从MySQL中的另一个查询调用查询并仅获取1个JSON(Call query from another query in MySQL and fetch into only 1 JSON)

我想调用并从另一个表中选择查询到另一个查询并仅发送一个JSON。我有2张桌子（ towing_list和towing_info ）

json get方法

[{"towing_id":"51","towing_username":"tow","towing_latitude":"3.7310769", "towing_longitude":"103.1240930","distance":"0"}, {"towing_id":"56","towing_username":"tow1","towing_latitude":"3.7311311", "towing_longitude":"103.1239854","distance":"0.013374089073083037"}]

我想使用“towing_username”并从另一个表中调用他们的详细信息，希望是他们的“towing_fullname”和“towing_contactnumber”，因此它将获得以下json结果：

[{"towing_id":"51","towing_username":"tow","towing_fullname":"tow_name", "towing_contactnumber":"0123456789","towing_latitude":"3.7310769", "towing_longitude":"103.1240930","distance":"0"}, {"towing_id":"56","towing_username":"tow1","towing_fullname":"tow1_name", "towing_contactnumber":"01518191904","towing_latitude":"3.7311311", "towing_longitude":"103.1239854","distance":"0.013374089073083037"}]

我的表

towing_list : (towing_id,towing_username,towing_latitude,towing_longitude) towing_info : (towing_id,towing_username,towing_fullname,towing_contactnumber)

这是我的代码的一部分

$q = " SELECT * , ( 6371 * acos ( cos ( radians($lat) ) * cos( radians( towing_latitude ) ) * cos( radians( towing_longitude ) - radians($lon) ) + sin ( radians($lat) ) * sin( radians( towing_latitude ) ) ) ) AS distance FROM towing_list WHERE `towing_status`='$status' HAVING distance < $total_dis_miles ORDER BY distance LIMIT 0 , 20 "; $r = mysql_query($q); while ($row=mysql_fetch_object($r)) { $array[]=$row; } echo json_encode($array);

有可能吗？我是JSON的新手。请帮忙..

I want to call and select query to another query from another table and send in only one JSON. I got 2 tables (towing_list and towing_info)

json get method

[{"towing_id":"51","towing_username":"tow","towing_latitude":"3.7310769", "towing_longitude":"103.1240930","distance":"0"}, {"towing_id":"56","towing_username":"tow1","towing_latitude":"3.7311311", "towing_longitude":"103.1239854","distance":"0.013374089073083037"}]

i want to use "towing_username" and call their detail from another table wish are their "towing_fullname" and "towing_contactnumber" so it will get this json result below:

[{"towing_id":"51","towing_username":"tow","towing_fullname":"tow_name", "towing_contactnumber":"0123456789","towing_latitude":"3.7310769", "towing_longitude":"103.1240930","distance":"0"}, {"towing_id":"56","towing_username":"tow1","towing_fullname":"tow1_name", "towing_contactnumber":"01518191904","towing_latitude":"3.7311311", "towing_longitude":"103.1239854","distance":"0.013374089073083037"}]

My Table

towing_list : (towing_id,towing_username,towing_latitude,towing_longitude) towing_info : (towing_id,towing_username,towing_fullname,towing_contactnumber)

this is part of my code

$q = " SELECT * , ( 6371 * acos ( cos ( radians($lat) ) * cos( radians( towing_latitude ) ) * cos( radians( towing_longitude ) - radians($lon) ) + sin ( radians($lat) ) * sin( radians( towing_latitude ) ) ) ) AS distance FROM towing_list WHERE `towing_status`='$status' HAVING distance < $total_dis_miles ORDER BY distance LIMIT 0 , 20 "; $r = mysql_query($q); while ($row=mysql_fetch_object($r)) { $array[]=$row; } echo json_encode($array);

Is it possible to do?. I m new to JSON. Please help..

最满意答案

你可以使用join

$q = "SELECT * , ( 6371 * acos ( cos ( radians($lat) ) * cos( radians( towing_latitude ) ) * cos(radians( towing_longitude ) - radians($lon) ) + sin ( radians($lat) ) * sin( radians( towing_latitude ) ) ) ) AS distance FROM towing_list INNER JOIN towing_info ON towing_info.towing_id = towing_list.towing_id WHERE towing_status='$status' HAVING distance < $total_dis_miles ORDER BY distance LIMIT 0 , 20 ";

you can use join

$q = "SELECT * , ( 6371 * acos ( cos ( radians($lat) ) * cos( radians( towing_latitude ) ) * cos(radians( towing_longitude ) - radians($lon) ) + sin ( radians($lat) ) * sin( radians( towing_latitude ) ) ) ) AS distance FROM towing_list INNER JOIN towing_info ON towing_info.towing_id = towing_list.towing_id WHERE towing_status='$status' HAVING distance < $total_dis_miles ORDER BY distance LIMIT 0 , 20 ";

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