如何在php中有两个select语句答案

【问题标题】：How to have two select statements in php如何在php中有两个select语句
【发布时间】：2016-06-10 22:04:55
【问题描述】：

我试图在我的php 文件中包含两个选择语句，但我遇到了一些困难。

基本上，我正在尝试从两个查询中获取详细信息并在我的 android 应用中使用它们。

这是我想要的JSON 格式：

[
    userstype:[
    {
        "usertypename" : name

    }
    ]
    userdetails:[
    {
        "forename" : forename
        "surname" : surname
        "age" : age
    }]
]

这是我的php 文件：

<?php
require "init.php";
$stmt = "SELECT userstypename FROM tbluserstype";
$result = mysqli_query($conn, $stmt);
$outcome = array();
if(mysqli_num_rows($result)){
    while($row = mysqli_fetch_assoc($result)){
        $outcome[] = array
        (
            "userstypename" => $row["userstypename"]
        );
    }
    echo json_encode($outcome); 
}
$stmt2 = "SELECT forename, surname, age FROM users";
$result = mysqli_query($conn, $stmt2);
$outcome = array();
if(mysqli_num_rows($result)){
    while($row = mysqli_fetch_assoc($result)){
        $outcome[] = array
        (
            "forename" => $row["forename"],
            "surname" => $row["surname"],
            "age" => $row["age"]
        );
    }
    echo json_encode($outcome); 
}
else{
    echo json_encode("Failed");
}

?>

我想使用userstype 和userdetails 作为标签

【问题讨论】：

标签： php android sql json

【解决方案1】：

 <?php
  require "init.php";
  $stmt = "SELECT userstypename FROM tbluserstype";
  $result = mysqli_query($conn, $stmt);
  $outcome = array();
  if(mysqli_num_rows($result)){
     while($row = mysqli_fetch_assoc($result)){
     // following line modified slightly
        $outcome['userstype'] = array(
          "userstypename" => $row["userstypename"]
        );
     }
     // take this line outecho json_encode($outcome); 
  }
  $stmt2 = "SELECT forename, surname, age FROM users";
  $result = mysqli_query($conn, $stmt2);
  //$outcome = array(); <- take this out
  if(mysqli_num_rows($result)){
      while($row = mysqli_fetch_assoc($result)){
      // following line modified slightly
          $outcome['userdetails'] = array(
            "forename" => $row["forename"],
            "surname" => $row["surname"],
            "age" => $row["age"]
         );
      }
    //take this out too, move it outside of if loop echo      json_encode($outcome); 
  }else{
     echo json_encode("Failed");
  }
  echo json_encode($outcome);

【讨论】：

【解决方案2】：

我希望这个解决方案能帮助您生成您在问题中定义的 json 格式。

<?php
require "init.php";
$stmt = "SELECT userstypename FROM tbluserstype";
$result = mysqli_query($conn, $stmt);
$outcome = array();
if(mysqli_num_rows($result)){
    while($row = mysqli_fetch_assoc($result)){
       // following line modified slightly
       $outcome['userstype'] = array(
            "userstypename" => $row["userstypename"]
       );
    }
    // take this line outecho json_encode($outcome); 
 }
 $stmt2 = "SELECT forename, surname, age FROM users";
 $result = mysqli_query($conn, $stmt2);
 //$outcome = array(); <- take this out
 if(mysqli_num_rows($result)){
     while($row = mysqli_fetch_assoc($result)){
        // following line modified slightly
        $outcome['userdetails'] = array(
           "forename" => $row["forename"],
           "surname" => $row["surname"],
           "age" => $row["age"]
        );
     }
     //take this out too, move it outside of if loop echo               json_encode($outcome); 
  }else{
        echo json_encode("Failed");
  }
  // to represent your outcome with json array, write this line.
  $mergedArray[] = $outcome;
  echo json_encode($mergedArray); 
?>

【讨论】：