json_decode file_get_contents帮助

Question

<?php
$json = file_get_contents('http://tiny.cc/ttrhelp');

$obj = json_decode($json);
$example = $obj->rooms->displayName;
?>

Name: <?php echo $example; ?>

尝试显示'displayName'的值，但不显示

Answer 1

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;
}

?>

Answer 2

您可能想要$obj->rooms[0]->displayName。

CodePad

Answer 3

echo $obj->rooms[2]->displayName;

Answer 4

尝试

echo $obj->rooms[0]->displayName;