PHP合并2数组然后将其解析为json

Question

我在PHP中创建了两个函数，这个函数返回一个数组，我想要合并数组并返回一个JSON。

示例代码：

 // random JS code followed by Google Analytics code

Answer 1

您可以使用array_merge和json_encode：

执行此类功能

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

根据您在评论中向我展示的内容，我认为您应该遵循，与数组合并无关：

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

使用第一个array_merge（）;

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

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

然后是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为一个数组。

<强> json_encode

  

json_encode（）用于转换为json数据

示例：

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

或

示例：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; 
?>

  

例如见......   https://eval.in/817282

Answer 4

使用array_merge

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