Question

$ 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));

我想传递单个数组。

Ans喜欢：

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

Answer 1

您可以通过将SQL查询改进为：

来实现此目的

  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)"

将其存储为$sql变量中的字符串，最后执行查询：

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

如何从单个数组中的两个查询传递json

1 个答案: