "stats": {
        "items": 10
      },
      "translations": null
    }, {
      "id": "22",
      "name": "sub 2",
      "stats": {
        "items": 10
      },
      "translations": null
    }],
    "image": null
  }]
};


var thCount = data.cars.reduce(function (collector, carItem) {
  var
    thSubs = carItem.thsub;

//if (Array.isArray(thSubs)) { ... }
  if ((thSubs != null) && (thSubs.length >= 1) && Number.isFinite(thSubs.length)) {

    collector.push(
      thSubs.reduce(function (count, subItem) {

        return (count + subItem.stats.items);

      }, 0)
    );
  }
  return collector;

}, []);


console.log(thCount);

【讨论】：