Php Tumblr API文件\u获取\u内容

我正在尝试使用TumblrAPI，我相信它是JSON。我正在尝试使用Php在下面的URL中获取“posts total”，并将其回显出来。我该怎么做

多谢各位

Update:

我正在尝试使用

$result = json_decode(file_get_contents('http://example.tumblr.com/api/read/json?num=0'));
$print_r($result); 
echo $result[1]; 

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我有500个错误。当我尝试回显$result时，什么也得不到。

简单php-

$result = json_decode(file_get_contents('http://example.tumblr.com/api/read/json?num=0'));
print_r($result); //will show contents return in an araay format
echo $result[1]; //or whatever the element is

简单php-

$result = json_decode(file_get_contents('http://example.tumblr.com/api/read/json?num=0'));
print_r($result); //will show contents return in an araay format
echo $result[1]; //or whatever the element is

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我有一个类似的问题，我想我解决了

var tumblr_api_read = {"tumblelog":{"title":"Ex-Sample","description":"A sample document used to provide examples of the various [and <i>varying<\/i>] HTML tags Tumblr puts out. Quote, specifically, has a lot of variants.","name":"example","timezone":"US\/Eastern","cname":false,"feeds":[]},"posts-start":0,"posts-total":"20","posts-type":false,"posts":[]};

---

我有一个类似的问题，我想我解决了

var tumblr_api_read = {"tumblelog":{"title":"Ex-Sample","description":"A sample document used to provide examples of the various [and <i>varying<\/i>] HTML tags Tumblr puts out. Quote, specifically, has a lot of variants.","name":"example","timezone":"US\/Eastern","cname":false,"feeds":[]},"posts-start":0,"posts-total":"20","posts-type":false,"posts":[]};

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我有500个错误。当我只回显$result时，我什么也没有得到。我得到了一个500的错误。当我只回显$result时，我什么也没有得到。我输入了一个错别字-应该是print\r-这可能会抛出你的500错误。。问题是url返回一个带有json的变量，而不仅仅是一个纯jsonI的输入错误-应该是print\r-这可能会引发500错误。。问题是url返回一个带有json的变量，而不仅仅是一个纯json