PHP merge 2数组，然后将其解析为json

我在PHP中创建了两个函数，都是返回数组的函数，我想合并数组并返回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);
}

---

您可以使用和执行类似的功能：

根据您在评论中向我展示的内容，我认为您应该执行以下操作，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;
}

使用第一个数组_merge（）

$result=array\u merge（$array1，$array2）


然后进行json_编码

$json=json\u encode（$result）

阵列合并

函数的作用是：将一个或多个数组合并到一个数组中

json_编码

用于转换为json数据的json_encode（）

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，看到这个。。。

使用

显示代码。你试过什么。提示：阅读并理解与$mergedArray=$array1+$array相比的行为如果它包含键和值，则没有问题，，，它存储在变量中。请检查此。。。例如如果它包含键和值..我的数组有键和值请在问题中举例说明。您的示例数组+您期望的结果函数get_data（$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）}函数likedislike（$conn，$Post_id，$myArray）{$qry1.$post_id.“'AND likes=1”；$res1=mysqli_query（$conn，$qry1）；$count=mysqli_num_rows（$res1）；全局$myArray；全局$username；$userlist=array（）；而（$row=mysqli_fetch\u assoc（$res1））{$Usrname=$row['username']；$userlist[]=array（$username'=>Usrname）}$newarray=array_merge（$myArray，$userlist）；echo json_encode（$newarray）；}@JitenderSingh…请检查我的ans…并且我的数组包含键和values@JitenderSingh…请检查我的Ans，，我希望它对您有所帮助，，，例如，请访问这里…这太棒了，Rishikesh，还有一个快速问题，在while循环中向数组中插入元素的最佳做法是什么。@Jitender..Okkkzz，等等，我给您发送消息用适当的例子链接…以便更好地理解…如果它能解决您的问题..请接受答案..谢谢！！
<?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; 
?>

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