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Php 2在一个json对象上选择不同数组中的查询_Php_Mysql_Json - Fatal编程技术网

Php 2在一个json对象上选择不同数组中的查询

Php 2在一个json对象上选择不同数组中的查询,php,mysql,json,Php,Mysql,Json,如何对从数据库中的2个不同表检索数据的两个数组进行编码,并在1个json响应中对其进行编码 Here is my 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); $re

如何对从数据库中的2个不同表检索数据的两个数组进行编码,并在1个json响应中对其进行编码

 Here is my 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"=>$row[1], "teamtwo"=>$row[2], "s_name"=>$row[10]));
 }



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





mysqli_close($con);
?>
我想要的是这样的东西

{
    "server_response": [{
        "n_name": null,
        "start": "2016-11-09 00:00:00",
        "end": "2016-11-16 00:00:00",
        "venue": "aaaaaa",
        "teamone": "aaa",
        "teamtwo": "bbb",
        "s_name": ""
    }]
    }
相反,我得到了这样的东西

{
        "server_response": [{
            "n_name": null,
            "start": "2016-11-09 00:00:00",
            "end": "2016-11-16 00:00:00",
            "venue": "aaaaaa"
        }],
        "0": [{
            "teamone": "aaa",
            "teamtwo": "bbb",
            "s_name": ""
        }]
    }
有人能帮我吗。谢谢

echo json_encode(数组(“服务器响应”=>array_merge($response,$data))

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


while($row=mysqli_fetch_array($result1))
{
    $response['server_response']["teamone"] = $row[1];
    $response['server_response']["teamtwo"] = $row[2];
    $response['server_response']["s_name"] = $row[10];
}

echo json_encode ($response);

echo json_编码(数组(“服务器_响应”=>$response+$data));//如果您在两个数组中都存在相同密钥的风险

echo json_encode(数组(“服务器响应”=>array_merge($response,$data))


echo json_编码(数组(“服务器_响应”=>$response+$data));//如果两个数组中存在相同密钥的风险

表2中是否有任何关系?是的,先生。如果它们没有关系,是否不能将它们合并?可能但不能使用
JOIN
show 2 table所有列名将显示该表?在2 table中是否有任何关系?是的,先生。如果它们没有关系,就不能合并它们吗?可以,但不能使用
JOIN
show 2 table所有列名称我会显示该表吗?非常感谢Razib先生!非常感谢你,拉兹布先生!主席先生,阵列合并游戏我{“计划响应”:[{“开始”:“2016-11-10 00:00:00”,“结束”:“2016-11-04 00:00:00”,“地点”:“bbbb”},{“团队一号”:“aaa”,“团队二号”:“bbb”,“s_名称”:“}}}}}我需要的是{“计划响应”:[{“开始”:“2016-11-10 00:00”,“结束”:“2016-11-04 00:00:00”,“地点”:“BBBBBBBB“teamone”:“aaa”,“teamtwo”:“bbb”,“s_name”:“}]谢谢你的方式:)先生数组合并游戏我{“schedule_response”:[{“start”:“2016-11-10 00:00:00”,“end”:“2016-11-04 00:00:00”,“地点”:“BBBBB”},{“teamone”:“aaa”,“teamtwo”:“bbb”,“s_name”:“}}我需要的是{“schedule_response”:[{“start”:2016-11-10 00:00:00,“结束”:“2016-11-04 00:00:00”,“地点”:“bbbb”}“团队一号”:“aaa”,“团队二号”:“bbb”,“s_名称”:“}]顺便说一句谢谢:)