在php中为json创建对象和数组

Question

如何创建一个带有响应的php json，如下所示，其中包含一个对象，一个包含3个项目的数组？

{"files": [
    {
        "name": "picture1.jpg",
        "size": 902604,
        "deleteType": "DELETE"
    }
]}

---

现在我提出如下：

 $info= (array ('file'=> array (array(  
       'name' => $name,
       'size' => $data['file_size'],
       'deleteType' => 'DELETE' ) ) ) ); 

echo json_encode（array（$ info））;

，响应如下

[
    {
        "file": [
            {
                "name": "1391845241happycny.jpg",
                "size": 233.41,
                "deleteType": "DELETE"
            }
        ]
    }
]

这不是我想要的，我不想在'文件'之前的[]

我想要的结果应该是

{
    "file": [
        {
            "name": "1391845241happycny.jpg",
            "size": 233.41,
            "deleteType": "DELETE"
        }
    ]
}

谁可以帮忙？

Answer 1

Visualize it properly :)

$array = array('files' => array(array('name' => 'picture1.jpg', 'size' => '902604', 'deleteType' => 'DELETE')));
echo json_encode($array);

输出：

{"files":[{"name":"picture1.jpg","size":"902604","deleteType":"DELETE"}]}

Answer 2

$array = 
    array("files"=>
        array(
            array(
                "name"=>"picture1.jpg",
                "size"=>"902604",
                "deleteType"=>"DELETE"
            )
        )
    );
echo json_encode($array);

Answer 3

//{"files": [ { "name": "picture1.jpg", "size": 902604, "deleteType": "DELETE" } ]}


$array  =   array('files'=>( array(array("name"=>"picture1.jpg", "size"=>902604, "deleteType"=>"DELETE" ) )));


echo json_encode($array);

Answer 4

您可以使用json_encode($array);

Answer 5

$阵列= （阵列        （＆＃34;文件＆＃34; = GT;            排列                 （阵列（                       ＆＃34;名称＆＃34; = GT;＆＃34;图片1，JPG＆＃34 ;,                       ＆＃34;大小＆＃34; = GT; 902604，                       ＆＃34; deleteType＆＃34; = GT;＆＃34; DELETE＆＃34;                      ）                 ）       ） ）; json_encode（$阵列）;

Answer 6

<?php //{"files": [ { "name": "picture1.jpg", "size": 902604, "deleteType": "DELETE" } ]}

$name = "picture1.jpg";
$size = "902604";
$deleteType = "DELETE";
header('Content-type: application/json');

?>
{
    "files": [{
        "name" : "<?php echo $name; ?>",
        "size" : "<?php echo $size; ?>",
        "deletetype": "<?php echo $deleteType; ?>"
   } ]
}

Answer 7

为此你需要

echo json_encode(array($your_array));