Difference between revisions of "Maximum"

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  min(x,y) = (x + y - ABS(x-y))/2
 
  min(x,y) = (x + y - ABS(x-y))/2
 
</div>
 
</div>
<source lang='sql' class='def'>SELECT id, x, y, (x+y+ABS(x-y))/2  
+
<source lang='sql' class='def'>
FROM t</source>
+
SELECT id, x, y, (x+y+ABS(x-y))/2  
 +
  FROM t</source>
 
<div class="ecomm e-mysql" style="display: none">In MySQL you can also use the <code>GREATEST</code> or <code>LEAST</code> functions to do this.</div>
 
<div class="ecomm e-mysql" style="display: none">In MySQL you can also use the <code>GREATEST</code> or <code>LEAST</code> functions to do this.</div>
 
<div class="ecomm e-oracle" style="display: none">In Oracle you can also use the <code>GREATEST</code> or <code>LEAST</code> functions to do this.</div>
 
<div class="ecomm e-oracle" style="display: none">In Oracle you can also use the <code>GREATEST</code> or <code>LEAST</code> functions to do this.</div>

Latest revision as of 15:04, 2 August 2012

In this example you are shows how to use the MAX or MIN functions to find the maximum or minimum over two fields instead of just one.

schema:scott
DROP TABLE t
 CREATE TABLE t(
  id VARCHAR(10),
  x INTEGER,
  y INTEGER );
INSERT INTO t VALUES ('A',1,2);
INSERT INTO t VALUES ('B',4,3);
INSERT INTO t VALUES ('C',5,5);

The function for using MAX over two fields is

max(x,y) = (x + y + ABS(x-y))/2

The function for using MIN over two fields is

min(x,y) = (x + y - ABS(x-y))/2
SELECT id, x, y, (x+y+ABS(x-y))/2 
  FROM t
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