Question

我正在尝试将JSON数据从Javascript Ajax解析为PHP，我在控制台中收到错误消息。如何将json编码为字符串格式。我做错了什么，这是我的附加代码。

function datasend(obj)
{
var flickr = {"name": "vivek", "age":"18"};
var data = JSON.stringify(flickr);

alert(data);
var request = new XMLHttpRequest();
request.open("POST", "getvalue.php", true);

request.setRequestHeader("Content-Type", "application/json");

request.onreadystatechange=function() {
      if (request.readyState == 4 && request.status == 200) {
    // Success!
           alert("yes");
       var resp = request.responseText;
       document.getElementById("sum").innerHTML=resp;
       } 
    }
request.send(data);
}

<?php

header('Content-type: application/json');
$json = file_get_contents('php://input');

$json_decode = json_decode($json, true);

$getnme=$json_decode->{'name'};
$json_response = json_encode($json_decode);
echo $getnme;

include 'db.php';

$sql= "UPDATE user1 SET name='".$getnme."' WHERE id=1";

if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully";
} 
?>

我想通过更新查询将我的名字发送到数据库，这里我在解码json数据后没有得到我的名字。

Answer 1

试试这种方式

写一个$getnme=$json_decode['name'];而不是$getnme=$json_decode->{'name'};

Answer 2

var request = new XMLHttpRequest();
request.open("POST", "getvalue.php", true);
xhttp.open("POST", "ajax_test.asp", true);
xhttp.setRequestHeader("Content-type", "application/JSON");
xhttp.send(JSON.stringify({"name": "vivek", "age":"18"}));

并在php方面

<?php
 $obj = json_decode($json);
    $getnme= $obj->name;
    echo $getnme;
    ?>

将json数据从javascript Ajax发送到php

2 个答案: