Question

我有如下的JSON文件，我想提取placeName＆＃39;的valus，这是＆＃34; Ince Minaret Medrese＆＃34;和＆＃34; Karatay，Konya＆＃34;和纬度和长期值为37.8728和32.49以及＆＃34; Karatay，Konya＆＃34; lat的值为37.8667，long的值为32.5289

{
    "head": {
        "link": [],
        "vars": ["placeName", "lat", "long"]
    },
    "results": {
        "distinct": false,
        "ordered": true,
        "bindings": [{
                "placeName": {
                    "type": "literal",
                    "xml:lang": "en",
                    "value": "Ince Minaret Medrese"
                },
                "lat": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "37.8728"
                },
                "long": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "32.49"
                }
            },
            {
                "placeName": {
                    "type": "literal",
                    "xml:lang": "en",
                    "value": "Karatay, Konya"
                },
                "lat": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "37.8667"
                },
                "long": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "32.5289"
                }
            }
        ]
    }
}

我试着这样做

    <p id="demo"></p>

    <script>

    var obj =    {
        "head": {
            "link": [],
            "vars": ["placeName", "lat", "long"]
        },
        "results": {
            "distinct": false,
            "ordered": true,
            "bindings": [{
                    "placeName": {
                        "type": "literal",
                        "xml:lang": "en",
                        "value": "Ince Minaret Medrese"
                    },
                    "lat": {
                        "type": "typed-literal",
                        "datatype": "http://www.w3.org/2001/XMLSchema#float",
                        "value": "37.8728"
                    },
                    "long": {
                        "type": "typed-literal",
                        "datatype": "http://www.w3.org/2001/XMLSchema#float",
                        "value": "32.49"
                    }
                },
                {
                    "placeName": {
                        "type": "literal",
                        "xml:lang": "en",
                        "value": "Karatay, Konya"
                    },
                    "lat": {
                        "type": "typed-literal",
                        "datatype": "http://www.w3.org/2001/XMLSchema#float",
                        "value": "37.8667"
                    },
                    "long": {
                        "type": "typed-literal",
                        "datatype": "http://www.w3.org/2001/XMLSchema#float",
                        "value": "32.5289"
                    }
                }
            ]
        }
    }
;
    var myJSON = JSON.parse(obj);
    document.getElementById("demo").innerHTML = myJSON.results.bindings[0].placeName.value;

    </script>

但是我没有得到任何结果，我该如何从嵌套对象中提取数据呢？

Answer 1

分别为：

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [0] [＆＃39; PLACENAME＆＃39;] [＆＃39;值＆＃39;]

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [1] [＆＃39; PLACENAME＆＃39;] [＆＃39;值＆＃39;]

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [1] [＆＃39; LAT＆＃39;] [＆＃39;值＆＃39;]

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [1] [＆＃39;长＆＃39;] [＆＃39;值＆＃39;]

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [0] [＆＃39; LAT＆＃39;] [＆＃39;值＆＃39;]

OBJ [＆＃39;结果＆＃39;] [＆＃39;绑定＆＃39;] [0] [＆＃39;长＆＃39;] [＆＃39;值＆＃39;]

您也可以使用点表示法（例如obj.results.bindings [1] .long.value），如果您在访问索引之前知道了索引，建议使用。

仅供参考：您不需要解析该对象，因为它已经是JSON。

＆＃13;

function localFunc() {
   const self = exports;
   console.log(self.myVar);
}

exports.myVar = 'foo';

exports.myFunc = function () {
    localFunc();
}

＆＃13;

Answer 2

您的代码实际上是正确的，只需删除JSON.parse()，因为您传递给parse方法的参数不是json字符串，它是一个实际的对象
所以按照以下方式进行。
var myJSON = obj; 它将开始工作。

Answer 3

您可以使用过滤器

var result = data.results.bindings.filter(function( obj ) {
    return obj.placeName.value == "Ince Minaret Medrese";
});

参见MDN Docs on Array.prototype.filter()

＆＃13;

var data = {
    "head": {
        "link": [],
        "vars": ["placeName", "lat", "long"]
    },
    "results": {
        "distinct": false,
        "ordered": true,
        "bindings": [{
                "placeName": {
                    "type": "literal",
                    "xml:lang": "en",
                    "value": "Ince Minaret Medrese"
                },
                "lat": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "37.8728"
                },
                "long": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "32.49"
                }
            },
            {
                "placeName": {
                    "type": "literal",
                    "xml:lang": "en",
                    "value": "Karatay, Konya"
                },
                "lat": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "37.8667"
                },
                "long": {
                    "type": "typed-literal",
                    "datatype": "http://www.w3.org/2001/XMLSchema#float",
                    "value": "32.5289"
                }
            }
        ]
    }
};
var result = data.results.bindings.filter(function( obj ) {
  return obj.placeName.value == "Ince Minaret Medrese";
});
console.log(result)

＆＃13;

如何从json嵌套对象中提取数据？

3 个答案: