本文介绍了在PHP和SHA256中Java的的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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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08-21 15:36