将加密的PHP字符串解密为Android Java(decrypt the encrypted PHP string to Android Java)

将加密的PHP字符串解密为Android Java(decrypt the encrypted PHP string to Android Java)

我想用PHP加密学生的最终成绩，并在Android Java中解密。 我在这里引用了我的代码，但它返回了错误的值。

这是我的PHP加密功能

function encode5t($value1){ for($i=0;$i<3;$i++) { $value1=base64_encode(strrev($value1)); } return $value1; }

调用函数：

foreach ($rows as $row){ $post["cSemester"] = $row["cSemester"]; $post["cSchoolYear"] = $row["cSchoolYear"]; $post["cSubjectCode"] = $row["cSubjectCode"]; $post["cDescription"] = $row["cDescription"]; $post["nFGrade"] = encode5t($row["nFGrade"]); $post["nCGrade"] = $row["nCGrade"]; $post["nCredit"] = $row["nCredit"]; //update our response JSON data array_push($response["posts"], $post); } echo json_encode($response);

这是我的Java代码。

vGrades = json.getJSONArray(TAG_POSTS); for (int i = 0; i < vGrades.length(); i++) { JSONObject c = vGrades.getJSONObject(i); String cSemester = c.getString(TAG_SEMESTER); String cSchoolYear = c.getString(TAG_SCHOOLYEAR); String cSubjectCode = c.getString(TAG_SUBJECTCODE); String cDescription = c.getString(TAG_DESCRIPTION); String encrypted_string = c.getString(TAG_FINALGRADE); String nCGrade = c.getString(TAG_COMPLETIONGRADE); String nCredit = c.getString(TAG_CREDIT); HashMap<String, String> map = new HashMap<String, String>(); try{ byteArray = Base64.decode(encrypted_string, Base64.DEFAULT); decrypt = new String(byteArray, "UTF-8"); }catch (UnsupportedEncodingException e) { e.printStackTrace(); } // this is where I want to decrypt it. nFGrade = decrypt; map.put(TAG_SEMESTER, cSemester); map.put(TAG_SCHOOLYEAR, cSchoolYear); map.put(TAG_SUBJECTCODE, cSubjectCode); map.put(TAG_DESCRIPTION, cDescription); map.put(TAG_FINALGRADE, nFGrade); map.put(TAG_COMPLETIONGRADE, nCGrade); map.put(TAG_CREDIT, nCredit); ViewGrades.add(map); }

PHP加密正在运行..但是当我解密它时，系统返回另一个加密值..例如fGrade是1.0。

PHP encypted String值为：“PT1RVERSRGU =”

Java解密值为：“== QTDRDe”

我哪里做错了？ 我需要帮助..谢谢你们！

I want to encrypt the final grades of students in PHP and decrypt it in Android Java. I referred my codes here but it returns wrong value.

This is my PHP encryption function

function encode5t($value1){ for($i=0;$i<3;$i++) { $value1=base64_encode(strrev($value1)); } return $value1; }

Call the function:

foreach ($rows as $row){ $post["cSemester"] = $row["cSemester"]; $post["cSchoolYear"] = $row["cSchoolYear"]; $post["cSubjectCode"] = $row["cSubjectCode"]; $post["cDescription"] = $row["cDescription"]; $post["nFGrade"] = encode5t($row["nFGrade"]); $post["nCGrade"] = $row["nCGrade"]; $post["nCredit"] = $row["nCredit"]; //update our response JSON data array_push($response["posts"], $post); } echo json_encode($response);

And this is my Java code.

vGrades = json.getJSONArray(TAG_POSTS); for (int i = 0; i < vGrades.length(); i++) { JSONObject c = vGrades.getJSONObject(i); String cSemester = c.getString(TAG_SEMESTER); String cSchoolYear = c.getString(TAG_SCHOOLYEAR); String cSubjectCode = c.getString(TAG_SUBJECTCODE); String cDescription = c.getString(TAG_DESCRIPTION); String encrypted_string = c.getString(TAG_FINALGRADE); String nCGrade = c.getString(TAG_COMPLETIONGRADE); String nCredit = c.getString(TAG_CREDIT); HashMap<String, String> map = new HashMap<String, String>(); try{ byteArray = Base64.decode(encrypted_string, Base64.DEFAULT); decrypt = new String(byteArray, "UTF-8"); }catch (UnsupportedEncodingException e) { e.printStackTrace(); } // this is where I want to decrypt it. nFGrade = decrypt; map.put(TAG_SEMESTER, cSemester); map.put(TAG_SCHOOLYEAR, cSchoolYear); map.put(TAG_SUBJECTCODE, cSubjectCode); map.put(TAG_DESCRIPTION, cDescription); map.put(TAG_FINALGRADE, nFGrade); map.put(TAG_COMPLETIONGRADE, nCGrade); map.put(TAG_CREDIT, nCredit); ViewGrades.add(map); }

The PHP encryption is running .. but when i decrypt it the system returns another encrypted value .. for example the fGrade is 1.0.

the PHP encypted String value is: "PT1RVERSRGU="

the Java decrypted value is: "==QTDRDe"

where did I go wrong? I need help please ..thanks guys!

最满意答案

Base64没有密钥，任何人都可以对其进行解码，更改并替换它。 但如果您只想阻止临时用户，那么这可能就是您所需要的。 您需要定义您要保护的人，以免知道每个方案都可能受到损害。

要加密数据，请使用AES 。 加密不容易纠正。 PHP mcrypt加密函数存在缺陷，请勿使用它。 考虑RNCryptor-php ，它提供了完整的解决方案，包括身份验证和密钥派生。

Base64 has no key, anyone can decode it, change it and replace it. But that may be all you need if you just want to thwart the casual user. You need to define who you are protecting against knowing that every scheme can be compromised.

To encrypt data use AES. Encryption is not easy to get correct. The PHP mcrypt encryption function is flawed, do not use it. Consider RNCryptor-php, it provides a full solution including authentication and key derivation.