提问人：Даниел Димитров 提问时间：12/23/2014 更新时间：12/23/2014 访问量：106
使用 php 从 URL 获取 JSON 对象
Get JSON object from URL with php
问：
我有来自url这个
[{"user_id":"3932131","username":"DanielDimitrov","count300":"1677134","count100":"239025","count50":"41207","playcount":"17730","ranked_score":"1413977663","total_score":"7355146958","pp_rank":"35848","level":"95.5852","pp_raw":"1582.26","accuracy":"97.88556671142578","count_rank_ss":"42","count_rank_s":"337","count_rank_a":"120","country":"BG","events":[]}]
我的PHP是这个
$json = file_get_contents('url');
$obj = json_decode($json);
echo $obj->user_id;
echo 'Username: '.$obj['username'].'<br>';
echo 'PP: '.(int)$obj['pp_raw'].'<br>';
echo 'Level: '.(int)$obj['level'].'<br>';
echo 'Play count: '.$obj['playcount'].'<br>';
我尝试从 url 中删除括号并运行代码，但我用括号得到它......我怎样才能删除它们？
答：
0赞
Halcyon
12/23/2014
#1
$obj是一个具有 1 个元素的数组。
简短的解决办法是，先做。$obj = $obj[0]
然后你可以做等等。$obj["user_id"]
0赞
Eugen
12/23/2014
#2
<?php
$json = '[{"user_id":"3932131","username":"DanielDimitrov","count300":"1677134","count100":"239025","count50":"41207","playcount":"17730","ranked_score":"1413977663","total_score":"7355146958","pp_rank":"35848","level":"95.5852","pp_raw":"1582.26","accuracy":"97.88556671142578","count_rank_ss":"42","count_rank_s":"337","count_rank_a":"120","country":"BG","events":[]}]';
$in = array("[{", "}]");
$out = array("{", "}");
$obj = json_decode(str_replace($in, $out, $json));
echo $obj->user_id;
echo 'Username: '.$obj->username.'<br>';
echo 'PP: '.(int)$obj->pp_raw.'<br>';
echo 'Level: '.(int)$obj->level.'<br>';
echo 'Play count: '.$obj->playcount.'<br>';
?>
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