我有一个小问题,当通过前端的ajax而不是来自用户表的userName时,JSON只返回来自一个表的数据。用var_dump检查后端看起来没问题:
/home/maciek/Workspace/Communic/public/admin/privMessage.php:11:
array (size=3)
0 =>
object(Privatemessage)[6]
private 'id' => string '4' (length=1)
private 'senderId' => string '2' (length=1)
private 'receiverId' => string '1' (length=1)
private 'creationDate' => string '2017-06-28 23:49:15' (length=19)
private 'text' => string 'asdasdasda' (length=10)
private 'readStatus' => string '1' (length=1)
**private 'userName' => string 'stefan' (length=6)**
MySQL查询(单独正确执行并返回所需的结果-结果中包含用户名):
SELECT p.*, u.username FROM PrivateMessage p RIGHT JOIN Users u ON p.sender_id=u.id WHERE receiver_id=:receiver_id
类Privatemessage中使用该查询的方法:
static public function loadAllRcvdPrvMsgsByUserId(PDO $pdo, $receiverId) {
$stmt = $pdo->prepare("SELECT p.*, u.username FROM PrivateMessage p RIGHT JOIN Users u ON p.sender_id=u.id WHERE receiver_id=:receiver_id");
$result = $stmt->execute([
'receiver_id' => $receiverId
]);
$rcvdPrvMsgsArray = [];
if ($result === true && $stmt->rowCount() > 0) {
while ($row = $stmt->fetchAll(PDO::FETCH_OBJ)) {
foreach ($row as $dbPrvMessage) {
$loadedPrvMsg = new Privatemessage($pdo);
$loadedPrvMsg->id = $dbPrvMessage->id;
$loadedPrvMsg->senderId = $dbPrvMessage->sender_id;
$loadedPrvMsg->receiverId = $dbPrvMessage->receiver_id;
$loadedPrvMsg->creationDate = $dbPrvMessage->privatemessage_datetime;
$loadedPrvMsg->text = $dbPrvMessage->privatemessage_text;
$loadedPrvMsg->readStatus = $dbPrvMessage->privatemessage_readstatus;
$loadedPrvMsg->userName = $dbPrvMessage->username;
$rcvdPrvMsgsArray[] = $loadedPrvMsg;
}
}
return $rcvdPrvMsgsArray;
}
return null;
}
js ajax:
function getReceivedPrivateMsg() {
$
.ajax({
url: '../../../rest/rest.php/privateMessage',
type: 'GET'
})
.done(function (response) {
console.log(response.success);
})
.fail(function (error) {
console.log('Create sent private message error', error);
});
}
console.log(response.success);in ajax在Chrome开发控制台中返回以下内容(同样,缺少userName ):
任何帮助都是非常感谢的!
编辑:我在Privatemessage类中实现了JsonSerializable,但忘记了在类中的jsonSerialize()方法中返回userName。
发布于 2018-07-23 03:03:45
您的类Privatemessage可能需要有一个公共$username属性才能正常工作。这取决于您是如何实现JSON转换的。从注释中可以看出,您似乎正在使用JsonSerializable,因此您需要确保在jsonSerialize方法中考虑到所有当前字段。
看一下您的代码,如果目标只是生成一个JSON响应,我不明白在这里创建Privatemessage实例的意义。为什么不直接使用从PDO返回的现有对象呢?
return $stmt->fetchAll(PDO::FETCH_OBJ);
https://stackoverflow.com/questions/51468335
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