如何在 API 调用期间让我的代码在 JavaScript 中变干

Question

所以我在我的 js 文件中调用了 5 个 API，除了 URL 和我传递的表单数据之外，所有代码都是相同的。并且相同的代码行重复了 5 次，我认为这不是写作的好处。我想要的是让这段代码变干，但不知道我应该做出什么改变

var formdata = new FormData();
        formdata.append("market", "KSE100");

        var requestOptions = {
            method: "POST",
            body: formdata,
            redirect: "follow",
        };

        fetch(
            "api_url here_1",
            requestOptions
        )
        .then((response) => response.json())
        .then((stockData) => console.log('aasfs',stockData ))
        .catch((error) => console.log("error", error));



var formdata = new FormData();
            formdata.append("symbol", "SYS");
    
            var requestOptions = {
                method: "POST",
                body: formdata,
                redirect: "follow",
            };
    
            fetch(
                "api_url here_2",
                requestOptions
            )
            .then((response) => response.json())
            .then((stockData) => console.log('aasfs',stockData ))
            .catch((error) => console.log("error", error));

Answer 1

你可以像这样定义一个函数

const postData = (url, data) => {
        const formdata = new FormData();
        Object.entries(data).forEach(([k, v]) => {
          formdata.append(k, v);
        }
        

        var requestOptions = {
            method: "POST",
            body: formdata,
            redirect: "follow",
        };

        return fetch(
            url,
            requestOptions
        )
        .then((response) => response.json())
        .then((stockData) => console.log('aasfs',stockData ))
        .catch((error) => console.log("error", error));

}

Answer 2

将通用代码包装在函数中，通过变量传递表单数据和url

const sendFormData = (url, formData) => {
  var requestOptions = {
    method: "POST",
    body: formData,
    redirect: "follow",
  };

  fetch(
      url,
      requestOptions
    )
    .then((response) => response.json())
    .then((stockData) => console.log('aasfs', stockData))
    .catch((error) => console.log("error", error));
}

var formdata1 = new FormData();
formdata.append("market", "KSE100");
sendFormData("api_url here_1", formdata1);

var formdata2 = new FormData();
formdata.append("symbol", "SYS");
sendFormData("api_url here_2", formdata2);

如何在 API 调用期间让我的代码在 JavaScript 中变干

How to make my code dry in JavaScript during API Call

javascript

fetch