PHP合并2个数组然后解析它到json

Question

我已经在PHP中创建了两个函数，这两个函数都返回一个数组，我想合并这个数组并返回一个JSON。

示例代码：

function get_data($conn, $post_id) { 
    $qry = "SELECT * FROM post WHERE Post_id='".$post_id."'"; 
    $res = mysqli_query($conn, $qry); 
    global $myArray; 
    if (mysqli_num_rows($res) > 0) 
    { while($row = mysqli_fetch_assoc($res)) { 
     $Post_id = $row['Post_id']; 
     $Post_title = $row['Post_title']; 
     $Post_body = $row['Post_body']; 
     $myArray[] = array('Post_id' => $Post_id, 'Post_title' => $Post_title, 'Post_body' => $Post_body); 
    } 
    likedislike($conn, $post_id, $myArray); } 
} 


function likedislike($conn, $post_id, $myArray) 
{ 
    $qry1 = "SELECT * FROM likeunlike WHERE postid = '".$post_id."' AND likes=1"; 
    $res1 = mysqli_query($conn, $qry1); 
    $count = mysqli_num_rows($res1); 
    global $myArray; 
    global $username; 
    $userlist = array(); 
    while($row = mysqli_fetch_assoc($res1)) { 
     $Usrname = $row['username']; 
     $userlist[] = array('username' => $Usrname); 
    } 
    $newarray = array_merge($myArray, $userlist); 
    echo json_encode($newarray); 
} 

Answer 1

您可以通过使用array_merge和json_encode做这样的功能：

function getMergedJson($array1, $array2) { 
    $result = array_merge($array1, $array2); 
    return json_encode($result); 
} 

与你给我的评论，我认为你应该做以下，必须有与阵列合并无关：

function get_data($conn, $post_id) { 
    $qry = "SELECT * FROM post WHERE Post_id='".$post_id."'"; 
    $res = mysqli_query($conn, $qry); 
    global $myArray; 
    if (mysqli_num_rows($res) > 0) { 
     while($row = mysqli_fetch_assoc($res)) { 
      $Post_id = $row['Post_id']; 
      $Post_title = $row['Post_title']; 
      $Post_body = $row['Post_body']; 
      $myArray[] = array('Post_id' => $Post_id, 'Post_title' => $Post_title, 'Post_body' => $Post_body, 'Username' => likedislike($conn, $post_id, $myArray);); 

     }   
    } 
    echo json_encode($newarray); 
} 

function likedislike($conn, $post_id, $myArray) 
{ 
    $qry1 = "SELECT * FROM likeunlike WHERE postid = '".$post_id."' AND likes=1"; 
    $res1 = mysqli_query($conn, $qry1); 
    $count = mysqli_num_rows($res1); 
    global $myArray; 
    global $username; 
    $userlist = array(); 
    while($row = mysqli_fetch_assoc($res1)) { 
     $Usrname = $row['username']; 
     $userlist[] = array('username' => $Usrname); 
    } 
    return $userlist; 
} 

Answer 2

使用冷杉t array_merge（）;

$result = array_merge($array1, $array2);

http://php.net/manual/en/function.array-merge.php

然后json_encode

$json = json_encode($result);

http://php.net/manual/en/function.json-encode.php

Answer 3

array_merge

array_merge（）函数将一个或多个数组合并到一个数组中。

json_encode

json_encode（），用于转换为JSON数据

实施例：

<?php 
$a1=array("red","green"); 
$a2=array("blue","yellow"); 
$myarray = array_merge($a1,$a2); 
$json_data = json_encode($myarray); 
echo json_data; 
?> 

或者

例如：如果键和值

<?php 
$a1=array("Volvo"=>"XC90","BMW"=>"X5"); 
$a2=array("abc"=>"XxX","xyz"=>"Xz"); 

$myarray = array_merge($a1,$a2); 

$json_data = json_encode($myarray); 
echo $json_data; 
?> 

例如，参见本... https://eval.in/817282

Answer 4

使用array_merge

$result = array_merge($array1, $array2); 
echo json_encode($result);