在PHP和SHA256中Java的 [英] SHA256 in PHP & Java

问题描述

我将一些Java代码移植到PHP代码中。在Java中，我有一个哈希SHA256代码如下：

  public static String hashSHA256（String input）
 throws NoSuchAlgorithmException { 
 MessageDigest mDigest = MessageDigest.getInstance（SHA-256）; 
 
 byte [] shaByteArr = mDigest.digest（input.getBytes（Charset.forName（UTF-8）））; 
 StringBuilder hexStrBuilder = new StringBuilder（）; 
 for（int i = 0; i< shaByteArr.length; i ++）{
 hexStrBuilder.append（Integer.toHexString（0xFF& shaByteArr [i]））; 
} 
 
 return hexStrBuilder.toString（）; 
} 
  

在PHP中，我将哈希表示如下：

  $ hash = hash（sha256，utf8_encode（$ input））; 
  

我使用 input =test 即可。然而，我得到了2个不一样的哈希字符串：

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ b

有人可以向我解释为什么以及如何让他们相互匹配？请注意，我无法修改Java实现代码，只能修改PHP。

真的很感谢！

解决方案

PHP版本正确; test 的SHA-256校验和为 9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08 。

Java版本正在返回与两个 0 s相同的校验和被剥离出来。这是因为你将字节转换为十六进制。用& 来替代 0xFF ，而不是 String.format（），如这个答案所示：

  hexStrBuilder.append（String.format（％02x，shaByteArr [i]））; 
  

我知道你说你不能修改Java代码，但它是错误的！

I'm porting some Java code to PHP code. In Java I have a hash SHA256 code as below:

public static String hashSHA256(String input)
        throws NoSuchAlgorithmException {
    MessageDigest mDigest = MessageDigest.getInstance("SHA-256");

byte[] shaByteArr = mDigest.digest(input.getBytes(Charset.forName("UTF-8")));
    StringBuilder hexStrBuilder = new StringBuilder();
    for (int i = 0; i < shaByteArr.length; i++) {
        hexStrBuilder.append(Integer.toHexString(0xFF & shaByteArr[i]));
    }

    return hexStrBuilder.toString();
}

In PHP, I hash as below:

$hash = hash("sha256", utf8_encode($input));

I run the sample code with both input = "test". However, I got 2 hash strings which are not the same:

Java: 9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2bb822cd15d6c15b0f0a8
PHP: 9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08

Can someone explain to me why and how to get them match each other? Please note that I cannot modify the Java implementation code, only to modify PHP.

Really appreciate!

解决方案

The PHP version is correct; the SHA-256 checksum of test is 9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08.

The Java version is returning the same checksum with two 0s stripped out. This is because of the way you're converting bytes into hex. Instead of &ing them with 0xFF, use String.format(), as in this answer:

hexStrBuilder.append(String.format("%02x", shaByteArr[i]));

I realise you say you cannot modify the Java code, but it is incorrect!

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