这就是 print_r($query->getResult()) :
print_r($query->getResult())
Array ( [0] => stdClass Object ( [id] => 9 [user_id] => [title] => [message] => [{"title":"Nice","option":"Text 1"},{"title":"nice 2","option":"text 2"},{"title":"nice 3","option":"text 3"}] [created_at] => 2020-06-21 16:59:49 ) )
我试图用以下方式回应它:
$query = $db->query("SELECT * from user_msg"); foreach (json_decode($query->getResult()) as $key => $additional_field) { echo $additional_field->title; }
但不幸的是,我遇到了这个错误:
json_decode()要求参数1为字符串,给定数组
我不知道CodeIgnitor,但基于结果数组和JSON,你需要做这样的事情:
foreach ($query->getResult() as $row) { foreach(json_decode($row->message) as $msg) { echo $msg->title; } }
循环结果中的每一行,解码 message 然后循环它以获得 title s
message
title