Php 从MYSQL数据库创建JSON字符串

我试图以json字符串的形式从MySQL数据库获取数据

我读到这个答案：

但这仅限于一张桌子。如果我想从多个表中获取数据（userDetails中的name、UserPurchases中的purchase数据等），该怎么办？如何创建自定义字符串，从多个表中获取数据，并像只从一个表中创建json字符串一样创建json字符串

 $query = "SELECT * FROM `thing` WHERE `id` = :thingId";
    $stmt = $dbh->prepare ( $query );
    $stmt->bindParam ( ":thingId" , $_GET['thingId']  );
    $stmt->execute ( );
    $rslt  =  $stmt->fetch ( );
    $thingName  = $rslt['name'];
    $thingOwnerId = $rslt['userId'];
    $thingDescription = $rslt['thingDescription'];

// Getting the thing owner details
    $query = "SELECT * from `user` WHERE ( `id` = :id ) ";    
    $stmt = $dbh->prepare( $query );
    $stmt->bindParam ( ":id" , $thingOwnerId );
    $stmt->execute(  );
    $rslt = $stmt->fetch ( );
    $thingOwnerName = $rslt['firstName']." ".$rslt['lastName'];

现在，我们来看看如何从单独的表中使用这些数据来创建一个json。

---

字符串应该有thingName、thingOwnerId、thingDescription、thingOwnerName。

如果它们是不同的查询，您可以合并结果并对该数组进行如下编码：

$merged = array();

 $query = "SELECT * FROM `thing` WHERE `id` = :thingId";
    $stmt = $dbh->prepare ( $query );
    $stmt->bindParam ( ":thingId" , $_GET['thingId']  );
    $stmt->execute ( );
    $rslt  =  $stmt->fetch ( );
$merged['thing'] = $rslt;    

// Getting the thing owner details
    $query = "SELECT * from `user` WHERE ( `id` = :id ) ";    
    $stmt = $dbh->prepare( $query );
    $stmt->bindParam ( ":id" , $thingOwnerId );
    $stmt->execute(  );
    $rslt = $stmt->fetch ( );
$merged['user'] = $rslt;

echo json_encode($merged);

---

您还可以在PHP中创建一个类，将数据库值设置为该类并编码为JSON，例如：

<?php

class MyCustomJson
{
    public $userId;
    public $thingId;
    //and go on...
}

//set db values to this class
$myCustomJson = new MyCustomJson();
//..read data from User table and set table values to your custom object
$myCustomJson->userId = $row["id"];
//..read data from Thing table and set table values to your custom object
$myCustomJson->thingId = $row["id"];

//and encode your custom object to json
echo json_encode($myCustomJson);
?>

---

从数组中的查询中收集所需数据，然后以JSON编码格式将该数组输出到浏览器。记住在任何输出之前设置

内容类型：application/json

头

PHP

---

你能告诉我如何返回json格式的数据吗。不仅仅是echo，jSON格式？当您执行jSON_encode（）时，它已经返回一个jSON。你可以做$var=json\u编码（$something）；而是使用echo。没关系吧？我想没有。我想知道twitter的功能。echo json_编码（$something）；应该像这个twitter示例那样打印json。如果没有打印json，那么在此之前您就出了问题。你能用实际的代码更新你的问题来解决吗？我正在得到结果。只是我在firefox中查看json数据的插件没有将其识别为json。也许这个插件有问题。谢谢<代码>内容类型帮助！

//collect your data in an array
$data=array(
    'name'=>$thingOwnerName,
    'description'=>$thingDescription,
    'otherField'=>$someValue
    ...
);

//send JSON header to browser
header('Content-Type: application/json');

//echo JSON encoded data
echo json_encode($data);