From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=-1.9 required=5.0 tests=BAYES_00,FREEMAIL_FROM autolearn=unavailable autolearn_force=no version=3.4.4 X-Received: by 10.13.232.198 with SMTP id r189mr21971708ywe.35.1437223733982; Sat, 18 Jul 2015 05:48:53 -0700 (PDT) X-Received: by 10.50.126.35 with SMTP id mv3mr42714igb.17.1437223733935; Sat, 18 Jul 2015 05:48:53 -0700 (PDT) Path: eternal-september.org!reader01.eternal-september.org!reader02.eternal-september.org!news.eternal-september.org!mx02.eternal-september.org!feeder.eternal-september.org!usenet.blueworldhosting.com!feeder01.blueworldhosting.com!border2.nntp.dca1.giganews.com!nntp.giganews.com!z61no1810315qge.0!news-out.google.com!a16ni21457ign.0!nntp.google.com!pg9no1854417igb.0!postnews.google.com!glegroupsg2000goo.googlegroups.com!not-for-mail Newsgroups: comp.lang.ada Date: Sat, 18 Jul 2015 05:48:53 -0700 (PDT) In-Reply-To: Complaints-To: groups-abuse@google.com Injection-Info: glegroupsg2000goo.googlegroups.com; posting-host=112.203.251.230; posting-account=7jTbjwoAAADuqLe9GO_M9l_Emer5tqxR NNTP-Posting-Host: 112.203.251.230 References: <53a16384-fa63-479c-9421-553683dd6009@googlegroups.com> User-Agent: G2/1.0 MIME-Version: 1.0 Message-ID: <4b621d18-91e2-4596-b064-cd7727ab12f7@googlegroups.com> Subject: Re: How to check if letters are in a string? From: Trish Cayetano Injection-Date: Sat, 18 Jul 2015 12:48:53 +0000 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Xref: news.eternal-september.org comp.lang.ada:26874 Date: 2015-07-18T05:48:53-07:00 List-Id: On Saturday, July 18, 2015 at 7:35:13 PM UTC+8, bj=F6rn lundin wrote: > Looks like I managed to mail this to the op instead of post it to c.l.a .= .. >=20 >=20 > On 2015-07-18 11:00, Trish Cayetano wrote: > > Is there another way how to check if (exact number of) letters are in > a string? > > Thank you very much! >=20 > declare > My_Letters: string :=3D "HID" > My_String: string :=3D "HIDDEN" > Is_In_String : boolean :=3D False; > begin > for L in My_Letters'range loop > Is_In_String :=3D False; > for S in My_String'range loop > if My_Letters(L) =3D My_String(S) then > Is_In_String :=3D True; > My_String(S) :=3D ' '; --set to space, so letter is not reused > exit; -- inner loop > end if; > end loop; > exit when not Is_In_String; > end loop; > Text_io.put_Line("Is_In_String =3D "& Is_In_String'Img); > end; >=20 >=20 >=20 > not compiled, not tested but the spirit of it > should do what you want >=20 > -- -- Bj=F6rn >=20 > --=20 > -- > Bj=F6rn Thank you Nasser and Bj=F6rn! Bj=F6rn nailed it! You are one genius, Bj=F6rn! It's working fine after a few tweaks (really m= inor like adding Ada and adding packages). Thank you so much!