如何用PHP中的几个对象解码JSON字符串？

Question

我知道如何在此示例的帮助下使用一个对象解码JSON字符串如何解码JSON字符串

但现在我想用几个对象改进解码JSON字符串,我无法理解如何做到这一点.

这是一个例子:

{ "inbox": [
  { "firstName": "Brett", "lastName":"McLaughlin" },
  { "firstName": "Jason", "lastName":"Hunter" },
  { "firstName": "Elliotte", "lastName":"Harold" }
 ],
"sent": [
  { "firstName": "Isaac", "lastName": "Asimov" },
  { "firstName": "Tad", "lastName": "Williams" },
  { "firstName": "Frank", "lastName": "Peretti" }
 ],
"draft": [
  { "firstName": "Eric", "lastName": "Clapton" },
  { "firstName": "Sergei", "lastName": "Rachmaninoff" }
 ]
}

1. 如何使一个foreach()解码上面的JSON字符串？
2. 如何检测对象的名称:此foreach()的收件箱,已发送或草稿？

Answer 1

新答案

修改你的修改过的问题:foreach实际上可以使用属性以及多值项(数组),详情请参见此处.例如,在您的问题中使用JSON字符串:

$data = json_decode($json);
foreach ($data as $name => $value) {
    // This will loop three times:
    //     $name = inbox
    //     $name = sent
    //     $name = draft
    // ...with $value as the value of that property
}

在属性的主循环中,您可以使用内部循环来遍历每个属性指向的数组条目.因此,例如,如果您知道每个顶级属性都有一个数组值,并且每个数组条目都有一个"firstName"属性,则此代码:

$data = json_decode($json);
foreach ($data as $name => $value) {
    echo $name . ':'
    foreach ($value as $entry) {
        echo '  ' . $entry->firstName;
    }
}

...将会呈现:

inbox:
  Brett
  Jason
  Elliotte
sent:
  Issac
  Tad
  Frank
draft:
  Eric
  Sergei

旧答案

开始编辑 重新评论:

现在我想知道如何用几个对象解码JSON字符串!

您发布的示例确实有几个对象,它们只包含在一个包装器对象中.这是JSON的要求; 你不能(例如)这样做:

{"name": "I'm the first object"},
{"name": "I'm the second object"}

那个JSON无效.有有是一个顶级对象.它可能只包含一个数组:

{"objects": [
    {"name": "I'm the first object"},
    {"name": "I'm the second object"}
]}

...当然,您可以给出单个对象的名称:

{
    "obj0": {"name": "I'm the first object"},
    "obj1": {"name": "I'm the second object"}
}

结束编辑

您的示例是一个包含三个属性的对象,每个属性的值都是一个对象数组.实际上,它与您链接的问题中的示例没有太大区别(它还有一个具有数组值的属性的对象).

所以:

$data = json_decode($json);
foreach ($data->programmers as $programmer) {
    // ...use $programmer for something...
}
foreach ($data->authors as $author) {
    // ...use $author for something...
}
foreach ($data->musicians as $musician) {
    // ...use $musician for something...
}

Answer 2

您可以使用该json_decode函数解码JSON字符串:

$json = <<<JSON
{ "programmers": [
  { "firstName": "Brett", "lastName":"McLaughlin" },
  { "firstName": "Jason", "lastName":"Hunter" },
  { "firstName": "Elliotte", "lastName":"Harold" }
 ],
"authors": [
  { "firstName": "Isaac", "lastName": "Asimov" },
  { "firstName": "Tad", "lastName": "Williams" },
  { "firstName": "Frank", "lastName": "Peretti" }
 ],
"musicians": [
  { "firstName": "Eric", "lastName": "Clapton" },
  { "firstName": "Sergei", "lastName": "Rachmaninoff" }
 ]
}
JSON;

$data = json_decode($json);

然后,要查看数据的外观,您可以转储它:

var_dump($data);

你会看到你有一个包含三个数组的对象,每个数组包含其他子对象:

object(stdClass)[1]
  public 'programmers' => 
    array
      0 => 
        object(stdClass)[2]
          public 'firstName' => string 'Brett' (length=5)
          public 'lastName' => string 'McLaughlin' (length=10)
      1 => 
        object(stdClass)[3]
          public 'firstName' => string 'Jason' (length=5)
          public 'lastName' => string 'Hunter' (length=6)
      ...
  public 'authors' => 
    array
      0 => 
        object(stdClass)[5]
          public 'firstName' => string 'Isaac' (length=5)
          public 'lastName' => string 'Asimov' (length=6)
      ...

这意味着您知道如何访问您的数据.

例如,要显示程序员列表,您可以使用:

foreach ($data->programmers as $programmer) {
  echo $programmer->firstName . ' ' . $programmer->lastName . '<br />';
}

哪个会得到以下输出:

Brett McLaughlin
Jason Hunter
Elliotte Harold