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;",
-1, &statement, 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, index, 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, NULL );
sqlite3_exec( database, hank, NULL, NULL, NULL );
sqlite3_exec( database, bill, NULL, NULL, NULL );
sqlite3_exec( database, fred, NULL, NULL, 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, 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');";
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.