Question

我正在尝试用c学习sqlite3 api，然后我创建了一个用于存储名称和电话的表，称为议程。然后我用3行填充它。我的下一步是创建以下c代码：

#include <stdio.h>
#include <sqlite3.h>

int main(int argc, char **argv)
{
  sqlite3 *db;
  sqlite3_stmt *res;
  const char *tail;
  int count = 0;

  if(sqlite3_open("agenda.db", &db))
  {
    sqlite3_close(db);
    printf("Can't open database: %s\n", sqlite3_errmsg(db));
    return(1);
  }

  printf("Database connection okay!\n");

  if(sqlite3_prepare_v2(db, "SELECT phone,name FROM agenda ORDER BY name", 128, &res, &tail) != SQLITE_OK)
  {
    sqlite3_close(db);
    printf("Can't retrieve data: %s\n", sqlite3_errmsg(db));
    return(1);
  }

  printf("Reading data...\n");

  printf("%16s | %32s\n", "Phone", "Name");

  while(sqlite3_step(res) != SQLITE_ROW)
  {
    printf("%16s | %32s\n",
           sqlite3_column_text(res, 0),
           sqlite3_column_text(res, 1));

    count++;
  }

  printf("Rows count: %d\n", count);

  sqlite3_finalize(res);

  sqlite3_close(db);

  return(0);
}

然后用gcc -o agenda agenda.c -lsqlite3 -Wall -ggdb编译它。但我得到的结果总是：

Database connection okay!
Reading data...
           Phone |                             Name
Rows count: 0

但实际上agenda.db文件中有3行。我做错了什么？

Answer 1

我相信你想要while (sqlite3_step(res) == SQLITE_ROW) {而不是!=

用c从sqlite中检索数据

1 个答案: