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Cyndi, In SQL you need to use the LIKE clause and the % sign eg. Select * from File1 where FIELD1 like 'Gra%' The above statement would return all records where FIELD1 contains the string 'Gra' starting in position 1 eg. Graham Smith Gradual Decline Grateful Dead but not The Great Graham is an egotistical SOB Not So Gradual Decline To find records where the letter 'Gra' apppears somewhere in the string use Select * from File1 where FIELD1 like '%Gra%' In the above example this should return only Graham Smith Gradual Decline Grateful Dead The Great Graham is an egotistical SOB Not So Gradual Decline Can be used in embedded SQL but should be noted that when using a substitution variable (ie :Var1) that the variable should be padded with % signs to fill out its length. Eg if Var1 is 30,A Select * from File1 where FIELD1 like :Var1 works as in the first example if Var1 = 'Gra%%%%%%%%%%%%%%%%%%%%%%%%%%%' but doesn't work if Var1 = 'Gra% ' because then the SQL is looking for strings where the first 3 chars are 'Gra' the 4th char is anything and the last 26 chars are blank. Regards Graham -----Original Message----- From: Cyndi Bradberry [mailto:CyndiB@IHFA.ORG] Sent: 02 January 2003 17:08 To: 'MIDRANGE-L@midrange. Com' (E-mail) Subject: SQL Question A co-worker has just asked what to use as a variable replacement character in a select statement. I can't find anything in any manual. Is there such a thing ? For example in dos you could type DEL *.* and it would delete everything. What do you use in SQL to imitate the * in the previous dos statement ? Thanks, Cyndi B. Boise, ID _______________________________________________ This is the Midrange Systems Technical Discussion (MIDRANGE-L) mailing list To post a message email: MIDRANGE-L@midrange.com To subscribe, unsubscribe, or change list options, visit: http://lists.midrange.com/cgi-bin/listinfo/midrange-l or email: MIDRANGE-L-request@midrange.com Before posting, please take a moment to review the archives at http://archive.midrange.com/midrange-l.
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