Question

我正在使用你可能不熟悉的tableToJson将表转换为JSON（如果你不知道，我已经在这里包含了代码）。很简单。输出完全符合预期，看起来像：

reader=[
        readerType=UCIFastReader
        file=$DataDir$/Train.txt
        randomize=None
        features=[
            dim=4
            start=0
        ]
        labels=[
            dim=2
            start=4
            labelDim=10
            labelMappingFile=$DataDir$/labelsmap.txt
        ]
    ]

但是，我需要扩展它以获取每个{"data":[ {"Time":"6:30am - 8:30am","Monday":"","Tuesday":"Maths","Wednesday":"","Thursday":"","Friday":""}, {"Time":"11:15am - 12:45pm","Monday":"Maths","Tuesday":"","Wednesday":"English","Thursday":"","Friday":""}, {"Time":"3:00pm - 5:00pm","Monday":"","Tuesday":"","Wednesday":"English","Thursday":"","Friday":""}, {"Time":"5:00pm - 7:00pm","Monday":"","Tuesday":"Science","Wednesday":"","Thursday":"","Friday":""}, {"Time":"7:00pm - 9:00pm","Monday":"","Tuesday":"Science","Wednesday":"","Thursday":"","Friday":""}] }的#id，以便JSON看起来像:(假设表格HTML为<tr>等。）

<tr id="session 1"></tr>

供参考tableToJson的代码如下：

{
"data":{
"session 1":{"Time":"6:30am - 8:30am","Monday":"","Tuesday":"sdfsdf","Wednesday":"","Thursday":"","Friday":""},
"session 2":{"Time":"11:15am - 12:45pm","Monday":"sdfsdfsdf","Tuesday":"","Wednesday":"","Thursday":"","Friday":""},
"session 3":{"Time":"3:00pm - 5:00pm","Monday":"","Tuesday":"","Wednesday":"","Thursday":"","Friday":""},
"session 4":{"Time":"5:00pm - 7:00pm","Monday":"","Tuesday":"","Wednesday":"","Thursday":"","Friday":""},
"session 5":{"Time":"7:00pm - 9:00pm","Monday":"","Tuesday":"","Wednesday":"","Thursday":"","Friday":""}
}
}

我可能遗漏了一些非常简单的东西，但老实说却看不出怎么做。任何帮助将非常感激。提前谢谢大家！

Answer 1

请重新考虑为什么需要更改它。这已经在JSON数组中提供了有效的有序内容（可以通过foo.data[0]，foo.data[1]等访问，也可以使用for(var i; i<foo.data.length;i++){}进行迭代）。

你想要的东西会给你一些没有保证顺序的东西（对象中的属性顺序是无关紧要在JSON中）并反对流程。

修改：愚蠢的API 。

如果必须符合某些晦涩难看的标准，请在创建后编辑数据。您不能指望$.fn.tableToJSON不符合JSON＆amp; javascript标准......

var foo = // output from $.fn.tableToJSON
var out = {data:{}}; //declare an *object* instead of an array
for(var i = 0; i < foo.data.length; i++){
  out.data['Session ' + (i + 1)] =  foo.data[i];
}
//debug output
console.log(JSON.stringify(out))

表到JSON - ＆gt;从ids添加其他对象标识符

1 个答案: