Question

这是我们想要使用jquery推方括号的json ..

{"ProductionAnticipationId":{"title":"","type":"hidden","required":true,"options":"","defaultValue":""},"ProductionType": {"title": "Production Type","type": "Combobox","required": true,"options": "/BgtProduction/GetProductionType","unique": true},"Grade": {"title": "Grade","type": "Combobox","required": true,"options": "/BgtProduction/GetGrades","unique": true},"Quantity": {"title": "Quantity ","type": "","required": true,"align":"right","defaultValue": "0"} },{"Quantity":"total"}

以下是我们用来将方括号推到JSON以上的代码..

    var ar = [];
    if (!$.isArray(l_schemaString))

        ar = ar.push($.parseJSON(l_schemaString));
    else


        ar = $.parseJSON(l_schemaString);

在View中应用此代码后，在控制台中显示错误， 未捕获的SyntaxError：意外的令牌，

请提供解决方案以解决此问题

Answer 1

我将您的JSON粘贴到格式化程序中：http://jsonformatter.curiousconcept.com/

结果是：错误：多个JSON根元素

{
   "ProductionAnticipationId":{
      "title":"",
      "type":"hidden",
      "required":true,
      "options":"",
      "defaultValue":""
   },
   "ProductionType":{
      "title":"Production Type",
      "type":"Combobox",
      "required":true,
      "options":"/BgtProduction/GetProductionType",
      "unique":true
   },
   "Grade":{
      "title":"Grade",
      "type":"Combobox",
      "required":true,
      "options":"/BgtProduction/GetGrades",
      "unique":true
   },
   "Quantity":{
      "title":"Quantity ",
      "type":"",
      "required":true,
      "align":"right",
      "defaultValue":"0"
   }
},                                 <-- the problem is here
{
   "Quantity":"total"
}

看起来问题是JSON中有多个对象，而不仅仅是一个。因此，当服务器写出它时，它应该被写为[ ... objects here ... ]，以便它们被包装在一个数组中。

在其他代码示例中，这里存在一个问题：

ar = ar.push($.parseJSON(l_schemaString));

push method会返回数组的新长度，因此ar最终会被设置为1而不是数组，这会导致以后出现问题。

如何使用Jquery在没有方括号的情况下在JSON中推方括号？

1 个答案: