php json_decode file_get_contents 帮助
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/7239326/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
提示：将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间：2020-08-26 02:18:16 来源:igfitidea点击:
json_decode file_get_contents help
提问by Martin
<?php
$json = file_get_contents('http://tiny.cc/ttrhelp');
$obj = json_decode($json);
$example = $obj->rooms->displayName;
?>
Name: <?php echo $example; ?>
Trying to show the value for 'displayName' but its not showing
尝试显示“displayName”的值但未显示
回答by Tudor Constantin
Untested code:
未经测试的代码：
<?php
$json = file_get_contents('http://pub.tapulous.com/tapplications/coresocial/v1/chat/api/index.php?method=room_list');
$obj = json_decode($json);
foreach($obj->rooms as $room){
$example = $room->displayName;
echo $example;
}
?>
回答by Quamis
echo $obj->rooms[2]->displayName;
echo $obj->rooms[2]->displayName;
回答by stealthyninja
Try
尝试
echo $obj->rooms[0]->displayName;
