如何使用PHP中的fetch()API POST方法获取数据？

Question

我正在尝试使用fetch()API POST方法来获取PHP中的POST数据.

这是我尝试过的:

var x = "hello";
fetch(url,{method:'post',body:x}).then(function(response){
    return response.json();
});

PHP:

<?php
if(isset($_GET['x']))
{
    $get = $_GET['x'];
    echo $get;
}
?>

它是否正确？

Answer 1

这取决于:

如果需要$_GET['x'],您需要在查询字符串中发送数据:

var url = '/your/url?x=hello';

fetch(url)
.then(function (response) {
  return response.text();
})
.then(function (body) {
  console.log(body);
});

如果您愿意$_POST['x'],您需要将数据发送为FormData:

var url = '/your/url';
var formData = new FormData();
formData.append('x', 'hello');

fetch(url, { method: 'POST', body: formData })
.then(function (response) {
  return response.text();
})
.then(function (body) {
  console.log(body);
});

Answer 2

显然,当使用Fetch API将数据发送到PHP服务器时,您必须处理与您习惯的请求略有不同的请求.

您正在"POST"或"GETting"的数据在超级全局变量中不可用,因为此输入不是来自 多部分数据表单或application/x-www-form-urlencoded

您可以通过阅读特殊文件获取数据:php://input例如,使用file_get_contents('php://input')然后尝试解码该输入json_decode().

希望它有所帮助.

你可以在这里阅读更多相关内容:

https://codepen.io/dericksozo/post/fetch-api-json-php

Answer 3

我使用MDN 中的 postData 函数：

/**
 * send_body___receive_response.js
 *
 * Can of-course be in <script> tags in HTML or PHP
 */

async function postData( url='', data={ } ) {
  // *starred options in comments are default values
  const response = await fetch(
    url,
    {
      method: "POST", // *GET, POST, PUT, DELETE, etc.
      mode: "same-origin", // no-cors, *cors, same-origin
      cache: 'no-cache', // *default, no-cache, reload, force-cache, only-if-cached
      credentials: "same-origin", // include, *same-origin, omit
      headers: {
        "Content-Type": "application/json",  // sent request
        "Accept":       "application/json"   // expected data sent back
      },
      redirect: 'follow', // manual, *follow, error
      referrerPolicy: 'no-referrer', // no-referrer, *no-referrer-when-downgrade, origin, origin-when-cross-origin, same-origin, strict-origin, strict-origin-when-cross-origin, unsafe-url
      body: JSON.stringify( data ), // body data type must match "Content-Type" header
    },
  );

  return response.json( ); // parses JSON response into native JavaScript objects
}

const data = {
  'key1': 'value1',
  'key2': 2
};

postData( 'receive_body___send_response.php', JSON.stringify( data ) )
  .then( response => {
    // Manipulate response here
    console.log( "response: ", response ); // JSON data parsed by `data.json()` call
    // In this case where I send entire $decoded from PHP you could arbitrarily use this
    console.log( "response.data: ", JSON.parse( response.data ) );
  } );

你可以只发布数据，但我喜欢收到它成功的响应。

/**
 * receive_body___send_response.php
 */

/* Get content type */
$contentType = trim($_SERVER["CONTENT_TYPE"] ?? ''); // PHP 8+
// Otherwise:
// $contentType = isset($_SERVER["CONTENT_TYPE"]) ? trim($_SERVER["CONTENT_TYPE"]) : '';

/* Send error to Fetch API, if unexpected content type */
if ($contentType !== "application/json")
  die(json_encode([
    'value' => 0,
    'error' => 'Content-Type is not set as "application/json"',
    'data' => null,
  ]));

/* Receive the RAW post data. */
$content = trim(file_get_contents("php://input"));

/* $decoded can be used the same as you would use $_POST in $.ajax */
$decoded = json_decode($content, true);

/* Send error to Fetch API, if JSON is broken */
if(! is_array($decoded))
  die(json_encode([
    'value' => 0,
    'error' => 'Received JSON is improperly formatted',
    'data' => null,
  ]));

/* NOTE: For some reason I had to add the next line as well at times, but it hadn't happen for a while now. Not sure what went on */
// $decoded = json_decode($decoded, true);

/* Do something with received data and include it in response */
// dumb e.g.
$response = $decoded['key2'] + 1; // 3

/* Perhaps database manipulation here? */
// query, etc.

/* Send success to fetch API */
die(json_encode([
  'value' => 1,
  'error' => null,
  'data' => null, // or ?array of data ($response) you wish to send back to JS
]));

Answer 4

请记住，$_POST在 PHP 中仅获取formData()或urlSearchParams()数据，对于其他所有类型的数据，尤其是从其他文件导入的数据或外部 api 数据，您必须遵循以下步骤。 脚步：

1. 用来file_get_contents(php://input)接收php中的数据
2. json_decode($data) 在读/写数据库后使用 and finally 对其进行解码
3. 使用 发送响应json_encode($response)。

这很容易 ：-））

Answer 5

如果碰巧您需要使用现有的服务器，它是用 $_POST 编码的，并且期望来自普通表单的参数，而不是编码为 JSON，您可以使用 formdata，这正是模拟表单。

let fd = new FormData();
fd.append("var1", val)
fd.append("var2", "Hello World");
fetch('/servers/server.php', {method: "POST", body: fd})

这样，您的服务器将收到 POST 字段，就像它们来自普通表单输入字段一样。

$var1 = $_POST['var1']; $var2 = $_POST['var2'];