php 文件:
$email=$_POST['email']; $passworda=$_POST['passworda']; $sql="SELECT * FROM user WHERE email='".$email."'AND passworda='".$password."' "; $result=mysqli_query($db,$sql); $count=mysqli_num_rows($result); if($count>=1){ echo json_encode("success"); } else { echo json_encode("error"); }
登录页面 Flutter:
class Login extends StatelessWidget { TextEditingController email = TextEditingController(); TextEditingController password = TextEditingController(); Future login(BuildContext cont) async { if (email.text == "" || password.text == "") { Fluttertoast.showToast( msg: "please complete!", toastLength: Toast.LENGTH_SHORT, gravity: ToastGravity.CENTER, fontSize: 16.0, ); } else { var url = "http://192.168.43.150/v1_flutter/lib/php/connection.php"; var response = await http.post(Uri.parse(url), body: { "email": email.text, "pass": password.text, }, headers: {"Accept":"applicarion/json"}); var data = jsonDecode(response.body); if (data == "success") { Navigator.pop(cont); Navigator.pushNamed(cont, "/registre"); } else { Fluttertoast.showToast( msg: "The user and password does not exist!", toastLength: Toast.LENGTH_SHORT, gravity: ToastGravity.CENTER, fontSize: 16.0, ); } }}
控制台:
E/flutter (6084): [错误:flutter/lib/ui/ui_dart_state.cc(198)] 未处理的异常:FormatException:意外的字符(位于 字符 1) E/颤振(6084):
E/颤动(6084):^ E/颤振(6084): E/颤振(6084): #0 _ChunkedJsonParser.fail (dart:convert-patch/convert_patch.dart:1383:5) E/颤振(6084): #1 _ChunkedJsonParser.parseNumber (dart:convert-patch/convert_patch.dart:1250:9) E/颤振(6084): #2 _ChunkedJsonParser.parse (dart:convert-patch/convert_patch.dart:915:22) E/颤振(6084): #3 _parseJson (dart:convert-patch/convert_patch.dart:35:10) E/颤振(6084): #4 JsonDecoder.convert (dart:convert/json.dart:612:36) E/颤振(6084): #5 JsonCodec.decode (dart:convert/json.dart:216:41) E/颤振(6084): #6 jsonDecode (dart:convert/json.dart:155:10) E/颤振(6084): #7 Login.login (package:mes_v1/pages/Authentification/login.dart:25:18) E/颤振(6084): E/颤振(6084):
P粉2447306252024-03-27 00:10:26
您的代码看起来确实容易出错,但重点关注您的问题:异常看起来很清楚,您的响应解析正在中断。 (我还想说,你应该用 try/catch 包装你的代码,以防止任何代码破坏和正确捕获问题)。
让我们解决您的问题:
var data = jsonDecode(response.body);
这会将您的字符串 (response.body) 转换为 json,这意味着 data
是 Map
或 List
if(数据==“成功”){
现在,让我们看一下您的 php 代码:
echo json_encode("成功");
我不是 php 专家,但从文档来看它应该像下面这样使用:
$response = array("result" => "Success"); echo json_encode($response);
现在让我们回到您的 dart 代码:
var success = false; try{ final baseUrl = "http://192.168.43.150"; // Use final wherever you can final url = "$baseUrl/v1_flutter/lib/php/connection.php"; // Improving flexibility final body ={ "email": email.text, "pass": password.text, }; // Decouple in variables for readability final headers = {"Accept":"application/json"}; // Fix typo final response = await http.post(Uri.parse(url), body: body, headers: headers); final data = jsonDecode(response.body); success = data["result"] == "Success"; } catch(e) { print("Catched an error!"); print(e); success = false; } if(success) { ...