我有一个示例程序来熟悉mysqlclient API。但是,当我静态地将其编译并链接到mysqlclient库(.a
文件)时,链接器会抱怨它找不到文件,尽管它存在于我的路径中。链接到共享库(在Mac上为.dylib
文件)有效。请帮助我解决这个问题。非常感谢!
这是我的驱动程序client.c
,它调用mysqlclient库。
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <mysql.h>
int main(int argc, char **argv)
{
MYSQL *mysql = NULL;
if (mysql_library_init(argc, argv, NULL)) {
fprintf(stderr, "could not initialize MySQL client library\n");
exit(1);
}
mysql = mysql_init(mysql);
if (!mysql) {
puts("Init faild, out of memory?");
return EXIT_FAILURE;
}
if (!mysql_real_connect(mysql, /* MYSQL structure to use */
NULL, /* server hostname or IP address */
NULL, /* mysql user */
NULL, /* password */
NULL, /* default database to use, NULL for none */
0, /* port number, 0 for default */
NULL, /* socket file or named pipe name */
CLIENT_FOUND_ROWS /* connection flags */ )) {
puts("Connect failed\n");
} else {
const char *query = "SELECT VERSION()";
if (mysql_real_query(mysql, query, strlen(query))) {
printf("Query failed: %s\n", mysql_error(mysql));
} else {
puts("Query OK");
}
}
mysql_close(mysql);
mysql_library_end();
return EXIT_SUCCESS;
}
这是我的编译方式
gcc -I /usr/local/Cellar/mysql/8.0.16/include/mysql client.c -L /usr/local/Cellar/mysql/8.0.16/lib/ -l mysqlclient.a
ld: library not found for -lmysqlclient.a
clang: error: linker command failed with exit code 1 (use -v to see invocation)
没有.a
的编译成功,因为它链接到共享库,而不是静态库。
最后,这是我的库文件:
ls /usr/local/Cellar/mysql/8.0.16/lib/libmysqlclient*
/usr/local/Cellar/mysql/8.0.16/lib/libmysqlclient.21.dylib /usr/local/Cellar/mysql/8.0.16/lib/libmysqlclient.a /usr/local/Cellar/mysql/8.0.16/lib/libmysqlclient.dylib
-l mysqlclient.a
使链接程序查找名为libmysqlclient.a.a
的文件。相反,您需要类似:
gcc -I /usr/local/Cellar/mysql/8.0.16/include/mysql client.c /usr/local/Cellar/mysql/8.0.16/lib/mysqlclient.a