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Php 如何将响应转换为适当的形式?_Php_Request - Fatal编程技术网

Php 如何将响应转换为适当的形式?

Php 如何将响应转换为适当的形式?,php,request,Php,Request,我使用二进制API。 PHP函数获取url的文件内容: https://api.binance.com/api/v1/klines?symbol=BTCUSDT&interval=4h&limit=1 我收到这样的答复: [[1528545600000,"7635.01000000","7650.00000000","7566.00000000","7621.15000000","3246.04484200",1528559999999,"24674635.12120312",

我使用二进制API。 PHP函数获取url的文件内容:

https://api.binance.com/api/v1/klines?symbol=BTCUSDT&interval=4h&limit=1
我收到这样的答复:

[[1528545600000,"7635.01000000","7650.00000000","7566.00000000","7621.15000000","3246.04484200",1528559999999,"24674635.12120312",57351,"1932.28202000","14685353.70584041","0"]]

如何获得第二个值?(7635.01000000)

如果要获取第二个值,请使用以下命令:

$receivedData = '[[1528545600000,"7635.01000000","7650.00000000","7566.00000000","7621.15000000","3246.04484200",1528559999999,"24674635.12120312",57351,"1932.28202000","14685353.70584041","0"]]';

$jsonDecodeData = json_decode($receivedData); 

$secondValue = jsonDecodeData[0][1];

echo $secondValue; //7635.01000000

使用图书馆?这就是我问这个问题的原因。我尝试了json_解码,但问题是json不接受“[]”括号。所以我得到了这样一个错误:注意:C:\xampp\htdocs\index.php中的未定义偏移量:1,第64行,JSON不接受“[]”括号是什么意思?请用一个例子解释一下。jsonDecodeData只返回在receivedData中写入的内容。但是我用str_replace和regex解决了这个问题。