在PHP中打印出一个大json对象的特定位？

Question

我有一个很大的JSON对象，里面有很多数据。 我想从中打印出特定的数据，但是似乎找不到适合打印语句的语法。

这是json对象的第一位，因此您可以看到结构：

{"response":{"status":"OK","advertisers":[{"id":26243,"code":null,"name":"Tego","state":"inactive","default_brand_id":null,"remarketing_segment_id":null,"lifetime_budget":null,"lifetime_budget_imps":null,"daily_budget":null,"daily_budget_imps":null,"enable_pacing":null,"profile_id":508487,"control_pct":0,"timezone":"PST8PDT","last_modified":"2011-08-11 04:35:39","stats":null,"billing_internal_user":null,"billing_address1":"","billing_address2":"","billing_city":"","billing_state":"","billing_country":"","billing_zip":"","default_category":{"id":"5","name":"Computers and Electronics"},"default_currency":"USD","labels":[{"id":"1","name":"Salesperson","value":"Kurt Mackey"},{"id":"3","name":"Account Manager",}],"use_insertion_orders":false,"time_format":"12-hour","default_brand":null},

我想返回第一个广告客户的ID（应该为26243）。

我正在尝试使用以下语句将其打印出来：

打印$ result-> response-> advertisers [0]-> id

但是它似乎没有用。 我得到空白页。

我究竟做错了什么？

Answer 1

您需要先使用json_decode($jsonString)解码字符串。 之后，您将能够通过引用的打印语句访问不同的属性。

$json = '{"response":{"status":"OK","advertisers":[{"id":26243,"code":null,"name":"Tego","state":"inactive","default_brand_id":null,"remarketing_segment_id":null,"lifetime_budget":null,"lifetime_budget_imps":null,"daily_budget":null,"daily_budget_imps":null,"enable_pacing":null,"profile_id":508487,"control_pct":0,"timezone":"PST8PDT","last_modified":"2011-08-11 04:35:39","stats":null,"billing_internal_user":null,"billing_address1":"","billing_address2":"","billing_city":"","billing_state":"","billing_country":"","billing_zip":"","default_category":{"id":"5","name":"Computers and Electronics"},"default_currency":"USD","labels":[{"id":"1","name":"Salesperson","value":"Kurt Mackey"},{"id":"3","name":"Account Manager",}],"use_insertion_orders":false,"time_format":"12-hour","default_brand":null}'
$jsonObj = json_decode($json);
print_r($jsonObj);

Answer 2

您需要重新验证您的json数据。 因为它是在对json数据进行一些修改后为我找到的。

“ labels”：[{“ id”：“ 1”，“名称”：“销售员”，“值”：“ Kurt Mackey”}，{“ id”：“ 3”，“名称”：“客户经理”， }]

请参阅“标签1 ”。 它没有“ labels [0]”具有的值。

试试Json Schema PHP Validator

    <?php
$json = '{"response":{"status":"OK","advertisers":[{"id":26243,"code":null,"name":"Tego","state":"inactive","default_brand_id":null,"remarketing_segment_id":null,"lifetime_budget":null,"lifetime_budget_imps":null,"daily_budget":null,"daily_budget_imps":null,"enable_pacing":null,"profile_id":508487,"control_pct":0,"timezone":"PST8PDT","last_modified":"2011-08-11 04:35:39","stats":null,"billing_internal_user":null,"billing_address1":"","billing_address2":"","billing_city":"","billing_state":"","billing_country":"","billing_zip":"","default_category":{"id":"5","name":"Computers and Electronics"},"default_currency":"USD","labels":[{"id":"1","name":"Salesperson","value":"Kurt Mackey"},{"id":"3","name":"Account Manager","value":null}],"use_insertion_orders":false,"time_format":"12-hour","default_brand":null}]
}}';

$result = json_decode($json);
print $result->response->advertisers[0]->id;