SQLCipher C++ API 版本

Question

我的 C++ 项目需要 SQLCipher 版本 4.x。我的 Ubuntu 中确实有 SQLCipher 4.6.0 社区，但它在我的 C++ 程序中显示 3.4.1。我尝试编译并使用 SQLCipher 4.6.0 和 4.6.1 来编译我的源代码，但它只给了我相同的结果。

我尝试处理 SQLCipher 3 (HMAC_SHA1) 加密数据库，它有效！但我确实需要处理一些使用 SQLCipher 4 (HMAC_SHA512) 加密的数据库。

我的目标是我的C++程序应该与SQLCipher 4.6.0社区一起工作并成功查询加密数据库。

源代码：

#define DSQLITE_HAS_CODEC = 1
#include <iostream>
#include <sqlcipher/sqlite3.h>
using namespace std;

void executeSQL(sqlite3* db, const char* sql) 
{
    char* errMsg = nullptr;
    int rc = sqlite3_exec(db, sql, nullptr, nullptr, &errMsg);
    
    cout << "SQL CMD: " << sql << endl;
    
    if (rc != SQLITE_OK) 
    {
        cerr << "SQL error: " << errMsg << endl;
        sqlite3_free(errMsg);
    } 
    else 
    {
        cout << "SQL command executed successfully." << endl;
    }
}

int main() {
    sqlite3 *db;
    sqlite3_stmt *stmt;
    int rc;
    
    rc = sqlite3_open("enc_Dev.db", &db);

    const char *sql_cipher_key = "PRAGMA key = '<64-byte key>';";
    executeSQL(db, sql_cipher_key);
    
    const char *sql_cipher_ver = "PRAGMA cipher_default_compatibility = 4;";
    executeSQL(db, sql_cipher_ver);

    const char *sql_cipher_page_size = "PRAGMA cipher_page_size = 4096;";
    executeSQL(db, sql_cipher_page_size);

    const char *sql_cipher_kdf_iter = "PRAGMA kdf_iter = 256000;";
    executeSQL(db, sql_cipher_kdf_iter);
    
    const char *sql_cipher_hmac_algo = "PRAGMA cipher_hmac_algorithm = HMAC_SHA512;";
    executeSQL(db, sql_cipher_hmac_algo);
    
    const char *sql_cipher_kdf_algo = "PRAGMA cipher_kdf_algorithm = PBKDF2_HMAC_SHA512;";
    executeSQL(db, sql_cipher_kdf_algo);
    
    const char *sql_cipher_pt_hsize = "PRAGMA cipher_plaintext_header_size = 0;";
    executeSQL(db, sql_cipher_pt_hsize);

    if (sqlite3_exec(db, "SELECT count(*) FROM input_records;", NULL, NULL, NULL) != SQLITE_OK)
    {
        cerr << "Open encrypted database fail!" << endl;
    }

    const char *sql_ck = "PRAGMA cipher_version;";
    rc = sqlite3_prepare_v2(db, sql_ck, -1, &stmt, 0);
    int i = 0;
    while (sqlite3_step(stmt) == SQLITE_ROW) 
    {
        const unsigned char *output = sqlite3_column_text(stmt, i);
        cout << "PRAGMA cipher_version;\nResult: " << output;
        i++;
    }
    cout <<endl;

    const char *sql_prvdck = "PRAGMA cipher_provider_version;";
    rc = sqlite3_prepare_v2(db, sql_prvdck, -1, &stmt, 0);
    i = 0;
    while (sqlite3_step(stmt) == SQLITE_ROW)
    {
        const unsigned char *output = sqlite3_column_text(stmt, i);
        cout << "PRAGMA cipher_provider_version;\nResult: " << output;
        i++;
    }
    cout <<endl;

    sqlite3_finalize(stmt);
    sqlite3_close(db);

    return 0;
}

输出：

SQL CMD: PRAGMA key = '<64-byte key>';
SQL command executed successfully.
SQL CMD: PRAGMA cipher_default_compatibility = 4;
SQL command executed successfully.
SQL CMD: PRAGMA cipher_page_size = 4096;
SQL command executed successfully.
SQL CMD: PRAGMA kdf_iter = 256000;
SQL command executed successfully.
SQL CMD: PRAGMA cipher_hmac_algorithm = HMAC_SHA512;
SQL command executed successfully.
SQL CMD: PRAGMA cipher_kdf_algorithm = PBKDF2_HMAC_SHA512;
SQL command executed successfully.
SQL CMD: PRAGMA cipher_plaintext_header_size = 0;
SQL command executed successfully.
Open encrypted database fail!
PRAGMA cipher_version;
Result: 3.4.1
PRAGMA cipher_provider_version;
Result: OpenSSL 3.0.0 7 sep 2021

我的 Ubuntu SQLCipher 版本：

SQLite version 3.45.3 2024-04-15 13:34:05 (SQLCipher 4.6.0 community)
Enter ".help" for usage hints.
Connected to a transient in-memory database.
Use ".open FILENAME" to reopen on a persistent database.
sqlite>

Answer 1

修好！！

1. 删除sqlcipher
2. 重新启动
3. 重新安装sqlcipher 4.6.1
4. 重新启动
5. 编译
6. 完成！