我們的登錄代碼有問題。我們不斷收到 SyntaxError: JSON Parse Error:我們縮小了在 .then(response) 行之一或 php 代碼中發生的響應錯誤。我不確定我在這里做錯了什么。有什么幫助嗎？！loginScreen.jslogin = () =>{    const { UserEmail }  = this.state ;    const { UserPassword }  = this.state ;    fetch('http://localhost:65535/login.php', {      method: 'POST',      headers: {        'Accept': 'application/json',        'Content-Type': 'application/json',      },      body: JSON.stringify({        user_email: UserEmail,        user_pass: UserPassword      })     //Error within line 59-61 or php    })    .then((response) => response.json())    .then((responseJson) => {        // If server response message same as Data Matched        if(responseJson === 'Data Matched'){            alert("Correct");             } else{            alert("Incorrect");        }     }).catch((error) => {         console.error(error);     });}  render() {    return (      <View style={styles.container}>        <ScrollView          style={styles.container}          contentContainerStyle={styles.contentContainer}>          <View style={styles.welcomeContainer}>            <Image              source={                __DEV__                  ? require('../assets/images/HootLogo.png')                  : require('../assets/images/robot-prod.png')              }              style={styles.welcomeImage}            />          </View>login.php 似乎一切都正確布局并且可以正常工作。我嘗試將 ' 更改為 ` 和所有內容。<?php // Importing DBConfig.php file. include 'DBConfig.php'; // Creating connection. $con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName); // Getting the received JSON into $json variable. $json = file_get_contents('php://input'); // decoding the received JSON and store into $obj variable. $obj = json_decode($json,true); // Populate User email from JSON $obj array and store into $email. $user_email = $obj['user_email']; // Populate Password from JSON $obj array and store into $password. $user_pass = $obj['user_pass'];