php - JsonArray 返回空值

Question

下面是我的 php 代码:

public function getMainChatList($myPhoneNo){

    $stmt = $this->conn->prepare("SELECT receiverPhoneNo,name FROM users,friend WHERE users.phoneNo=friend.receiverPhoneNo AND senderPhoneNo=? AND chatted = 'y' ORDER BY update_time DESC");
    $stmt->bind_param("s", $myPhoneNo);
    $stmt->execute();
    $stmt->store_result();

    $result = array();

    while($row = $stmt->fetch()){
         array_push($result,array('receiverPhoneNo'=>$row['receiverPhoneNo'],'name'=>$row['name'],));
     }              
     //echo json_encode(array("result"=>$result));
     echo json_encode($result);
     echo json_last_error();
     $stmt->close();        
    }

返回json

[{"receiverPhoneNo":null,"name":null},{"receiverPhoneNo":null,"name":null}]0

json_last_error() 返回 0。我不知道它为什么返回 null。而我直接在xampp MySQL服务器上执行sql语句。下面是结果。

result.jpg谢谢!

Answer 1

只需使用 bind_result()相反:

public function getMainChatList($myPhoneNo)
{
    $stmt = $this->conn->prepare("
        SELECT receiverPhoneNo, name FROM users, friend 
        WHERE users.phoneNo = friend.receiverPhoneNo 
        AND senderPhoneNo = ? 
        AND chatted = 'y' 
        ORDER BY update_time DESC
    ");
    $stmt->bind_param('s', $myPhoneNo);
    $stmt->execute();
    // bind
    $stmt->bind_result($receiverPhoneNo, $name);

    $result = array();

    while($stmt->fetch()){
        $result[] = array(
            'receiverPhoneNo' => $receiverPhoneNo,
            'name' => $name,
        );
    }

    echo json_encode($result);
}