By default, all the value in the selection list, so Select percent select list
HelloI created the application process to extract the name of the State in the selection list based on the name of the city.
Now I need if I don't have the name of the city and I select % in city name select list then any name should come in the select list.
Now if I select % in select list and then he shows me empty in state name select list.
How can I extract all the name of the State if I select % in select list.
My application process
BEGIN
OWA_UTIL.mime_header ('text/xml', FALSE);
HTP.p ('Cache-Control: no-cache');
HTP.p ('Pragma: no-cache');
OWA_UTIL.http_header_close;
htp.prn('<data>');
HTP.prn ('<select>');
FOR c IN (select S.NAME,S.CODE from STATE_MAS S,CITY_MAS C where C.STATE_CODE=S.CODE AND C.CODE =:SELECT5)
LOOP
HTP.prn ('<option value="' || c.CODE || '">' || c.NAME || '</option>');
END LOOP;
HTP.prn ('</select>');
htp.prn('</data>');
END;
My city name TableCREATE TABLE "CITY_MAS"
( "CODE" VARCHAR2(3) NOT NULL ENABLE,
"NAME" VARCHAR2(1000) NOT NULL ENABLE,
"STATE_CODE" VARCHAR2(3) NOT NULL ENABLE
)
My name status TableCREATE TABLE "STATE_MAS"
( "CODE" VARCHAR2(3) NOT NULL ENABLE,
"NAME" VARCHAR2(50) NOT NULL ENABLE
)
/
How can I do that.Thank you
Published by: 805629 on January 25, 2011 02:06
Change the cursor query loop to handle this
select S.NAME,S.CODE from STATE_MAS S,CITY_MAS C where C.STATE_CODE=S.CODE AND C.CODE =:SELECT5
TO
select S.NAME,S.CODE from STATE_MAS S,CITY_MAS C where C.STATE_CODE=S.CODE AND
( C.CODE =:SELECT5 OR :SELECT5 = '%' )
Tags: Database
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