獲取JSON內容

Question

我想從以下網址JSON數據：

https://chart.googleapis.com/chart?cht=gv&chl=digraph框架{1的color = red] 2; 3色=紅色]; 4; 5; 6; 7; 8; 9; 10; 11; 12 [顏色=紅色]; 13; 14 [顏色=紅色]; 15; 16 [色=紅色]; 17 [顏色=紅色]; 2 - > 1; 3 - > 1; 4 - > 1; 5 - > 1; 6 - > 2; 8 - > 9; 9 - > 8; 12 - > 2; 12→3; 12 - > 4; 12 - > 5; 7 - > 6; 7 - > 10; 7 - > 11; 8 - > 7; 13 - > 5; 14 - > 13; 15 - > 6; 16 - > 15; 17 - > 12; 10 - > 4; 11 - > 3; 14 - > 16; 16 - > 17; } & CHOF = JSON

我曾嘗試提出了其他問題，這幾種方法：

$jsonurl = "https://chart.googleapis.com/chart?cht=gv&chl=digraph framework { 1[color=red]; 2; 3[color=red]; 4; 5; 6; 7; 8; 9; 10; 11; 12[color=red]; 13; 14[color=red]; 15; 16[color=red]; 17[color=red]; 2 -> 1; 3 -> 1; 4 -> 1; 5 -> 1; 6 -> 2; 8 -> 9; 9 -> 8; 12 -> 2; 12 -> 3; 12 -> 4; 12 -> 5; 7 -> 6; 7 -> 10; 7 -> 11; 8 -> 7; 13 -> 5; 14 -> 13; 15 -> 6; 16 -> 15; 17 -> 12; 10 -> 4; 11 -> 3; 14 -> 16; 16 -> 17; }&chof=json" 

//Method 1 
$json = file_get_contents($jsonurl); 
// returns an error from the google page: 400. That’s an error. Your client has issued a malformed or illegal request. That’s all we know 

//Method 2 
$curlSession = curl_init(); 
curl_setopt($curlSession, CURLOPT_URL, $jsonurl); 
curl_setopt($curlSession, CURLOPT_BINARYTRANSFER, true); 
curl_setopt($curlSession, CURLOPT_RETURNTRANSFER, true); 

$json_decoded = json_decode(curl_exec($curlSession)); 
curl_close($curlSession); 
// returns NULL 
// This makes sense if the JSON isn't not returned correctly. I think json_decode() returns NULL if the input is not correct json data. 

//Method 3 
$json = file_get_contents(urlencode($jsonurl)); 
// Gives another error: [function.file-get-contents]: failed to open stream: File name too long 

如果我只是張貼在我的代碼的原始JSON數據我的算法的其餘部分工作正常。關鍵是，在我的實際代碼中，url是動態的，我需要能夠從url獲取JSON數據。

這似乎是谷歌的安全問題。但是，我不知道如何解決這個問題。

這是我在Stackoverflow上的第一個問題。請讓我知道，如果你想知道更多/如果我應該編輯我的問題。

在此先感謝！

Answer 1

您的URL需要通過URI轉義才能被cURL正確使用。如果我改變你的腳本是：

# note that in this string, all special characters (eg. spaces) are %-escaped 
$jsonurl = "https://chart.googleapis.com/chart?cht=gv&chl=digraph%20framework%20{%201[color=red];%202;%203[color=red];%204;%205;%206;%207;%208;%209;%2010;%2011;%2012[color=red];%2013;%2014[color=red];%2015;%2016[color=red];%2017[color=red];%202%20-%3E%201;%203%20-%3E%201;%204%20-%3E%201;%205%20-%3E%201;%206%20-%3E%202;%208%20-%3E%209;%209%20-%3E%208;%2012%20-%3E%202;%2012%20-%3E%203;%2012%20-%3E%204;%2012%20-%3E%205;%207%20-%3E%206;%207%20-%3E%2010;%207%20-%3E%2011;%208%20-%3E%207;%2013%20-%3E%205;%2014%20-%3E%2013;%2015%20-%3E%206;%2016%20-%3E%2015;%2017%20-%3E%2012;%2010%20-%3E%204;%2011%20-%3E%203;%2014%20-%3E%2016;%2016%20-%3E%2017;%20}&chof=json"; 

# your settings here are fine 
$curlSession = curl_init(); 
curl_setopt($curlSession, CURLOPT_URL, $jsonurl); 
curl_setopt($curlSession, CURLOPT_BINARYTRANSFER, true); 
curl_setopt($curlSession, CURLOPT_RETURNTRANSFER, true); 

# XXX: not entirely safe here 
# curl_exec() could dump out something other than JSON if it encounters an error 
# I recommend saving the output of curl_exec($c) separately in case of error 
$json = json_decode(curl_exec($c)); 

然後$json包含的數據對象我期望的那樣。我在Mac上用PHP 5.3.15測試了這個。