ByteBuffer 到 bigdecimal、二进制、字符串

发布于 2024-10-06 06:32:55 字数 673 浏览 6 评论 0原文

**请检查本文底部的编辑
我有一个字节缓冲区 [128 位] [其中有数字]，我需要将其转换为大十进制、二进制、字符串，因为这些是使用 jdbc 时相应的 sql 映射。

我可以使用库 API 来执行此操作吗？我看到 String.valueof() 不接受字节数组作为参数。所以我坚持做这样的事情：

BigDecimal bd = new BigDecimal(bigmyBuffer.asCharBuffer().toString());

这对我来说看起来像是一个黑客？有没有更好的方法来执行此操作，或者更确切地说以有效的方式执行 jdbc 部分。到目前为止，我专注于在相应的 sql 列中进行插入。

编辑：
我错了，字节缓冲区不仅仅是数字，而是各种位。所以现在我需要获取 128 位字节缓冲区并将其转换为 2 个 long，然后合并为一个 bigdecimal，以便数字保持其完整性。所以像这样： LongBuffer lbUUID = guid.asLongBuffer();

firstLong=      lbUUID.get();
secondLong =      lbUUID.get();

BigDecimal = firstLong + secondLong ;

谢谢。

原文

**Please check the edit at the bottom of this post
I have a bytebuffer[128 bits] [which has numbers] which I need to convert to bigdecimal, binary, string since these are the corresponding sql mapping while using jdbc.

Is there a library API that I can make use of to do this. I see String.valueof() does not take a byte array as a parameter.So I am stuck to doing something like this:

BigDecimal bd = new BigDecimal(bigmyBuffer.asCharBuffer().toString());

This looks like a hack to me ?. Is there a better way of doing this or rather doing the jdbc part in an efficient manner. I am focused on doing inserts in the respective sql columns as of now.

Edit:
I was wrong , the bytebuffers were not just numbers but all sort of bits. So now I need to take the 128 bit byte buffer and convert it to 2 longs and then merge to a bigdecimal so that the numbers maintain their sanity. So something like this:
LongBuffer lbUUID = guid.asLongBuffer();

firstLong=      lbUUID.get();
secondLong =      lbUUID.get();

BigDecimal = firstLong + secondLong ;

Thanks.

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时光清浅 2024-10-13 06:32:55

最好绕道 BigInteger？使用 BigInteger，您可以将字节数组解释为正大数，并且从 BigInteger 到 BigDecimal 只迈出了很小的一步：

byte[] data= new byte[] { 0x12, 0x04, 0x07, 0x05, 0x08, 0x11, 0x38, 0x44, 0x77, 0x33};
BigInteger bi =new BigInteger(1,data);
BigDecimal bd = new BigDecimal(bi);

May it is better to detour over BigInteger? Using BigInteger you can interpretate a byte-array as positive large number and from BigInteger it is a very small step to BigDecimal:

byte[] data= new byte[] { 0x12, 0x04, 0x07, 0x05, 0x08, 0x11, 0x38, 0x44, 0x77, 0x33};
BigInteger bi =new BigInteger(1,data);
BigDecimal bd = new BigDecimal(bi);

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笔芯 2024-10-13 06:32:55

您最大的障碍是 String 仅在内部与 char 一起操作，因此您需要以某种方式进行转换。一个稍微便宜的方法是

BigDecimal bd = new BigDecimal(new String[bigmyBuffer]);

因为你只有数字，所以你不必担心这里的字符集。不幸的是，这仍然会创建一个临时 String 对象。

唯一的选择是手动解析字节缓冲区（即逐字节遍历它，并以这种方式初始化 BigDecimal） - 这将避免分配任何临时对象，但最终会导致更多的函数调用，因此您可能不这样做除非您真的想避免创建该字符串，否则不想走这条路。

我不知道更多关于您的应用程序的上下文，所以我不确定您的 BigDecimal 到底是如何使用的。

Your biggest hurdle is that String ONLY operates with char internally, so you'll need to convert somehow. A slightly cheaper way is

BigDecimal bd = new BigDecimal(new String[bigmyBuffer]);

Since you only have digits, you won't have to worry about charsets here. Unfortunately, this will still create a temporary String object.

The only alternative is to parse the byte buffer manually (i.e. go through it, byte by byte, and initialize your BigDecimal that way) - that will avoid allocating any temporary objects, but ends up in a lot more function calls, so you probably don't want to go that route unless you're really trying to avoid creating that String.

I don't know more about the context of your application, so I'm not sure how exactly your BigDecimal is used.

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