如何生成标题JSON格式

Question

// Check connection 
if (mysqli_connect_errno($con)) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 


$result = mysqli_query($con,"SELECT * FROM recipe"); 


$encode = array(); 

while($allRow = mysqli_fetch_array($result)) 
    { 
     $new = array(
        'id' => $allRow['id']); 


     $encode[] = $new; 

    } 
echo json_encode($encode); 

mysqli_close($con); 
?> 

我有我的JSON格式错误..

[{"id":"1"},{"id":"2"},{"id":"3"},{"id":"4"},{"id":"5"},{"id":"6"},{"id":"7"}] 

我需要有这样的事情
enter link description here

Answer 1

好每一个你的食谱你会添加你想要显示的属性。

$new = array(
       'id' => $allRow['id'], 
       'title'=>$allRow['title'], 
       'ingredients'=>$allRows['ingredients']); 
    $encode[] = $new; 

等

然后你会创建一个包装角落找寻当时的编码对象调用json_encode

$json = json_encode(array("recipes"=>$encode));