我如何 post 使用 fetch api 形成数据?

How do I post form data with fetch api?

我的代码:

fetch("api/xxx", {
    body: new FormData(document.getElementById("form")),
    headers: {
        "Content-Type": "application/x-www-form-urlencoded",
        // "Content-Type": "multipart/form-data",
    },
    method: "post",
}

我尝试 post 使用 fetch api 我的表单,它发送的正文如下:

-----------------------------114782935826962
Content-Disposition: form-data; name="email"

test@example.com
-----------------------------114782935826962
Content-Disposition: form-data; name="password"

pw
-----------------------------114782935826962--

(不知道为什么boundary里的数字每次发送都变...)

我想用"Content-Type"发送数据:"application/x-www-form-urlencoded",我该怎么办?或者如果我只需要处理它,我该如何解码控制器中的数据?


谁回答我的问题,我知道我可以做到:

fetch("api/xxx", {
    body: "email=test@example.com&password=pw",
    headers: {
        "Content-Type": "application/x-www-form-urlencoded",
    },
    method: "post",
}

我想要的是 jQuery 中的 $("#form").serialize() (w/o 使用 jQuery)或解码方式 mulitpart/form-data 在控制器中。谢谢你的回答。

您可以将 body 设置为 URLSearchParams 的实例,并将查询字符串作为参数传递

fetch("/path/to/server", {
  method:"POST"
, body:new URLSearchParams("email=test@example.com&password=pw")
})

document.forms[0].onsubmit = async(e) => {
  e.preventDefault();
  const params = new URLSearchParams([...new FormData(e.target).entries()]);
  // fetch("/path/to/server", {method:"POST", body:params})
  const response = await new Response(params).text();
  console.log(response);
}
<form>
  <input name="email" value="test@example.com">
  <input name="password" value="pw">
  <input type="submit">
</form>

客户端

不设置 content-type header.

// Build formData object.
let formData = new FormData();
formData.append('name', 'John');
formData.append('password', 'John123');

fetch("api/SampleData",
    {
        body: formData,
        method: "post"
    });

服务器

使用 FromForm 属性指定绑定源是表单数据。

[Route("api/[controller]")]
public class SampleDataController : Controller
{
    [HttpPost]
    public IActionResult Create([FromForm]UserDto dto)
    {
        return Ok();
    }
}

public class UserDto
{
    public string Name { get; set; }
    public string Password { get; set; }
}

引用MDN on FormData(强调我的):

The FormData interface provides a way to easily construct a set of key/value pairs representing form fields and their values, which can then be easily sent using the XMLHttpRequest.send() method. It uses the same format a form would use if the encoding type were set to "multipart/form-data".

因此,当您使用 FormData 时,您将自己锁定在 multipart/form-data 中。无法发送 FormData object 作为 body 和 而不是 multipart/form-data 格式发送数据。

如果要将数据作为 application/x-www-form-urlencoded 发送,则必须将 body 指定为 URL-encoded 字符串,或者传递 URLSearchParams object. The latter unfortunately cannot be directly initialized from a form element. If you don’t want to iterate through your form elements yourself (which you could do using HTMLFormElement.elements),您也可以从 FormData object:

创建 URLSearchParams object
const data = new URLSearchParams();
for (const pair of new FormData(formElement)) {
    data.append(pair[0], pair[1]);
}

fetch(url, {
    method: 'post',
    body: data,
})
.then(…);

请注意,您不需要自己指定 Content-Type header。


如评论中 monk-time 所述,您还可以创建 URLSearchParams 并直接传递 FormData object,而不是在循环中附加值:

const data = new URLSearchParams(new FormData(formElement));

尽管浏览器仍对它提供一些实验性支持,因此请务必在使用前对其进行正确测试。

使用FormDatafetch抓取和发送数据

fetch(form.action, {method:'post', body: new FormData(form)});

function send(e,form) {
  fetch(form.action, {method:'post', body: new FormData(form)});

  console.log('We send post asynchronously (AJAX)');
  e.preventDefault();
}
<form method="POST" action="myapi/send" onsubmit="send(event,this)">
    <input hidden name="csrfToken" value="a1e24s1">
    <input name="email" value="a@b.com">
    <input name="phone" value="123-456-789">
    <input type="submit">    
</form>

Look on chrome console>network before/after 'submit'

到 post 使用 fetch api 形成数据, 试试这个对我有用的代码^_^

function card(fileUri) {
let body = new FormData();
let formData = new FormData();
formData.append('file', fileUri);

fetch("http://X.X.X.X:PORT/upload",
  {
      body: formData,
      method: "post"
  });
 }

要添加上面的好答案,您还可以避免在 HTML 中明确设置操作并在 javascript 中使用事件处理程序,使用“this”作为创建“FormData”的表单" 对象

Html 形式 :

<form id="mainForm" class="" novalidate>
<!--Whatever here...-->
</form>

在你的 JS 中:

$("#mainForm").submit(function( event ) {
  event.preventDefault();
  const formData = new URLSearchParams(new FormData(this));
  fetch("http://localhost:8080/your/server",
    {   method: 'POST',
        mode : 'same-origin',
        credentials: 'same-origin' ,
        body : formData
    })
    .then(function(response) {
      return response.text()
    }).then(function(text) {
        //text is the server's response
    });
});

‍这些可以帮到你:

let formData = new FormData();
            formData.append("name", "John");
            formData.append("password", "John123");
            fetch("https://yourwebhook", {
              method: "POST",
              mode: "no-cors",
              cache: "no-cache",
              credentials: "same-origin",
              headers: {
                "Content-Type": "form-data"
              },
              body: formData
            });
            //router.push("/registro-completado");
          } else {
            // doc.data() will be undefined in this case
            console.log("No such document!");
          }
        })
        .catch(function(error) {
          console.log("Error getting document:", error);
        });

使用 fetch api 结果表明您不必包含 headers "Content-type": "multipart/form-data".

因此以下工作:

let formData = new FormData()
formData.append("nameField", fileToSend)

fetch(yourUrlToPost, {
   method: "POST",
   body: formData
})

请注意,对于 axios,我必须使用 content-type。

@KamilKiełczewski 如果您同意表单数据格式采用表单多部分样式,那么答案很好,但是如果您需要以查询参数样式提交的表单:

You can also pass FormData directly to the URLSearchParams constructor if you want to generate query parameters in the way a would do if it were using simple GET submission.

        form = document.querySelector('form')
        const formData = new FormData(form);
        formData["foo"] = "bar";
        const payload = new URLSearchParams(formData)
        fetch(form.action, payload)