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Flutter 颤振-参数类型';字符串';can';不能分配给参数类型';Uri';_Flutter_Dart - Fatal编程技术网

Flutter 颤振-参数类型';字符串';can';不能分配给参数类型';Uri';

Flutter 颤振-参数类型';字符串';can';不能分配给参数类型';Uri';,flutter,dart,Flutter,Dart,因此,我试图在dart中创建一个登录/注册表单: // SERVER LOGIN API URL var url = 'https://www.mywebsite.com/apicall.php'; // Store all data with Param Name. var data = {'email': email, 'password': password}; // Starting Web API Call. var response

因此,我试图在dart中创建一个登录/注册表单:

    // SERVER LOGIN API URL
    var url = 'https://www.mywebsite.com/apicall.php';

    // Store all data with Param Name.
    var data = {'email': email, 'password': password};

    // Starting Web API Call.
    var response = await http.post(url, body: json.encode(data));
但我得到了这个错误:

Error: The argument type 'String' can't be assigned to the parameter type 'Uri'.
 - 'Uri' is from 'dart:core'.
    var response = await http.post(url, body: json.encode(data));


有没有办法解决这个问题?

http.post需要一个Uri而不是字符串。试试这个:

var uri = Uri.parse('https://www.mywebsite.com/apicall.php');
var response = await http.post(uri , body: json.encode(data));

这里的错误非常明确,您使用的参数
url
不应该是字符串,而应该是对象Uri。

您应该参考以了解如何轻松使用此库:)

正确方法:
var response=wait http.post(Uri.parse(url),body:json.encode(data))var url=Uri.parse(“”https://www.mywebsite.com/apicall.php)
;http.post()需要一个对象。“string”是另一种对象。parse()是从“字符串”初始化“Uri”的一种方法。这是否回答了您的问题?顺便说一句,搜索关于特定错误消息的现有问题非常容易。