Javascript 当引用位于数组中时，如何从分层对象获取值？

我有以下目标

{
"locations": {
    "Base 1": {
        "title": "This is base 1",
        "Suburb 1": {
            "title": "Suburb 1 in Base 1",
            "Area A": {
                "title": "Title for Area A",
                "Street S1": {
                    "title": "Street S1 title"
                },
                "Street C4": {
                    "title": "Street C4 title"
                },
                "Street B7": {
                    "title": "Street B7 title"
                }
            },
            "Another Area": {
                "title": "Title for Area A",
                "Street S1": {
                    "title": "Street S1 title"
                },
                "Street C4": {
                    "title": "Street C4 title"
                },
                "Street B7": {
                    "title": "Street B7 title"
                }
            }
        },
        "Another Suburb": {
            "title": "Suburb 1 in Base 1",
            "Area A": {
                "title": "Title for Area A",
                "Street S1": {
                    "title": "Street S1 title"
                },
                "Street C4": {
                    "title": "Street C4 title"
                },
                "Street B7": {
                    "title": "Street B7 title"
                }
            },
            "Another Area": {
                "title": "Title for Area A",
                "Street S1": {
                    "title": "Street S1 title"
                },
                "Street C4": {
                    "title": "Street C4 title"
                },
                "Street B7": {
                    "title": "Street B7 title"
                }
            }
        }
    },
    "Base2": {}
}
}

我得到了从“位置”对象获取“标题”的数组，每个数组可以不同。我知道我可以访问个人值，如下所示：

locations["Base 1"]["title"]
locations["Base 1"]["Another Suburb"]["title"]
locations["Base 1"]["Another Suburb"]["Area A"]["title"]
etc etc.

AnArray = ["Base 1", "title"];
AnArray = ["Base 1", "Another Suburb", "title"];
AnArray = ["Base 1", "Another Suburb", "Area A", "title"];
AnArray = ["Base 1", "Another Suburb", "Another Area", "title"];

但是，如果给我这样的数组，我不确定如何获得title的值：

locations["Base 1"]["title"]
locations["Base 1"]["Another Suburb"]["title"]
locations["Base 1"]["Another Suburb"]["Area A"]["title"]
etc etc.

AnArray = ["Base 1", "title"];
AnArray = ["Base 1", "Another Suburb", "title"];
AnArray = ["Base 1", "Another Suburb", "Area A", "title"];
AnArray = ["Base 1", "Another Suburb", "Another Area", "title"];

是否有办法解析/使用这些数组，以便每个数组都从locations对象返回正确的标题值

我必须在每种情况下获取标题的值，我甚至不知道从哪里开始。我尝试加入数组，然后获取“title”值，但这似乎不起作用

这里没有问题，所以请不要介意这个问题听起来是否愚蠢/或毫无意义

---

所以问题是，当引用位于数组中时，如何从分层对象中获取值？

我想您需要这样的东西：

• 遍历每个数组
• 遍历数组的值
• 对于每个值，在对象中嵌套一层
• 将最终值存储在数组中

试试这个：

var arrays = [
    ["Base 1", "title"],
    ["Base 1", "Another Suburb", "title"],
    ["Base 1", "Another Suburb", "Area A", "title"],
    ["Base 1", "Another Suburb", "Another Area", "title"]
],
array, i, j,
obj = {
    "locations": {
        "Base1": {
            "title": "Thisisbase1",
            "Suburb1": {
                "title": "Suburb1inBase1",
                "AreaA": {
                    "title": "TitleforAreaA",
                    "StreetS1": {
                        "title": "StreetS1title"
                    },
                    "StreetC4": {
                        "title": "StreetC4title"
                    },
                    "StreetB7": {
                        "title": "StreetB7title"
                    }
                },
                "AnotherArea": {
                    "title": "TitleforAreaA",
                    "StreetS1": {
                        "title": "StreetS1title"
                    },
                    "StreetC4": {
                        "title": "StreetC4title"
                    },
                    "StreetB7": {
                        "title": "StreetB7title"
                    }
                }
            },
            "AnotherSuburb": {
                "title": "Suburb1inBase1",
                "AreaA": {
                    "title": "TitleforAreaA",
                    "StreetS1": {
                        "title": "StreetS1title"
                    },
                    "StreetC4": {
                        "title": "StreetC4title"
                    },
                    "StreetB7": {
                        "title": "StreetB7title"
                    }
                },
                "AnotherArea": {
                    "title": "TitleforAreaA",
                    "StreetS1": {
                        "title": "StreetS1title"
                    },
                    "StreetC4": {
                        "title": "StreetC4title"
                    },
                    "StreetB7": {
                        "title": "StreetB7title"
                    }
                }
            }
        },
        "Base2": {}
    }
},
current, key,
result = [];

for (i = 0; i < arrays.length; i++) {
    array = arrays[i];

    // reset current to the locations object
    current = obj.locations;

    for (j = 0; j < array.length; j++) {

        // remove spaces from the keys
        key = array[j].replace(' ', '', 'gi');

        if (current[key]) {
            // if there is such key then nest further
            current = current[key];
        } else {
            // there is no such key - log it
            console.log(key);
        }
    }

    if (typeof current === "string") {
        // if you have reached a string with the nesting add it to the result
        result.push(current);
    }
}

// output the final result
console.log(result);

---

我添加了额外的日志记录，这样就可以清楚地知道键是否出错。

如果数组是指向所需标题的“路径”，您可以这样做

请注意，在这种情况下需要“位置”，因为它是路径的起点，而不需要标题，因为我们一直在寻找它

var path = ["locations", "Base1", "AnotherSuburb"];

(function search(o, a){
    for(var p in o){
        if(typeof o[p] === 'object' && path.indexOf(p) !== -1){
            a.push(p);
            if(a.join('') === path.join('')){
                // match, print the title
                console.log(o[p].title);
            }
            else{
               search(o[p], a);
            }
        }
    }
})(object, []);

将在BASE1中打印郊区1

演示：

---

其中x是您的对象，ar_keys是一组键。

Wow这是问过并回答过的问题了吗，例如，和。您显示的不是有效的对象或数组定义，因此很难为您提供帮助，因为如何访问对象的细节在很大程度上取决于对象的实际结构。如果内存中有此项，您应该能够使用浏览器中内置的开发工具以有效格式显示它，然后将其复制并粘贴到此处。这些是什么类型的数组？我不明白你是如何或为什么以这种方式获得顶级位置阵列的。。。所以你说你得到了4个数组对象而不是一个？谢谢你的参考T.J。。。我刚刚用一个有效的JSON对象更新了这个问题。@Azurlake-我有数组来查询每种情况下的标题（对于每个数组），查询应该从locations对象返回正确的标题。这个问题中的数组只是一个例子。很好的递归，但是你可以遍历对象中的每个属性。您可以使用

hasOwnProperty

方法，也可以根本不迭代对象属性。OP在数组值中也有空格。。。我更新了问题来解决这个问题。感谢您提供的解决方案：）

 getNested(x['locations'], ar_keys);