Getting HTTP status code by cURL

I often use cURL command at the command line on Linux and OS X to fetch information by HTTP requests. Sometimes I just want to know only the HTTP status code. The option -v, which means verbose printing, prints too much information though I don't need.

-w "%{http_code}"

A more specific way to print out just the HTTP status code is something along the lines of:

$ curl -sI -o /dev/null -w "%{http_code}" http://www.example.org/

The option -w means to print assigned format below:

-w, --write-out 
        Make curl display information on stdout after a completed transfer. The format is a string that may
        contain plain text mixed with any number of variables. The format can be  specified  as  a  literal
        "string",  or  you  can  have curl read the format from a file with "@filename" and to tell curl to
        read the format from stdin you write "@-".

        The variables present in the output format will be substituted by  the  value  or  text  that  curl
        thinks  fit,  as  described  below. All variables are specified as %{variable_name} and to output a
        normal % you just write them as %%. You can output a newline by using \n, a carriage return with \r
        and a tab space with \t.

        NOTE: The %-symbol is a special symbol in the win32-environment, where all occurrences of % must be
        doubled when using this option.

Too simple and many format types are available. show man page to just type man curl. The option -I might be added to improve response load performance. This parameter just request for status headers of response, without download response body.

Can't access MySQL 5.7 as root - Error 1698 (28000)

MySQL 5.7+ isn't allowed to login as root user without sudo because that secure model has been changed. The results of login appears ERROR 1698 (28000) as following output;

$ mysql -uroot -p
Enter password:
ERROR 1698 (28000): Access denied for user 'root'@'localhost' (using password: YES)

Skip grant tables to login

Stop the MySQL service for starting mysqld_safe.

$ sudo service mysql stop

Start the MySQL as safe mode with --skip-grant-tables to login without verification.

$ sudo mysqld_safe --skip-grant-tables

Connect to the MySQL as root.

$ mysql -uroot
Welcome to the MySQL monitor.  Commands end with ; or \g.
Your MySQL connection id is 94
Server version: 5.7.14 MySQL Community Server (GPL)

Update mysql.user

It should be that the Plugin of root user sets auth_socket. The Plugin value will be modified to mysql_native_password for compatible former MySQL.

mysql> SELECT User, Host, Plugin FROM mysql.user;
+-----------+-----------+-----------------------+
| User      | Host      | Plugin                |
+-----------+-----------+-----------------------+
| root      | localhost | auth_socket           |
| mysql.sys | localhost | mysql_native_password |
+-----------+-----------+-----------------------+
2 rows in set (0.00 sec)

mysql> UPDATE mysql.user SET authentication_string = PASSWORD ('[root-password]'), Plugin = 'mysql_native_password' WHERE User = 'root';
Query OK, 1 row affected, 1 warning (0.01 sec)

mysql> FLUSH PRIVILEGES;
Query OK, 0 rows affected (0.01 sec)

Restart the MySQL

Restart the MySQL service and then enter via mysql-client after typed FLUSH PRIVILEGES. It should be solved.

$ mysql -uroot
Welcome to the MySQL monitor.  Commands end with ; or \g.
Your MySQL connection id is 94
Server version: 5.7.14 MySQL Community Server (GPL)

That's it!