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我正在尝试使用 Post 方法创建登录系统，但出现错误

我用邮递员对其进行了测试，使用它就像一个魅力。

我尝试了其他事情，但无法解决此错误。如果这里有人可以解决我的问题，将不胜感激。

飞镖代码

 Future userLogin() async {
    // var data = {'username': "admin", 'password': "admin12345"};
    var response = await http
        .post(loginAPI, body: {'username': 'admin', 'password': 'admin12345'});
    var message = json.decode(response.body);
    print(message);
  }

我用来支持数据的登录脚本。

<?php

 if($_SERVER['REQUEST_METHOD']=='POST'){

 include 'DatabaseConfig.php';
 
 $con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
 
 $username = $_POST['username'];
 $password = $_POST['password'];
 
 $Sql_Query = "select * from UserDetails where username = '$username' and password = '$password'";
 
 $check = mysqli_fetch_array(mysqli_query($con,$Sql_Query));
 
 if(isset($check)){
 
 echo "Success";
 }
 else{
 echo "Invalid";
 }
 
 }else{
 echo "Again";
 }
mysqli_close($con);

?>

每当我尝试登录时，我都会收到此错误。

E/flutter ( 3935): [ERROR:flutter/lib/ui/ui_dart_state.cc(177)] Unhandled Exception: FormatException: Unexpected character (at character 1)
E/flutter ( 3935): Success
E/flutter ( 3935): ^
E/flutter ( 3935):
E/flutter ( 3935): #0      _ChunkedJsonParser.fail (dart:convert-patch/convert_patch.dart:1404:5)
E/flutter ( 3935): #1      _ChunkedJsonParser.parseNumber (dart:convert-patch/convert_patch.dart:1271:9)
E/flutter ( 3935): #2      _ChunkedJsonParser.parse (dart:convert-patch/convert_patch.dart:936:22)
E/flutter ( 3935): #3      _parseJson (dart:convert-patch/convert_patch.dart:40:10)
E/flutter ( 3935): #4      JsonDecoder.convert (dart:convert/json.dart:505:36)
E/flutter ( 3935): #5      JsonCodec.decode (dart:convert/json.dart:156:41)
E/flutter ( 3935): #6      _BodyContainerState.userLogin
package:wish_coin/Screens/LoginScreen.dart:104
E/flutter ( 3935): <asynchronous suspension>
E/flutter ( 3935): #7      _BodyContainerState.loginButtonPressed.<anonymous closure>
4

1 回答 1

0

我修复了这个问题。

<?php

 include 'DatabaseConfig.php';
 
 $con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
 
 $username = $_POST['username'];
 $password = $_POST['password'];
 
 $Sql_Query = "select * from UserDetails where username = '$username' and password = '$password'";
 
 $check = mysqli_fetch_array(mysqli_query($con,$Sql_Query));
 
 if(isset($check)){
 
 echo "Success";
 }
 else{
 echo "Invalid";

 }else{
 echo "Again";
 }
mysqli_close($con);

?>
于 2021-01-24T14:59:21.263 回答