PHP JSON 获取键和值

Question

我有以下 JSON 格式：

{
    "S01": ["001.cbf", "002.cbf", "003.cbf", "004.cbf", "005.cbf", "006.cbf", "007.cbf", "008.cbf", "009.cbf"],
    "S02": ["001.sda", "002.sda", "003.sda"],
    "S03": ["001.klm", "002.klm"]
}

我尝试使用此代码：

$json = json_decode('{"S01":["001.cbf","002.cbf","003.cbf","004.cbf","005.cbf","006.cbf","007.cbf","008.cbf","009.cbf"],"S02":["001.sda","002.sda","003.sda"],"S03":["001.klm","002.klm"]}');

foreach($json as $key => $val) {
     if ($key) { echo 'KEY IS: '.$key; };
     if ($val) { echo 'VALUE IS: '.$value; };
     echo '<br>';
}

但我得到了空的结果......我需要得到这样的输出：

KEY IS: S01
VALUE IS: 001.cbf
VALUE IS: 002.cbf
VALUE IS: 003.cbf
VALUE IS: 004.cbf
VALUE IS: 005.cbf
VALUE IS: 006.cbf
VALUE IS: 007.cbf
VALUE IS: 008.cbf
VALUE IS: 009.cbf

KEY IS: S02
VALUE IS: 001.sda
VALUE IS: 002.sda
VALUE IS: 003.sda

KEY IS: S03
VALUE IS: 001.klm
VALUE IS: 002.klm

这是我需要的，以便我可以使用值和键名生成 ul 和 li 元素……这是存储在 mysql 数据库中的 JSON 格式，并被读取到需要解析上述输出中的 JSON 的 php 脚本，以便我可以使用输出创建 ul 和 li 元素。

我尝试执行 foreach 但结果为空？我知道当我获得价值时，我需要使用爆炸字符串，explode(', ', $value)但我无法在 PHP 中根据需要读取 $value 和 $key。

Answer 1

这解决了您的问题，您必须将 $json 强制转换为数组，因为它被视为 stdClass 对象；)

<?php 
$json = json_decode('{"S01":["001.cbf","002.cbf","003.cbf","004.cbf","005.cbf","006.cbf","007.cbf","008.cbf","009.cbf"],"S02":["001.sda","002.sda","003.sda"],"S03":["001.klm","002.klm"]}');
foreach($json as $key => $val) {
    echo "KEY IS: $key<br/>";
    foreach(((array)$json)[$key] as $val2) {
        echo "VALUE IS: $val2<br/>";
    }
}
?>

在线试用！

我建议您下次遇到麻烦时使用函数 var_dump($var)，它会帮助您找出问题所在。

Answer 2

$json = '{"S01":["001.cbf","002.cbf","003.cbf","004.cbf","005.cbf","006.cbf","007.cbf","008.cbf","009.cbf"],"S02":["001.sda","002.sda","003.sda"],"S03":["001.klm","002.klm"]}';

$array = json_decode($json, true);

foreach($array as $key => $val) {
  echo 'KEY IS:'.$key.'<br/>';
  foreach($val as $_key => $_val) {
   echo 'VALUE IS: '.$_val.'<br/>';
  }
}