在 React Native 的 TextInput 中获取 json 数据

Question

我只是一个初学者所以帮助我！

我想在本机 TextInput 中获得 JSON 响应。[我在本机应用程序中有 2 页。在第一页 ==> 我想要该 JSON 数据并导航到带有该 JSON 响应的第二页。

我使用 PHP 作为服务器端脚本语言。我的PHP代码是：

<?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 column from JSON $obj array and store into $coulmn.

$firstname = $obj['firstname'];

$lastname = $obj['lastname'];

//$mobileno = $obj['mobileno'];

$email = $obj['email'];

$profession = $obj['profession'];

$mobileno = '7874853188';

//Applying User Login query with mobile number match.

$Sql_Query = "select firstname,lastname,email,profession from member where mobileno = '$mobileno' ";

// Executing SQL Query.

$check = mysqli_fetch_array(mysqli_query($con,$Sql_Query));

if(isset($check)){

 $SuccessLoginMsg = 'Data Matched';

 // Converting the message into JSON format.

$SuccessLoginJson = json_encode($SuccessLoginMsg);

 $first_name = $check[0];

 $last_name = $check[1];

 $email = $check[2];

 $profession = $check[3];

// Echo the message.

 echo $SuccessLoginJson ; 

 }

 else{

 // If the record inserted successfully then show the message.

$InvalidMSG = 'Enter valid phone number' ;

// Converting the message into JSON format.

$InvalidMSGJSon = json_encode($InvalidMSG);

// Echo the message.

 echo $InvalidMSGJSon ;

 }

 mysqli_close($con);

?>

上面的代码是 100% 正确的。[在网页上测试]

在我的本机代码中，首先我检查手机号码 => 如果手机号码正确[存在于数据库中]，那么该用户可以使用该 JSON 响应转到下一页。

Answer 1

您收到的某些数据似乎不是string value。您只能设置string valuefor Textinput。检查接收到的数据的值并确保数据是string.如果你只是想摆脱错误，navigation.navigate("Profile",&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {EmailId: JSON.stringify($email),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Name: JSON.stringify($firstname+$lastname),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Profe : JSON.stringify($profession),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; mobileno : JSON.stringify(UserMNO),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; });