Question

我用c++编程，用mysql_real_query()执行了一条select查询语句，成功之后，

怎么才能看到查询的结果啊？就是如select * from tab where id<5;这符合条件的记录怎么才能显示出来啊？

Answer 1

天啦，你既然要用mysql，难道不看API手册吗？
给你个参考例子吧。
#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>
#include "mysql.h"

MYSQL *mysql;
MYSQL_RES *results;
MYSQL_ROW record;

static char *server_options[] = { "mysql_test", "--defaults-file=my.cnf" };
int num_elements = sizeof(server_options)/ sizeof(char *);

static char *server_groups[] = { "libmysqld_server", "libmysqld_client" };

int main(void)
{
mysql_server_init(num_elements, server_options, server_groups);
mysql = mysql_init(NULL);
mysql_options(mysql, MYSQL_READ_DEFAULT_GROUP, "libmysqld_client");
mysql_options(mysql, MYSQL_OPT_USE_EMBEDDED_CONNECTION, NULL);

mysql_real_connect(mysql, NULL,NULL,NULL, "database1", 0,NULL,0);

mysql_query(mysql, "SELECT column1, column2 FROM table1");

results = mysql_store_result(mysql);

while((record = mysql_fetch_row(results))) {
printf("%s - %s \n", record[0], record[1]);
}

mysql_free_result(results);
mysql_close(mysql);
mysql_server_end();

return 0;
}

Answer 2

//打印各行
MYSQL_ROW row = NULL;
row = mysql_fetch_row(result);// result 查询的结果集
while (NULL != row)
{
printf("获取到的结果： %s 、 %s 、 %s 、 %s、 %s\n", row[0], row[1], row[2], row[3], row[4]);//假设有5个字段的输出
row = mysql_fetch_row(result);
}
mysql_free_result(result);//结果集释放