我正在通过Javascript调用帖子请求,这是它的外观,

function syncDeviceId(deviceID, mod){

var request = new Request('url', {

method: 'POST',

body: JSON.stringify({

uuid: unique_id,

}),

mode: 'cors'

})

fetch(request).then(function(data) {

return

})

而我正试图改变这样的价值观,

$post['uuid'] = $_POST['uuid'];

?>

这将返回为空,如何从PHP中的fetch post请求中检索值.谢谢

解决方法:

这是因为您没有将Request的正文设置为正确的格式.

07001

body: Any body that you want to add to your request: this can be a

Blob, BufferSource, FormData, URLSearchParams, or USVString object.

Note that a request using the GET or HEAD method cannot have a body.

因此,将body设置为JSON字符串会使content-type标头成为text / plain.即使您将请求Content-Type设置为application / json也没关系,因为PHP默认情况下不知道如何解析传入的JSON请求有效负载(除非它最近在PHP 7中添加).

你可以在客户端做几件事

从对象创建一个FormData对象,并将其用作主体,将使用multipart / form内容类型

var data = {some:"data",even:"more"};

var fd = new FormData();

//very simply, doesn't handle complete objects

for(var i in data){

fd.append(i,data[i]);

}

var req = new Request("url",{

method:"POST",

body:fd,

mode:"cors"

});

创建一个URLSearchParams对象,将内容类型设置为application / x-www-form-urlencoded.注意：URLSearchParams不受广泛支持

//Similar to creating a simple FormData object

var data = {some:"data",even:"more"};

var params = new URLSearchParams();

for(i in data){

params.append(i,data[i]);

}

var req = new Request("url",{

method:"POST",

body:params,

mode:"cors"

});

创建一个查询字符串(即a = hello& b = world)并使用Headers对象将Content-Type设置为application / x-form-urlencoded

var data = {some:"data",even:"more"};

var headers = new Headers({

"Content-Type":"application/x-form-urlencoded"

});

var params = [];

for(i in data){

params.push(i + "=" + encodeURIComponent(data[i]));

}

var req = new Request("url",{

method:"POST",

body:params.join("&"),

headers:headers,

mode:"cors"

});

如果您仍想发送JSON有效负载而不是执行上述操作,则必须读取原始请求输入,然后使用json_decode获取数据

$json = file_get_contents('php://input');

$data = json_decode($json);

标签：php,javascript

来源： https://codeday.me/bug/20190609/1201578.html