Flutter调用本地API遇FormatException及断言失败问题求助

调用本地API地址127.0.0.1/app/sanjhapizza/lib/userLogin.php时收到FormatException错误，报错信息如下：

Exception:FormatException: //127.0.0.1/app/sanjhapizza/lib/userLogin，Api result:failed，Error: Assertion failed

以下是用户提供的API调用代码：

import 'dart:convert';
import "package:http/http.dart" as http;
Future<String> userLogin(String userid) async {
try {
String ulink = "http://127.0.0.1/app/sanjhapizza/lib/userLogin.php";
final response = await http.post(Uri.http(ulink),
body: {"userid": userid});
print('Response status: ${response.statusCode}');
print('Response body: ${response.body}');
} on Exception catch (e) {
print("Exception:$e");
return ("failed");
}
}

问题核心 #

Uri.http()方法的用法错误。这个方法需要分开传入主机地址和路径，不能直接把完整的http://开头的URL传进去，否则会导致URL解析格式混乱，触发FormatException。

修复方案 #

有两种简单的修复方式，选其一即可：

方式1：拆分主机和路径调用Uri.http()

import 'dart:convert';
import "package:http/http.dart" as http;

Future<String> userLogin(String userid) async {
  try {
    // 拆分主机和路径
    final host = "127.0.0.1";
    final path = "/app/sanjhapizza/lib/userLogin.php";
    final response = await http.post(Uri.http(host, path),
        body: {"userid": userid});
    print('Response status: ${response.statusCode}');
    print('Response body: ${response.body}');
    // 补充原代码缺失的返回值
    return response.body;
  } on Exception catch (e) {
    print("Exception:$e");
    return "failed";
  }
}

方式2：用Uri.parse()直接解析完整URL（更直观）

import 'dart:convert';
import "package:http/http.dart" as http;

Future<String> userLogin(String userid) async {
  try {
    String ulink = "http://127.0.0.1/app/sanjhapizza/lib/userLogin.php";
    // 用Uri.parse替代Uri.http，直接解析完整URL
    final response = await http.post(Uri.parse(ulink),
        body: {"userid": userid});
    print('Response status: ${response.statusCode}');
    print('Response body: ${response.body}');
    // 补充原代码缺失的返回值
    return response.body;
  } on Exception catch (e) {
    print("Exception:$e");
    return "failed";
  }
}

额外提示 #

• 如果在真机上测试，127.0.0.1指向的是设备自身，需要换成电脑的局域网IP（比如192.168.x.x）
• 确保本地API服务正常运行，若API不是用默认80端口，要在URL里加上端口号（比如http://127.0.0.1:8080/...）
• 原代码try块末尾缺少返回值，会触发编译错误，修复后的代码已补充返回逻辑

内容的提问来源于stack exchange，提问作者Raman Thakur