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[英]json_decode function php://input return Invalid JSON Format
[英]Decode php://input json
我有来自file_get_contents('php://input')
这段文字:
[{"email":"text@examlple.com","event":"processed","sg_event_id":"JJTr9qA","sg_message_id":"AjvX5L7IQeGOfxmJV-OCnw","smtp-id":"<AfxmJV-OCnw@ismon1.sendgrid.net>","timestamp":16813363}]
我需要获取单个元素,如电子邮件或事件,但无法获取json_decode
和其他json_decode
才能正常工作。
$obj = json_decode(file_get_contents('php://input'));
如何引用json数据中的单个元素?
您有一个数组,因此,需要获取第一个元素:
$json = '[{"email":"text@examlple.com","event":"processed","sg_event_id":"JJTr9qA","sg_message_id":"AjvX5L7IQeGOfxmJV-OCnw","smtp-id":"<AfxmJV-OCnw@ismon1.sendgrid.net>","timestamp":16813363}]';
$obj = json_decode($json);
echo $obj[0]->email; //output text@examlple.com
到这里,这是一个完全有效的答案:
<?php
$str = '[{"email":"text@examlple.com","event":"processed","sg_event_id":"JJTr9qA","sg_message_id":"AjvX5L7IQeGOfxmJV-OCnw","smtp-id":"<AfxmJV-OCnw@ismon1.sendgrid.net>","timestamp":16813363}]';
$obj = json_decode($str);
echo $obj[0]->email;
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