如何通过JavaScript打印多维JSON对象?

时间:2016-02-26 12:13:11

标签: javascript json

在我的项目中,有一个像这样的JSON对象:

{
"face": [
    {
        "attribute": {
            "age": {
                "range": 5,
                "value": 35
            },
            "gender": {
                "confidence": 99.9974,
                "value": "Male"
            },
            "glass": {
                "confidence": 98.3809,
                "value": "None"
            },
            "pose": {
                "pitch_angle": {
                    "value": 2e-05
                },
                "roll_angle": {
                    "value": -6.70428
                },
                "yaw_angle": {
                    "value": 0.005689
                }
            },
            "race": {
                "confidence": 61.7833,
                "value": "Black"
            },
            "smiling": {
                "value": 10.932
            }
        },
        "face_id": "b8b25a0fbb13f149a87438047fdd3e18",
        "position": {
            "center": {
                "x": 50.973236,
                "y": 55.583333
            },
            "eye_left": {
                "x": 36.641119,
                "y": 47.448167
            },
            "eye_right": {
                "x": 65.265207,
                "y": 45.143333
            },
            "height": 42.166667,
            "mouth_left": {
                "x": 41.970316,
                "y": 68.837667
            },
            "mouth_right": {
                "x": 65.547445,
                "y": 68.029667
            },
            "nose": {
                "x": 52.064964,
                "y": 58.712167
            },
            "width": 61.557178
        },
        "tag": ""
    }
],
"img_height": 629,
"img_id": "9ea3cab5193d005e89be75fb92f1bd88",
"img_width": 431,
"session_id": "8e57472a5d1542a4943237e0bfd8798f",
"url": null
}

还有一个像这样的JavaScript脚本,我希望显示年龄范围:

<script>
    var txt1 = '<?php echo $response['body']; ?>';
    objF = JSON.parse(txt1);
    document.getElementById("fac").innerHTML = objF.face[0].attribute.age.range;
</script>

所以请帮我解决这个问题。

2 个答案:

答案 0 :(得分:1)

只需移除'脚本代码周围的php符号,就无需解析JSON

var txt1 = <?php echo $response['body']; ?>;
document.getElementById("fac").innerHTML = objF.face[0].attribute.age.range;

答案 1 :(得分:0)

从第一行

中删除&#39;
var txt1 = <?php echo $response['body']; ?>;
objF = JSON.parse(txt1);
document.getElementById("fac").innerHTML = objF.face[0].attribute.age.range;

就是这样