c – 如何使用SOCI库将变量绑定到预准备语句?
发布时间:2020-12-16 07:03:56 所属栏目:百科 来源:网络整理
导读:目前我的应用程序只支持SQLite数据库,但我想支持SQLite和 MySQL数据库,所以我正在测试 SOCI library,看看它是否能满足我的需求.然而,尽管有 examples and documentation,我还是无法弄清楚SOCI如何处理准备好的陈述. 使用SQLite C API时,您准备声明: sqlite3
目前我的应用程序只支持SQLite数据库,但我想支持SQLite和
MySQL数据库,所以我正在测试
SOCI library,看看它是否能满足我的需求.然而,尽管有
examples and documentation,我还是无法弄清楚SOCI如何处理准备好的陈述.
使用SQLite C API时,您准备声明: sqlite3_stmt* statement; sqlite3_prepare_v2( database_handle_pointer,"SELECT * FROM table WHERE user_id=:id;",-1,&statement,NULL ); 然后将值绑定到:id占位符,执行语句并逐步执行结果: const sqlite3_int64 user_id = some_function_that_returns_a_user_id(); const int index = sqlite3_bind_parameter_index( statement,":id" ); sqlite3_bind_int64( statement,index,user_id ); while ( sqlite3_step( statement ) == SQLITE_ROW ) { // Do something with the row } 我如何使用SOCI进行此操作?看起来准备和绑定概念不像原生SQLite API那样分开.在使用soci :: use()准备期间是否必须发生绑定? 更新1:如果我没有充分解释这个问题:这是一个使用SQLite C API的小型工作C示例.如果我能看到这个使用SOCI重新实现,它将回答这个问题. #include <sqlite3.h> #include <iostream> // Tables and data const char* table = "CREATE TABLE test ( user_id INTEGER,name CHAR );"; const char* hank = "INSERT INTO test (user_id,name) VALUES(1,'Hank');"; const char* bill = "INSERT INTO test (user_id,name) VALUES(2,'Bill');"; const char* fred = "INSERT INTO test (user_id,name) VALUES(3,'Fred');"; // Create a SQLite prepared statement to select a user from the test table. sqlite3_stmt* make_statement( sqlite3* database ) { sqlite3_stmt* statement; sqlite3_prepare_v2( database,"SELECT name FROM test WHERE user_id=:id;",NULL ); return statement; } // Bind the requested user_id to the prepared statement. void bind_statement( sqlite3_stmt* statement,const sqlite3_int64 user_id ) { const int index = sqlite3_bind_parameter_index( statement,":id" ); sqlite3_bind_int64( statement,user_id ); } // Execute the statement and print the name of the selected user. void execute_statement( sqlite3_stmt* statement ) { while ( sqlite3_step( statement ) == SQLITE_ROW ) { std::cout << sqlite3_column_text( statement,0 ) << "n"; } } int main() { // Create an in-memory database. sqlite3* database; if ( sqlite3_open( ":memory:",&database ) != SQLITE_OK ) { std::cerr << "Error creating database" << std::endl; return -1; } // Create a table and some rows. sqlite3_exec( database,table,NULL,NULL ); sqlite3_exec( database,hank,bill,fred,NULL ); sqlite3_stmt* statement = make_statement( database ); bind_statement( statement,2 ); execute_statement( statement ); // Cleanup sqlite3_finalize( statement ); sqlite3_close( database ); return 1; } 使用SOCI部分实现相同的程序(注意标记为HELPME的两个存根函数) #include <soci/soci.h> #include <iostream> const char* table = "CREATE TABLE test ( user_id INTEGER,'Fred');"; soci::statement make_statement( soci::session& database ) { soci::statement statement = database.prepare << "SELECT name FROM test WHERE user_id=:id"; return statement; } void bind_statement( soci::statement& statement,const int user_id ) { // HELPME: What goes here? } void execute_statement( soci::statement& statement ) { // HELPME: What goes here? } int main() { soci::session database( "sqlite3",":memory:" ); database << table; database << hank; database << bill; database << fred; soci::statement statement = make_statement( database ); bind_statement( statement,2 ); execute_statement( statement ); } 更新2:当我找到cppdb library时,我最终放弃了SOCI.与SOCI不同,它只是一个非常薄的本地C API包装器,适合我的需求. 解决方法
该文档说明了如何使用
prepared statements with parameters:
int user_id; string name; statement st = (database.prepare << "SELECT name FROM test WHERE user_id = :id",use(user_id),into(name)); user_id = 1; st.execute(true); 请注意,user_id和name变量的生命周期必须至少与st的持续时间一样长. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |