SQL WHERE Field1> = 1或Field2> = 1允许（空）值

Question

I'm trying to query a database table and have the following WHERE clause on my query:

WHERE (QTY_ON_HAND >= 1 OR QTY_ON_ORDER >= 1);

I am wanting to return all results who have either quantity on hand OR quantity on order.

However, this is allowing records to return that have 0 QTY_ON_HAND and (null) QTY_ON_ORDER .

Scrolling throug the results, it's even allowing -1 (And other negatives) and NULL .

Here is the rest of the WHERE clauses:

    WHERE CHARGE_CODE = 'RETAIL' OR CHARGE_CODE = 'RETAILSN' AND
      LOCATION = 100 AND
      SPA_ITEM_ID LIKE '1%' AND
      (QTY_ON_HAND >= 1 OR QTY_ON_ORDER >= 1);

Solution:

Forgot the parenthesis on the first WHERE clause

CHARGE_CODE = 'RETAIL' OR CHARGE_CODE = 'RETAILSN'

needs to be

(CHARGE_CODE = 'RETAIL' OR CHARGE_CODE = 'RETAILSN')

Answer 1

AND operator has precedence over OR , so you need to use parentheses:

WHERE 
   ( CHARGE_CODE = 'RETAIL' OR CHARGE_CODE = 'RETAILSN' )
   AND LOCATION = 100
   AND SPA_ITEM_ID LIKE '1%'
   AND ( QTY_ON_HAND >= 1 OR QTY_ON_ORDER >= 1 )

Better yet using the IN operator:

WHERE 
   CHARGE_CODE IN( 'RETAIL', 'RETAILSN' )
   AND LOCATION = 100
   AND SPA_ITEM_ID LIKE '1%'
   AND ( QTY_ON_HAND >= 1 OR QTY_ON_ORDER >= 1 )

More on the matter explained in documentation .

Answer 2

Try:

WHERE
    CHARGE_CODE IN ('RETAIL','RETAILSN') AND
    LOCATION = 100 AND
    SPA_ITEM_ID LIKE '1%' AND
    (QTY_ON_HAND >= 1 OR QTY_ON_ORDER >= 1);

Hope It Helps!