Question

Iam遇到一个问题，其中预准备语句和绑定值适用于插入语句但不适用于选择语句.... 这是一段代码

 int main(int argc, CHAR* argv[])
{
 sqlite3 *db; // sqlite3 db struct
  char *zErrMsg = 0;
  char *szSQL;
  int rc;

  // Open the test.db file
   rc = sqlite3_open("test32.db", &db);

   if( rc )
   {
     // failed
     fprintf(stderr, "Can't open database: %s\n", 
          sqlite3_errmsg(db));
   }
    else
  {
// success
fprintf(stderr, "Open database successfully\n");
 }

  // create myTable
 //szSQL = "create table myTable1011 (FirstName varchar(30), LastName varchar(30), Age smallint not null);";

   // rc = sqlite3_exec(db, szSQL, callback, 0, &zErrMsg);

   // if( rc == SQLITE_OK )
    // {
   //insert 1 record into myTable
    // RunInsertParamSQL(db, "asyb", "com", 42);




// fetch records

sqlite3_stmt *stmt;
const char *pzTest;

 //Doesnt WORK 

szSQL = "select * from myTable1011 where age = ?";

 rc = sqlite3_prepare_v2(db, szSQL, strlen(szSQL), &stmt, &pzTest);

if( rc == SQLITE_OK ) 

  // bind the value 

  int test = 42;

sqlite3_bind_int(stmt, 1, test);//Doesnt work if i give 42 directly instead if test in bind statement

// commit 
 sqlite3_step(stmt);
 sqlite3_finalize(stmt);
}


 // rc = sqlite3_exec(db, szSQL, callback, 0, &zErrMsg);


// Close test.db file
 sqlite3_close(db);

 getchar();

  return 0;
}

// Insert rec
//WORKS WELL
void RunInsertParamSQL(sqlite3 *db, char *fn, char *ln, int age)
{
  if (!db)
    return;

  char *zErrMsg = 0;
  sqlite3_stmt *stmt;
  const char *pzTest;
  char *szSQL;

  // Insert data item into myTable
 szSQL = "insert into myTable1011 (FirstName, LastName, Age) values (?,?,?);";

  int rc = sqlite3_prepare(db, szSQL, strlen(szSQL), &stmt, &pzTest);

  if( rc == SQLITE_OK ) {
   // bind the value 
     sqlite3_bind_text(stmt, 1, fn, strlen(fn), 0);
    sqlite3_bind_text(stmt, 2, ln, strlen(ln), 0);
    sqlite3_bind_int(stmt, 3, age);

    // commit 
     sqlite3_step(stmt);
    sqlite3_finalize(stmt);
  }
}

除了成功打开数据库之外我没有输出

但是，如果我给 szSQL =“select * from myTable1011”; rc = sqlite3_exec（db，szSQL，callback，0，＆amp; zErrMsg）; //我得到了整张桌子......

我如何有选择地打印？为什么我的Where Condition失败?????

Answer 1

没有输出，因为代码甚至没有尝试输出任何内容。

您必须使用sqlite3_column_* functions：

读取列值

...
for (;;) {
    rc = sqlite3_step(stmt);
    if (rc != SQLITE_ROW)
        break;
    // this depends on what columns you actually have:
    int         first_column  = sqlite3_column_int (stmt, 0);
    const char *second_column = sqlite3_column_text(stmt, 1);
    ...
    printf("ID: %d, name: %s\n", first_column, second_column);
}
if (rc != SQLITE_DONE)
    printf("error: %s\n", sqlite3_errmsg(db));
sqlite3_finalize(stmt);
...

Answer 2

最后我明白了......

如果您使用Sprintf而不是准备/绑定语句，我可以获得输出....

sprintf(szSQL,"select * from Foldersonly WHERE Foldersonly.FolderName = \"%s\" ",Root_Path);
sqlite3_exec(db, szSQL, callback, 0, &zErrMsg);

这将有效...

SQLITE选择查询？无法打印

2 个答案: