Question

我需要使用PHP创建一个JSON对象，因为我需要为每个节点提供属性，就像XML一样，我只能创建一个PHP数组加载（我认为），所以我正在创建PHP对象和这样做。

问题是我可以正确地将JSON格式化。

这就是我的尝试：

$object = new stdClass();

$object->{'0'}['title'] = 'Home';
$object->{'0'}['entry'] = '123';

$object->{'1'}['title'] = 'About';
$object->{'1'}['entry'] = '123';

$object->{'2'}['title'] = 'Gallery';
$object->{'2'}['entry'] = '123';

$object->{'2'} = new stdClass();

$object->{'2'}->{'0'}['title'] = 'Past';
$object->{'2'}->{'0'}['entry'] = '1234';

$object->{'2'}->{'1'}['title'] = 'Present';
$object->{'2'}->{'1'}['entry'] = '1235';

$object->{'2'}->{'0'} = new stdClass();

$object->{'2'}->{'0'}->{'0'}['title'] = '1989';
$object->{'2'}->{'0'}->{'0'}['entry'] = '12345';

$object->{'2'}->{'0'}->{'1'}['title'] = '1990';
$object->{'2'}->{'0'}->{'1'}['entry'] = '12346';


$ob=json_encode($object);

echo $ob;

哪个输出：

{
"0":{"title":"Home","entry":"123"},
"1":{"title":"About","entry":"123"},
"2":{
"0":{
"0":{"title":"1989","entry":"12345"},
"1":{"title":"1990","entry":"12346"}},
"1":{"title":"Present","entry":"1235"}
}
}

我需要＆＃34; 2＆＃34;第一个节点具有属性＆＃34; title＆＃34;：＆＃34; Gallery＆＃34;，＆＃34; entry＆＃34;：＆＃34; 123＆＃34;但也包含过去和现在的子节点，多年来也是如此。

在XML中，它可能看起来像这样：

<0 title="Home" entry="123">
<0/>
<1 title="About" entry="123">
<1/>
<2 title="Gallery" entry="123">
  <0 title="Past" entry="1234">
     <0 title="1989" entry="12345"><0/>
     <1 title="1990" entry="12346"><1/>
  <0/>
  <1 title="Present" entry="1235">
  <1/>
<2/>

Answer 1

您正在使用对象创建删除它们：

交换这些线：

$object->{'2'}['title'] = 'Gallery';
$object->{'2'}['entry'] = '123';
//this line creating the new object is effectively erasing the previous 2 lines.
$object->{'2'} = new stdClass();

成为：

$object->{'2'} = new stdClass();

$object->{'2'}['title'] = 'Gallery';
$object->{'2'}['entry'] = '123';

Answer 2

在PHP中使用json的最简单方法是使用内置的json_encode（）和json_decode（）函数。

这非常好，因为你可以直接将php数组编码成json而无需做任何事情！

$array = array(
    array(
        "title" => "Home",
        "entry" => "123" 
    ),
    array(
        "title" => "About",
        "entry" => "123" 
    ),
    array(
        "title" => "Gallery",
        "entry" => "123",
    ),
);

然后继续嵌套，然后可以将其转换为json对象：

$json = json_encode($array);

输出如：

[{"title":"Home","entry":"123"},{"title":"About","entry":"123"},{"title":"Gallery","entry":"123"}]

然后，您可以通过执行json_decode并像对象一样移动它来使用php再次访问这些内容。

我做了一个操场让你在这里弄乱： http://codepad.viper-7.com/qzMJO3

希望有所帮助！

Answer 3

您使用$object->{'2'}设置$object->{'2'}->{'0'}和new stdClass()，丢失了之前设置的数据。

试试这个：

<?php
$object = new stdClass();

$object->{'0'}['title'] = 'Home';
$object->{'0'}['entry'] = '123';

$object->{'1'}['title'] = 'About';
$object->{'1'}['entry'] = '123';

$object->{'2'}['title'] = 'Gallery';
$object->{'2'}['entry'] = '123';

$object->{'2'}['0']['title'] = 'Past';
$object->{'2'}['0']['entry'] = '1234';

$object->{'2'}['1']['title'] = 'Present';
$object->{'2'}['1']['entry'] = '1235';

$object->{'2'}['0']['0']['title'] = '1989';
$object->{'2'}['0']['0']['entry'] = '12345';

$object->{'2'}['0']['1']['title'] = '1990';
$object->{'2'}['0']['1']['entry'] = '12346';


$ob=json_encode($object);

在PHP中创建JSON对象

3 个答案: