Question

我创建了一个包含对象的数组，其中一些属性也是对象。我已成功将其转换为JSON，需要将其转换回对象数组，或以某种方式从JSON对象的正确索引中提取正确的数据。

更新

这是我通过JSON.parse运行时获得的示例：

[{"Result":"Fail","Method":"T97E-v1","Beam1":{"BeamAge":"1","WidthUpper":1,"WidthCenter":1,"WidthLower":1,"WidthAverage":1,"DepthRight":1,"DepthCenter":1,"DepthLeft":1,"DepthAverage":1,"MaxLoad":1,"FS":18,"PSI":"18.00000","BreakOutside":"No"},"Beam2":{"BeamAge":"","WidthUpper":null,"WidthCenter":null,"WidthLower":null,"WidthAverage":null,"DepthRight":null,"DepthCenter":null,"DepthLeft":null,"DepthAverage":null,"MaxLoad":null,"FS":null,"PSI":"NaN"},"WaitForCuring":"No","AverageOfBeams":"NaN"}]

更新2

以下是我正在做的代码：

try {
    localStorage["flexuralStrengthSamples"] = JSON.stringify(JSON.stringify(t97Samples));
    var parsedObject = JSON.parse(localStorage["flexuralStrengthSamples"]);

    console.log(parsedObject);                    
    console.log(parsedObject[0].Beam1.MaxLoad);            
} catch (err) {
    alert(err.message);
}

Answer 1

您可以使用JSON.parse()解析JSON。

<强>更新
以下是JSON.parse()的数据示例。

[{"Result":"Fail","Method":"T97E-v1","Beam1":{"BeamAge":"1","WidthUpper":1,"WidthCenter":1,"WidthLower":1,"WidthAverage":1,"DepthRight":1,"DepthCenter":1,"DepthLeft":1,"DepthAverage":1,"MaxLoad":1,"FS":18,"PSI":"18.00000","BreakOutside":"No"}}]

为了获取数据，您需要为数组使用括号表示法，为对象使用点符号。因此，让data等于该JSON数组，然后您可以执行data[0].Result "Fail"或data[0].Beam1.MaxLoad，即1。

Answer 2

我找出原因：在将其存储在localstorage

之前，你要进行两次连字

try {
     localStorage["flexuralStrengthSamples"] = (JSON.stringify(t97Samples)); //Stringify only once, since localstorage values needs to be string
     var parsedObject = JSON.parse(localStorage["flexuralStrengthSamples"]); // should give the original object.


   console.log(parsedObject[0].Beam1.MaxLoad); // Since parsedObject is still string, this was failing. Now should work fine           
} catch (err) {
   alert(err.message);
}

请在此处查看工作小提琴：http://jsfiddle.net/sandenay/pnb88p4s/

将JSON解析为Object Array

2 个答案: