如何检查抓取的响应是否是javascript中的json对象

javascript json fetch-api

2022-08-30 02:35:37

我正在使用获取polyfill从URL中检索JSON或文本，我想知道如何检查响应是JSON对象还是仅文本

fetch(URL, options).then(response => {
   // how to check if response has a body of type json?
   if (response.isJson()) return response.json();
});

答案 1

您可以检查响应，如以下 MDN 示例所示：content-type

fetch(myRequest).then(response => {
  const contentType = response.headers.get("content-type");
  if (contentType && contentType.indexOf("application/json") !== -1) {
    return response.json().then(data => {
      // The response was a JSON object
      // Process your data as a JavaScript object
    });
  } else {
    return response.text().then(text => {
      // The response wasn't a JSON object
      // Process your text as a String
    });
  }
});

如果您需要绝对确定内容是有效的JSON（并且不信任标头），则始终可以接受响应并自行解析：text

fetch(myRequest)
  .then(response => response.text()) // Parse the response as text
  .then(text => {
    try {
      const data = JSON.parse(text); // Try to parse the response as JSON
      // The response was a JSON object
      // Do your JSON handling here
    } catch(err) {
      // The response wasn't a JSON object
      // Do your text handling here
    }
  });

异步/等待

如果你正在使用，你可以用更线性的方式编写它：async/await

async function myFetch(myRequest) {
  try {
    const reponse = await fetch(myRequest);
    const text = await response.text(); // Parse it as text
    const data = JSON.parse(text); // Try to parse it as JSON
    // The response was a JSON object
    // Do your JSON handling here
  } catch(err) {
    // The response wasn't a JSON object
    // Do your text handling here
  }
}

答案 2

您可以使用帮助程序函数干净利落地执行此操作：

const parseJson = async response => {
  const text = await response.text()
  try{
    const json = JSON.parse(text)
    return json
  } catch(err) {
    throw new Error("Did not receive JSON, instead received: " + text)
  }
}

然后像这样使用它：

fetch(URL, options)
.then(parseJson)
.then(result => {
    console.log("My json: ", result)
})

这将引发错误，因此您可以根据需要进行。catch