PHP中的特殊字符JSON返回空？

Question

我正在尝试从我的JSON中显示一些值，但在两个上失败，而其余的工作完美。 我怀疑它与内部的角色有关，但不知道如何继续。

尝试过技巧：

1. UTF-8-ing everything
2. 直接重新制作JSON
3. 解码，使用JSON_PRETTY_PRINT / JSON_UNESCAPED_UNICODE进行编码并再次解码
4. 还有一些我不能再记得......

关于如何解决这个问题的任何指示都非常感激：

$ json的var_dump

{
    "products":{
        "product":[{
            "@id":"1",
            "name":"First Name",
            "fetchers":{
                "fetcher":[{
                    "@fetcherId":"1",
                    "link1":"http:\/\/www.example.com\/pv\/?1&NAME=[[first-name\/1\/?refID=1\/first-name]]&denk5=[[1]]",
                    "link2":"http:\/\/www.example.com\/pc\/?1&NAME=[[first-name\/1\/?refID=1\/first-name]]&denk5=[[1]]"
                }]
            }
        },{
            "@id":"2",
            "name":"Second Name",
            "fetchers":{
                "fetcher":[{
                    "@fetcherId":"2",
                    "link1":"http:\/\/www.example.com\/pv\/?1&NAME=[[second-name\/2\/?refID=2\/second-name]]&denk5=[[2]]",
                    "link2":"http:\/\/www.example.com\/pc\/?1&NAME=[[second-name\/2\/?refID=2\/second-name]]&denk5=[[2]]"
                }]
            }
        }]
    }
}

PHP代码

<?php
$json = json_decode($json,true);
foreach($json['products']['product'] as $data) {
    echo $data['@id'].'<br/>';
    echo $data['name'].'<br/>';
    echo $data['fetchers']['fetcher']['@fetcherId'].'<br/>';
    echo $data['fetchers']['fetcher']['link1'].'<br/>';
    echo $data['fetchers']['fetcher']['link2'].'<br/>';
}

预期结果

1
First Name
1
http://www.example.com/pv/?1&NAME=[[first-name/1/?refID=1/first-name]]&denk5=[[1]]
http://www.example.com/pc/?1&NAME=[[first-name/1/?refID=1/first-name]]&denk5=[[1]]
2
Second Name
2
http://www.example.com/pv/?1&NAME=[[second-name/2/?refID=2/second-name]]&denk5=[[2]]
http://www.example.com/pc/?1&NAME=[[second-name/2/?refID=2/second-name]]&denk5=[[2]]

我得到了什么

1
First Name
1
 <- link1 empty
 <- link2 empty
2
Second Name
2
 <- link1 empty
 <- link2 empty

我错过了什么？

修改 @Brad解决方案非常有效。

Answer 1

您的问题与JSON无关，与字符编码无关。你没有选择正确的元素。 $data['fetchers']['fetcher']是一个数组。如果您想要第一个元素，请使用[0]。

foreach($json['products']['product'] as $data) {
    echo $data['@id'].'<br/>';
    echo $data['name'].'<br/>';
    echo $data['fetchers']['fetcher'][0]['@fetcherId'].'<br/>';
    echo $data['fetchers']['fetcher'][0]['link1'].'<br/>';
    echo $data['fetchers']['fetcher'][0]['link2'].'<br/>';
}

另外，不要忘记在HTML上下文中使用任意数据htmlspecialchars()。

Answer 2

它不是关于JSON数据，您可以通过这种方式正确访问它们。考虑这个例子：

$json_string = '{"products":{"product":[{"@id":"1","name":"First Name","fetchers":{"fetcher":[{"@fetcherId":"1","link1":"http:\/\/www.example.com\/pv\/?1&NAME=[[first-name\/1\/?refID=1\/first-name]]&denk5=[[1]]","link2":"http:\/\/www.example.com\/pc\/?1&NAME=[[first-name\/1\/?refID=1\/first-name]]&denk5=[[1]]"}]}},{"@id":"2","name":"Second Name","fetchers":{"fetcher":[{"@fetcherId":"2","link1":"http:\/\/www.example.com\/pv\/?1&NAME=[[second-name\/2\/?refID=2\/second-name]]&denk5=[[2]]","link2":"http:\/\/www.example.com\/pc\/?1&NAME=[[second-name\/2\/?refID=2\/second-name]]&denk5=[[2]]"}]}}]}}';
$json_data = json_decode($json_string, true);
$data = array();
foreach($json_data['products']['product'] as $key => $value) {
    echo '@id => ' . $value['@id'] . '<br/>';
    echo 'name => ' .$value['name'] . '<br/>';
    if(isset($value['fetchers'])) {
        $fetchers = reset($value['fetchers']['fetcher']); // <-- set pointer to first which is zero index
        echo 'fetcher_id => ' . $fetchers['@fetcherId'] . '<br/>';
        echo 'link1 => ' . $fetchers['link1'] . '<br/>';
        echo 'link2 => ' . $fetchers['link2'] . '<br/>';
    }
    echo '<br/>';
}


?>

应该产生这样的东西：

@id => 1
name => First Name
fetcher_id => 1
link1 => http://www.example.com/pv/?1&NAME=[[first-name/1/?refID=1/first-name]]&denk5=[[1]]
link2 => http://www.example.com/pc/?1&NAME=[[first-name/1/?refID=1/first-name]]&denk5=[[1]]

@id => 2
name => Second Name
fetcher_id => 2
link1 => http://www.example.com/pv/?1&NAME=[[second-name/2/?refID=2/second-name]]&denk5=[[2]]
link2 => http://www.example.com/pc/?1&NAME=[[second-name/2/?refID=2/second-name]]&denk5=[[2]]