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Php 从Mysql创建嵌套数组_Php_Mysql_Json - Fatal编程技术网

Php 从Mysql创建嵌套数组

Php 从Mysql创建嵌套数组,php,mysql,json,Php,Mysql,Json,我正在尝试从mysql获取正确的Json数据。我取得了很大的进步。看一看, 我有如下表格: name folder path RayMala 787 01.jpg,02.jpg,03.jpg,04.jpg,05.jpg... RayMala 788 01.jpg,02.jpg,03.jpg,04.jpg,05.jpg,06.jpg... Falitiko 332 01.jpg..

我正在尝试从mysql获取正确的Json数据。我取得了很大的进步。看一看,

我有如下表格:

name             folder    path
RayMala          787       01.jpg,02.jpg,03.jpg,04.jpg,05.jpg...
RayMala          788       01.jpg,02.jpg,03.jpg,04.jpg,05.jpg,06.jpg...
Falitiko         332       01.jpg...
Falitiko         333       01.jpg,02.jpg...
我当前的代码:从单个表单元格生成数组

$rows = array();
while($r = mysqli_fetch_assoc($result)) {   
    $rows[] = $r;
}

$nArray = array();
foreach($rows as $value){
    $nArray[] = array('name' => $value['name'], 'folder' => $value['folder'], 'path' => explode(",", $value['path']));
}

print json_encode($nArray);
当前JSON输出:请参阅名称值重复

[
    {
        "name": "RayMala",
        "folder": "787",
        "Paths": ["1.jpg", "2.jpg", "3.jpg"]
    },
    {
        "name": "RayMala",
        "folder": "788",
        "Paths": ["1.jpg", "2.jpg", "3.jpg"]
    },
    {
        "name": "Falitiko",
        "folder": "333",
        "Paths": ["1.jpg", "2.jpg", "3.jpg"]
    }
]
想要的JSON输出:

[
    {
        "name":"RayMala",
        "random": [
            {
                "folder": "787",
                "Paths": ["1.jpg", "2.jpg", "3.jpg"]
            },
            {
                "folder": "788",
                "Paths": ["1.jpg", "2.jpg", "3.jpg"]
            }
        ]
    },
    {
        "name":"Falitiko",
        "random": [
            {
                "folder": "332",
                "Paths": ["1.jpg", "2.jpg", "3.jpg"]
            },
            {
                "folder": "333",
                "Paths": ["1.jpg", "2.jpg", "3.jpg"]
            }
        ]
    }
]

这样做
$nArray[$value['name']]=
你的答案很完美,但我忘了添加名称键。因为这样我不能用Angularjs来称呼它。我编辑了我想要的Json输出。您可以编辑您的答案吗?我添加了一个额外的foreach循环,以将名称放入数组中。我不知道如何在一个循环中完成,因为您必须检查名称是否已经存在。
$nArray = array();
while($r = mysqli_fetch_assoc($result)) {   
   $nArray[$r["name"]][] = array('folder' => $r['folder'], 'path' => explode(",", $r['path']));
}
$output_arr = array();
foreach($nArray as $key=>$value){
   $output_arr[]=array("name"=>$key,"random"=>$value);
}
echo json_encode($output_arr);