如何从单个数组中的两个查询中传递 json答案

【问题标题】：how to pass json from two query in single array如何从单个数组中的两个查询中传递 json
【发布时间】：2018-01-25 07:10:42
【问题描述】：

$sql = "SELECT user_registration.user_id, user_registration.full_name, user_registration.username, user_profile.profile_picture FROM user_registration LEFT JOIN user_profile ON user_registration.user_id = user_profile.user_id ";

    $result = $con->query($sql);

    if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {
            $outaa[] = $row;

            $f_user_id = $row['user_id'];

        $sql1 = "SELECT status FROM user_follower WHERE (from_user_id = '$f_user_id' AND to_user_id = '$user_id' ) OR (from_user_id = '$user_id' AND to_user_id = '$f_user_id' )";

        $result1 = $con->query($sql1);
        if ($result1->num_rows > 0) {
        while($row1 = $result1->fetch_assoc()) {

            $outaa[] = $row1;
        } 
        }
        else {
            $outaa[] =  "No";
        }   

    }
    }
   $out = array_merge(array('result'=>'true','reason'=>'Data Fetching Succesfully','user_suggested_data' => $outaa));

我想传入单个数组。

像这样的回答：

"user_suggested_data": { “0”：{ "user_id": "121", "full_name": "Ankit Shah", “用户名”：“shah_ankit39”， “profile_picture”：空， “状态”：0 }, “1”：{ "user_id": "122", "full_name": "pooja", “用户名”：“pooja25”， “profile_picture”：空， “状态”：0
}, “2”：{ "user_id": "123", "full_name": "swapnil", “用户名”：“swapnil25”， “profile_picture”：空， “状态”：0 },

【问题讨论】：

标签： mysql arrays database

【解决方案1】：

您可以通过将 SQL 查询改进为：

  SELECT
    user_registration.user_id,
    user_registration.full_name,
    user_registration.username,
    user_profile.profile_picture,
    user_follower.status
  FROM
    user_registration LEFT JOIN user_profile USING(user_id)
  WHERE
    (user_follower.to_user_id = '$user_id'
      AND user_follower.from_user_id = user_registration.user_id)
    OR (user_follower.from_user_id = '$user_id'
      AND user_follower.to_user_id = user_registration.user_id)"

将其作为字符串存储在$sql 变量中，最后执行查询：

$result = $con->query($sql);

【讨论】：