Question

我已经定义了一个类电子邮件，其中包含以下详细信息：
Email：

String name;
String subject;
List<String> attachment;
String jsonContent;   
....

在上面的类中，jsonContent变量加载了一个strinigified json对象创建Email对象后，我将整个Email对象进行字符串化并发送给客户端。

我需要解析客户端中的Email对象并在UI中呈现它但它会在客户端抛出Email对象的解析错误，即

JSON.parse(emailString);

因为jsonContent字段中有双引号。
这是一个字符串化JAVA对象的问题，该对象具有已经字符串化的jsonContent变量。

修复它的一种方法是将jsonContent变量定义为对象而不是String。还有其他修复方法吗？

示例电子邮件JSON：

    {
    "id": "e4682ec0-a7c3-4f4d-abcd-f404f5fdb1eb",
    "entityType": "email",
    "subject": "Presentation 1",
    "from": "aaa <a@a.com>",
    "to": [
        "undisclosed-recipients:;"
    ],
    "cc": [],
    "bcc": [
        "jack.porter@forwardaccelerator.com"
    ],
    "recievedDate": 1423101398000,
    "recievedDateString": "Wed, 4 Feb 2015 12:26:38 -0800",
    "bodyText": " Please find the link to my recent presentation",
"jsonContent": "{
  "typeOfMail": "NormalMail",
  "normalMail": {
    "mailType": "NormalMail",
    "paragraphs": [
      "Pleasefindthelinktomyrecentpresentation"
    ]
  }
}"
}

Answer 1

你需要转义很多字符串才能把字符串作为字符串。

将json对象存储在需要转义它的json对象中。所以

  "jsonContent": "{
  "typeOfMail": "NormalMail",
  "normalMail": {
    "mailType": "NormalMail",
    "paragraphs": [
      "Pleasefindthelinktomyrecentpresentation"
    ]
  }
}"

变为

"jsonContent": "{\"typeOfMail\": \"NormalMail\",\"normalMail\":{\"mailType\":\"NormalMail\",\"paragraphs\":[\"Pleasefindthelinktomyrecentpresentation\"]}}"

现在，如果你想在java中编译它，如果你手动输入它作为Java字符串（执行片段），它应该是这样的。

var json =  {
    "id": "e4682ec0-a7c3-4f4d-abcd-f404f5fdb1eb",
    "entityType": "email",
    "subject": "Presentation 1",
    "from": "aaa <a@a.com>",
    "to": [
        "undisclosed-recipients:;"
    ],
    "cc": [],
    "bcc": [
        "jack.porter@forwardaccelerator.com"
    ],
    "recievedDate": 1423101398000,
    "recievedDateString": "Wed, 4 Feb 2015 12:26:38 -0800",
    "bodyText": " Please find the link to my recent presentation",
"jsonContent": "{\"typeOfMail\": \"NormalMail\",\"normalMail\":{\"mailType\":\"NormalMail\",\"paragraphs\":[\"Pleasefindthelinktomyrecentpresentation\"]}}"
}
console.log("This is the json object having a string with json");
console.log(json);
console.log("This is it parsed as string");
var x = {hello:JSON.stringify(json)};
console.log(JSON.stringify(x).substring(10,JSON.stringify(x).length-2));
document.getElementById('content').textContent = JSON.stringify(x).substring(10,JSON.stringify(x).length-2);

<div id="content"></div>

这就是在发送的JSON文件/请求答案中的样子

{
    "id": "e4682ec0-a7c3-4f4d-abcd-f404f5fdb1eb",
    "entityType": "email",
    "subject": "Presentation 1",
    "from": "aaa <a@a.com>",
    "to": [
        "undisclosed-recipients:;"
    ],
    "cc": [],
    "bcc": [
        "jack.porter@forwardaccelerator.com"
    ],
    "recievedDate": 1423101398000,
    "recievedDateString": "Wed, 4 Feb 2015 12:26:38 -0800",
    "bodyText": " Please find the link to my recent presentation",
"jsonContent": "{\"typeOfMail\": \"NormalMail\",\"normalMail\":{\"mailType\":\"NormalMail\",\"paragraphs\":[\"Pleasefindthelinktomyrecentpresentation\"]}}"
}

现在我不明白你为什么要把jsonContent作为一个字符串，因为你可以把它作为一个对象传递（删除它周围的引号让你得到

"jsonContent": {
  "typeOfMail": "NormalMail",
  "normalMail": {
    "mailType": "NormalMail",
    "paragraphs": [
      "Pleasefindthelinktomyrecentpresentation"
    ]
  }
}

如果你在javascript中需要它作为字符串，你可以JSON.stringify(json.jsonContent);来更轻松地获得相同的结果。

在JSON.parse Javascript中字符串化抛出错误的Java对象

1 个答案: