title: select multiple columns count one column and group by one column in one table
body: <p>I have a table named <code>[delivery]</code>, with columns <code>[ID],[Employee],[Post],[c_name],[Deli_Date],[note]</code></p>
  
  <p>I need to Select all columns and count <code>Employee</code> and then group by <code>Employee</code>.
  data in the table like this:</p>
  
  <pre><code>  ---------------------------------------------------
    | Employee| Post      |c_name  |Deli_Date   |Note |
    |---------|-----------|---------------------|-----|
    |  DAVID  |MANAGER    | a      |11/11/2018  |note |
    |  DAVID  |MANAGER    | b      |01/01/2015  |note |
    |  SAM    |ACOUNTS    | c      |10/10/2016  |note |
    |  DAVID  |IT         | a      |10/02/2015  |note |
    |  DAVID  |DOCTOR     | c      |20/02/2017  |note |
    |  JAMES  |DELIVERYMAN| b      |05/05/2015  |note |
    |  OLIVER |DRIVER     | b      |02/02/2014  |note |
    |  SAM    |ACOUNTS    | c      |02/02/2012  |note |
  </code></pre>
  
  <p>this code:</p>
  
  <pre><code>select Employee, count(Employee) as TIMES from delivery
  group by Employee
  </code></pre>
  
  <p>the result is: (but I need to show the other columns too).</p>
  
  <pre><code>| Employee| TIMES | 
  |---------|-------|
  |  DAVID  |   4   |
  |  JAMES  |   1   |
  |  OLIVER |   1   |
  |  SAM    |   2   |
  </code></pre>
  
  <p>I need my code show Like This:</p>
  
  <pre><code>| Employee| TIMES | Post      |c_name  |Deli_Date   |Note |
  |---------|-------|-----------|---------------------|-----|
  |  DAVID  |   4   |MANAGER    | a      |11/11/2018  |note |
  |  JAMES  |   1   |DELIVERYMAN| b      |05/05/2015  |note |
  |  OLIVER |   1   |DRIVER     | b      |02/02/2014  |note |
  |  SAM    |   2   |ACOUNTS    | c      |10/10/2016  |note |
  </code></pre>
  
  <p>what is the best Query could give me that result?</p>
  
  <p>see columns <code>[c_name],[Deli_Date]</code> they shows the last inserted data.
  or just give me the result without <code>[c_name],[Deli_Date]</code> it's ok.</p>