mysql在两个不同的表中选择Query

Question

如何使用一个php脚本从sql中的两个不同表进行查询并将其解析为一个json对象？谢谢！

这是我的php

<?php  



$sql = "select * from schedule;"; 
 $sql1 = "select * from matches;"; 
 $con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$db_name); 

 $result = mysqli_query($con,$sql);
 $result1 = mysqli_query($con,$sql1);


 $response = array();
while($row=mysqli_fetch_array($result))
{
array_push($response, array("n_name"=>$row[1],"start"=>$row[4],"end"=>$row[5],"venue"=>$row[6]));
 }

 $data = array();
while($row=mysqli_fetch_array($result1))
{
array_push($data, array("teamone"=>$row[1]));
 }

echo json_encode (array("schedule_response"=>$response, $data));

mysqli_close($con);
?>

这是输出

{"schedule_response":
    [{
        "n_name":null,
        "start":"2016-11-09 00:00:00",
        "end":"2016-11-16 00:00:00",
        "venue":"aaaaaa"
     }],
    "0":[{"teamone":"aaa"}]}

我想要的是这样的东西

{"schedule_response":
    [{
        "n_name":null,
        "start":"2016-11-09 00:00:00",
        "end":"2016-11-16 00:00:00",
        "venue":"aaaaaa",
        "teamone":"aaa"
    }]
}