只想从索引中删除键，而不是 php 中的值 - Want to remove only key from index not values in php

我在php中有数组，并在json中转换它。现在我有 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"

}采用 JSON 格式但是我想删除索引 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:)

array_values。

也是权威资源：

array_values($array);

是的，您可以使用 PHP 的array_values删除键并仅保留值

只需执行json_decode，然后使用array_values。

默认情况下，数组索引是数字索引。

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字数组。

我希望你想从 JSON 中删除索引，方法是这样的：

$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)));

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。

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

尝试使用 -

$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