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SELECT <span class="bold">COUNT(DISTINCT DEPARTMENT_ID)</span> "Number of Departments"
FROM EMPLOYEES;
</pre>
<p>Result:</p>
<pre xml:space="preserve" class="oac_no_warn">
Number of Departments
---------------------
                   11
</pre></div>

<p>The query in the following example uses several aggregate functions to show statistics for the salaries of each <code>JOB_ID</code>.</p>
<div class="example"><a id="CHDCEHHE" name="CHDCEHHE"></a><a id="TDDDG168" name="TDDDG168"></a>
<p class="titleinexample">Using Aggregate Functions for Statistical Information</p>
<pre xml:space="preserve" class="oac_no_warn">
SELECT JOB_ID,
<span class="bold">COUNT(*)</span> "#",
<span class="bold">MIN(SALARY)</span> "Minimum",
<span class="bold">ROUND(AVG(SALARY), 0)</span> "Average",
<span class="bold">MEDIAN(SALARY)</span> "Median",
<span class="bold">MAX(SALARY)</span> "Maximum",
<span class="bold">ROUND(STDDEV(SALARY))</span> "Std Dev"
FROM EMPLOYEES
<span class="bold">GROUP BY JOB_ID</span>
<span class="bold">ORDER BY JOB_ID</span>;
</pre>
<p>Result:</p>
<pre xml:space="preserve" class="oac_no_warn">
JOB_ID              #    Minimum    Average     Median    Maximum    Std Dev
---------- ---------- ---------- ---------- ---------- ---------- ----------
AC_ACCOUNT          1       8300       8300       8300       8300          0
AC_MGR              1      12000      12000      12000      12000          0
AD_ASST             1       4400       4400       4400       4400          0
AD_PRES             1      24000      24000      24000      24000          0
AD_VP               2      17000      17000      17000      17000          0
FI_ACCOUNT          5       6900       7920       7800       9000        766
FI_MGR              1      12000      12000      12000      12000          0
HR_REP              1       6500       6500       6500       6500          0
IT_PROG             5       4200       5760       4800       9000       1926
MK_MAN              1      13000      13000      13000      13000          0
MK_REP              1       6000       6000       6000       6000          0
 
JOB_ID              #    Minimum    Average     Median    Maximum    Std Dev
---------- ---------- ---------- ---------- ---------- ---------- ----------
PR_REP              1      10000      10000      10000      10000          0
PU_CLERK            5       2500       2780       2800       3100        239
PU_MAN              1      11000      11000      11000      11000          0
SA_MAN              5      10500      12200      12000      14000       1525
SA_REP             30       6100       8350       8200      11500       1524
SH_CLERK           20       2500       3215       3100       4200        548
ST_CLERK           20       2100       2785       2700       3600        453
ST_MAN              5       5800       7280       7900       8200       1066
 
19 rows selected.
</pre></div>

<p>To have the query return only rows where aggregate values meet specified conditions, use the <code>HAVING</code> clause.</p>
<p>The query in the following example shows how much each department spends annually on salaries, but only for departments for which that amount exceeds $1,000,000.</p>
<div class="example"><a id="CHDIDDDC" name="CHDIDDDC"></a><a id="TDDDG169" name="TDDDG169"></a>
<p class="titleinexample">Limiting Aggregate Functions to Rows that Satisfy a Condition</p>
<pre xml:space="preserve" class="oac_no_warn">
SELECT DEPARTMENT_ID "Department",
SUM(SALARY*12) "All Salaries"
FROM EMPLOYEES
<span class="bold">HAVING SUM(SALARY * 12) &gt;= 1000000</span>
GROUP BY DEPARTMENT_ID;
</pre>
<p>Result:</p>
<pre xml:space="preserve" class="oac_no_warn">
Department All Salaries
---------- ------------
        50      1876800
        80      3654000
</pre></div>

<p>The <a id="sthref141" name="sthref141"></a><code>RANK</code> function returns the relative ordered rank of a number, and the <a id="sthref142" name="sthref142"></a><code>PERCENT_RANK</code> function returns the percentile position of a number.</p>
<p>The query in the following example shows that a salary of $3,000 is the 20th highest, and is in the 42nd percentile, among all clerks.</p>
<div class="example"><a id="CHDCFCCE" name="CHDCFCCE"></a><a id="TDDDG170" name="TDDDG170"></a>
<p class="titleinexample">Showing the Rank and Percentile of a Number Within a Group</p>
<pre xml:space="preserve" class="oac_no_warn">
SELECT <span class="bold">RANK(3000) WITHIN GROUP</span>
  <span class="bold">(ORDER BY SALARY DESC)</span> "Rank",
ROUND(100 * <span class="bold">(PERCENT_RANK(3000) WITHIN GROUP</span>
  <span class="bold">(ORDER BY SALARY DESC)</span>), 0) "Percentile"
FROM EMPLOYEES
WHERE JOB_ID LIKE '%CLERK';
</pre>
<p>Result:</p>
<pre xml:space="preserve" class="oac_no_warn">
      Rank Percentile
---------- ----------
        20         42
</pre></div>

<p>The <code>DENSE_RANK</code> function is like the <code>RANK</code> function, except that the identical values have the same rank, and there are no gaps in the ranking. Using the <code>DENSE_RANK</code> function, $3,000 is the 12<sup>th</sup> highest salary for clerks, as the following example shows.</p>
<div class="example"><a id="CHDGIEAD" name="CHDGIEAD"></a><a id="TDDDG171" name="TDDDG171"></a>
<p class="titleinexample">Showing the Dense Rank of a Number Within a Group</p>
<pre xml:space="preserve" class="oac_no_warn">
SELECT <span class="bold">DENSE_RANK(3000) WITHIN GROUP (ORDER BY salary DESC)</span> "Rank"
FROM EMPLOYEES
WHERE JOB_ID LIKE '%CLERK';
</pre>
<p>Result:</p>
<pre xml:space="preserve" class="oac_no_warn">
      Rank
----------
        12
</pre></div>

<div class="helpinfonotealso">
<h2>Related Topics</h2>
<p><a href="javascript:open('http://www.oracle.com/pls/db112/lookup?id=SQLRF20035','newWindow').focus()"><span class="italic">Oracle Database SQL Language Reference</span></a></p>
<p><a href="tdddg_selecting010.htm#BCGCCGJF">Using Operators and Functions in Queries</a></p>
</div>
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