Question

我在php中有数组并在json.Now中转换它我有这个json格式的代码

{
    "2": "12:40 to 13:0",
    "3": "13:0 to 13:20",
    "4": "13:20 to 13:40",
    "5": "13:40 to 14:0",
    "6": "14:0 to 14:20",
    "7": "14:20 to 14:40",
    "8": "14:40 to 15:0",
    "9": "15:0 to 15:20",
    "10": "15:20 to 15:40",
    "11": "15:40 to 16:0",
    "12": "16:0 to 16:20",
    "13": "16:20 to 16:40",
    "14": "16:40 to 17:0",
    "15": "17:0 to 17:20",
    "16": "17:20 to 17:40",
    "17": "17:40 to 18:0"

} 以json格式但我想删除索引2,3,4,5表单代码并希望数据

    {
       "12:40 to 13:0",
         "13:0 to 13:20",
        "13:20 to 13:40",
        "13:40 to 14:0",
         "14:0 to 14:20",
        "14:20 to 14:40",
        "14:40 to 15:0",
        "15:0 to 15:20",
         "15:20 to 15:40",
         "15:40 to 16:0",
         "16:0 to 16:20",
        "16:20 to 16:40",
         "16:40 to 17:0",
        "17:0 to 17:20",
         "17:20 to 17:40",
          "17:40 to 18:0"
}

任何帮助:)

Answer 1

array_values

authoritative资源：

array_values($array);

Answer 2

是的，您可以使用PHP array_values

删除密钥并保留唯一值

只需json_decode，然后使用array_values。

默认情况下，数组索引是数字。

如果您生成数组，例如$arr = array('PHP', 'JAVA')，你会得到

这是数字数组

$arr = (0 => 'PHP', '1' => 'JAVA')

现在，如果我们创建一个新的关联数组。

$arr = ('scripting' => 'PHP', 'programming' => 'JAVA')

因此，array_values将数组从关联数组转换为数值数组。

Answer 3

尝试 -

$json = '{
    "2": "12:40 to 13:0",
    "3": "13:0 to 13:20",
    "4": "13:20 to 13:40",
    "5": "13:40 to 14:0",
    "6": "14:0 to 14:20",
    "7": "14:20 to 14:40",
    "8": "14:40 to 15:0",
    "9": "15:0 to 15:20",
    "10": "15:20 to 15:40",
    "11": "15:40 to 16:0",
    "12": "16:0 to 16:20",
    "13": "16:20 to 16:40",
    "14": "16:40 to 17:0",
    "15": "17:0 to 17:20",
    "16": "17:20 to 17:40",
    "17": "17:40 to 18:0"
}';

$data = (array)json_decode($json); //your json data
$data_values = array_values($data);//if you want array
$newStr = "{".implode(',', $data_values)."}"; // if you want string
$newJson = json_encode($data_values); //if you want json

Answer 4

我希望你想从JSON中删除索引，这是方法：

$json = '{
    "2": "12:40 to 13:0",
    "3": "13:0 to 13:20",
    "4": "13:20 to 13:40",
    "5": "13:40 to 14:0",
    "6": "14:0 to 14:20",
    "7": "14:20 to 14:40",
    "8": "14:40 to 15:0",
    "9": "15:0 to 15:20",
    "10": "15:20 to 15:40",
    "11": "15:40 to 16:0",
    "12": "16:0 to 16:20",
    "13": "16:20 to 16:40",
    "14": "16:40 to 17:0",
    "15": "17:0 to 17:20",
    "16": "17:20 to 17:40",
    "17": "17:40 to 18:0"
    }';
 $json = json_encode(array_values(json_decode($json,true)));

Answer 5

试试这个

<?php 
$json = '{
    "2": "12:40 to 13:0",
    "3": "13:0 to 13:20",
    "4": "13:20 to 13:40",
    "5": "13:40 to 14:0",
    "6": "14:0 to 14:20",
    "7": "14:20 to 14:40",
    "8": "14:40 to 15:0",
    "9": "15:0 to 15:20",
    "10": "15:20 to 15:40",
    "11": "15:40 to 16:0",
    "12": "16:0 to 16:20",
    "13": "16:20 to 16:40",
    "14": "16:40 to 17:0",
    "15": "17:0 to 17:20",
    "16": "17:20 to 17:40",
    "17": "17:40 to 18:0"

}';

echo json_encode(array_values(json_decode($json, true)));
?>

{"12:40 to 13:0","13:0 to 13:20",.....}当密钥不在JOSN中时，这不是有效的json格式。

正确的格式为["12:40 to 13:0","13:0 to 13:20","13:20 to 13:40".....]

想要从php中删除索引中的键而不是值

5 个答案: