我将选择从第三方网站返回值的数据，如下Json对象格式，并且该网站使用Java语言和tomcat数据库构建。
问题如何从该网站获取内容？
服务器描述
网址:http://103.5.126.24/PalmHallServer/coodraw/coodraw!queryAllProduct.action
返回数据json对象，luckDrawProductList为所有商品信息数组，
{
"LuckDrawProductList": [
{
"BaseTime": 20140119133000,
"CommissionRatio": 10,
"DeductBetCount": 1,
"Detail": "You can enter any amount between ...",
"Enable": 1,
"FreeResDiscount": 1,
"LotteryTime": 20151016164500,
"LuckDrawDesc": "DRAW1",
"LuckDrawDiscount": 1,
"MaxLuckNumLen": 5,
"MaxMoney": 2000000,
"MinLuckNumLen": 5,
"MinMoney": 100,
"NextStartTime": "2015-10-16 12:00:00",
"NextStopTime": "2015-10-16 16:15:00",
"OpenDrawType": -1,
"PeriodUnit": 1,
"PeriodUnitNum": 1,
"Price": 40,
"ProductCode": "DRAW1",
"ProductDesc": "Draw1",
"ProductEndTime": 20990119163000,
"ProductId": 11111,
"ProductType": 11,
"Rate": 0,
"ResTmplProductId": 11111,
"StopTime": 1800,
"ValidPeriod": 2
}
],
"Period": "201510161645"
}
最佳答案
请尝试下面的代码
$URL = "http://103.5.126.24/PalmHallServer/coodraw/coodraw!queryAllProduct.action";
$content = file_get_contents($URL); // get json data using file_get_content
$content_arr = json_decode($content); // json data to php array.
或者如果您想使用curl
$s = curl_init();
curl_setopt($s,CURLOPT_URL,$this->_url);
curl_setopt($s,CURLOPT_RETURNTRANSFER,true);
$result = curl_exec($s);
curl_close($s);
$content_arr = json_decode($result); // json data to php array.
然后您可以 print_r($content_arr) 并检查“LuckDrawProductList” key 或访问 $content_arr['LuckDrawProductList][0]
关于java - 如何使用Codeigniter 3从java网站获取Json数据？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33344193/