Question

我正在使用此代码传递userName变量...

return fetch('http://creat1vedesign.com/userTabs4.php', {
  method: 'POST',
  headers: {
    Accept: 'application/json',
            'Content-Type': 'application/json',
  },
}, {
  body: JSON.stringify({
      userName: 'carolf'
  }),

...此php脚本...

<?php
include 'DBConfig.php';
$con = new mysqli($HostName, $HostUser, $HostPass, $DatabaseName);
    $json = file_get_contents('php://input');
    $obj = json_decode($json,true);
    $userName = $obj['userName'];
$sql = "select * from Users where userName = '$userName'";
if ($result = mysqli_query($con,$sql)) { 
 while($row[] = $result->fetch_assoc()) {
    $tem = $row;
    $json = json_encode($tem);
 }
} else {
 echo "No Results Found.";
}
 echo $json;
$conn->close();
?>

...以检索特定条件的数据。但是或者React Native没有发送数据或者php脚本没有接收到数据或者语法不正确或者我在做其他错误的事情：（

有什么想法吗？

Answer 1

您可以通过在参数“？”之后添加参数来将参数传递给获取URL中的PHP脚本。像这样：

fetch("http://creat1vedesign.com/userTabs4.php?userName=carolf",{
        method:'POST',
        headers:{
          'Accept': 'application/json',
          'Content-Type': 'application/json',
        },
      })

      .then( (response) => {
            return response.json() })   
                .then( (json) => {
                    console.log(json)
                });

只需确保在您的php脚本中包含$ _GET：

<?php

$userName= $_GET["userName"];

include 'DBConfig.php';
$con = new mysqli($HostName, $HostUser, $HostPass, $DatabaseName);
$json = file_get_contents('php://input');
$obj = json_decode($json,true);

$sql = "select * from Users where userName = '$userName'";
if ($result = mysqli_query($con,$sql)) { 
 while($row[] = $result->fetch_assoc()) {
    $tem = $row;
    $json = json_encode($tem);
 }
} else {
 echo "No Results Found.";
}
 echo $json;
$conn->close();
?>

如何将数据从React Native应用传递到PHP脚本？

1 个答案: