我试图得到一个字符串的Andr​​oid SHA256。

下面是我想匹配的PHP code：

 回声BIN2HEX（mhash（MHASH_SHA256，ASDF））;
//输出f0e4c2f76c58916ec258f246851bea091d14d4247a2fc3e18694461b1816e13b
 

现在，在Java中，我试图做到以下几点：

 字符串密码=ASDF
            消息摘要消化= NULL;
    尝试 {
        摘要= MessageDigest.getInstance（SHA-256）;
    }赶上（抛出：NoSuchAlgorithmException E1）{
        // TODO自动生成的catch块
        e1.printStackTrace（）;
    }
       digest.reset（）;
       尝试 {
        Log.i（Eamorr，digest.digest（password.getBytes（UTF-8））的toString（））;
    }赶上（UnsupportedEncodingException E）{
        // TODO自动生成的catch块
        e.printStackTrace（）;
    }
 

但这种打印出：a42yzk3axdv3k4yh98g8

我做了什么错在这里？

解决方案感谢埃里克森：

  Log.i（Eamorr，BIN2HEX（getHash（ASDF）））;

 公共字节[] getHash（字符串密码）{
       消息摘要消化= NULL;
    尝试 {
        摘要= MessageDigest.getInstance（SHA-256）;
    }赶上（抛出：NoSuchAlgorithmException E1）{
        // TODO自动生成的catch块
        e1.printStackTrace（）;
    }
       digest.reset（）;
       返回digest.digest（password.getBytes（））;
 }
静态字符串BIN2HEX（byte []的数据）{
    返回的String.Format（％0+（data.length * 2）+X的，新的BigInteger（1，数据））;
}
 

解决方案

PHP函数 BIN2HEX 意味着需要字节串和连接codeS作为一个十六进制数。

在Java的code，你正在试图采取了一堆随机字节和DE code他们为使用平台的默认字符编码的字符串。这是不会工作，并且如果它没有，它不会产生相同的结果。

下面是一个快速和肮脏的二进制到十六进制转换为Java的：

 静态字符串BIN2HEX（byte []的数据）{
  返回的String.Format（％0+（data.length * 2）+'X'，新的BigInteger（1，数据））;
}
 

这是很快的写的，不一定快的执行。如果你是做了很多的这些，你应该重写功能，以更快地实现。

I'm trying to get the SHA256 of a string in Android.

Here is the PHP code that I want to match:

echo bin2hex(mhash(MHASH_SHA256,"asdf"));
//outputs "f0e4c2f76c58916ec258f246851bea091d14d4247a2fc3e18694461b1816e13b"

Now, in Java, I'm trying to do the following:

            String password="asdf"
            MessageDigest digest=null;
    try {
        digest = MessageDigest.getInstance("SHA-256");
    } catch (NoSuchAlgorithmException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
       digest.reset();
       try {
        Log.i("Eamorr",digest.digest(password.getBytes("UTF-8")).toString());
    } catch (UnsupportedEncodingException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

But this prints out: "a42yzk3axdv3k4yh98g8"

What did I do wrong here?

Solution thanks to erickson:

 Log.i("Eamorr",bin2hex(getHash("asdf")));

 public byte[] getHash(String password) {
       MessageDigest digest=null;
    try {
        digest = MessageDigest.getInstance("SHA-256");
    } catch (NoSuchAlgorithmException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
       digest.reset();
       return digest.digest(password.getBytes());
 }
static String bin2hex(byte[] data) {
    return String.format("%0" + (data.length*2) + "X", new BigInteger(1, data));
}

The PHP function bin2hex means that it takes a string of bytes and encodes it as a hexadecimal number.

In the Java code, you are trying to take a bunch of random bytes and decode them as a string using your platform's default character encoding. That isn't going to work, and if it did, it wouldn't produce the same results.

Here's a quick-and-dirty binary-to-hex conversion for Java:

static String bin2hex(byte[] data) {
  return String.format("%0" + (data.length * 2) + 'x', new BigInteger(1, data));
}

This is quick to write, not necessarily quick to execute. If you are doing a lot of these, you should rewrite the function with a faster implementation.