我正在尝试使用

mysql C API执行C程序,使用更新查询连接到mysql并且我没有得到任何编译或链接错误,但是在db表中没有更新行.

当我运行此代码时,我在emp中更新了空值.状态字段

#define STRING_SIZE 256

char* eStatus,myeStatus;

int myempid,empid;

int i;

for(i = 0; i < 5 ; i++){

const char* sqlQuery = "update employee_info set estatus = ? where empID = ?";

if (mysql_stmt_prepare(stmt, sqlQuery, strlen(sqlQuery))) {

fprintf(stderr, " mysql_stmt_prepare(), update failed\n");

fprintf(stderr, " %s\n", mysql_stmt_error(stmt));

return -1;

}

memset(param, 0, sizeof(param)); /* zero the structures */

if (info.state == 2)

eStatus = "present";

else

eStatus = "absent";

empid = i;

// Init param structure

// Select

param[0].buffer_type = MYSQL_TYPE_STRING;

param[0].buffer = (void *) &eStatus;

param[0].buffer_length = STRING_SIZE;

param[0].is_null = 0;

param[0].length = &str_length;

param[1].buffer_type = MYSQL_TYPE_SHORT;

param[1].buffer = (void *) &myempID;

param[1].buffer_length = STRING_SIZE;

param[1].is_null = 0;

param[1].length = 0;

myeStatus = eStatus;

myempid = empid;

if (mysql_stmt_bind_param(stmt, param) != 0) {

fprintf(stderr, " mysql_stmt_bind_param() failed\n");

fprintf(stderr, " %s\n", mysql_stmt_error(stmt));

return -1;

}

/* Execute the statement */

if (mysql_stmt_execute(stmt)) {

fprintf(stderr, " mysql_stmt_execute(), failed\n");

fprintf(stderr, " %s\n", mysql_stmt_error(stmt));

return -1;

}

} // end of for loop

mysql中的表模式

empid INT(11)

estatus varchar(10)

我无法弄清楚为什么状态没有在mysql表中更新.它是不匹配的数据类型,还是值没有正确绑定到sqlquery？

任何线索？谢谢.