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Message-ID: <22285.35517.qm@web113713.mail.gq1.yahoo.com>
Date: Mon, 22 Mar 2010 18:20:57 -0700 (PDT)
From: Neil Stlyz <neilstylz@yahoo.com>
Subject: string functions and operators
To: pgsql-novice@postgresql.org, pgsql-general@postgresql.org,
	pgsql-sql@postgresql.org
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Hello, =0A=0AI have a dilema and I was hoping someone here may offer guidan=
ce or assistance. I bet this is a very simple question for someone out ther=
e but I am having problems coming up with a solution. Here it is...=0A=0Asu=
ppose I have a field with the following values:=0A=0A77.1=0A77.2=0A134.1=0A=
134.2=0A134.3=0A5.1=0A5.2=0A=0AI need two seperate SELECT queries. One woul=
d return the following values (everything left of the decimal point)=0A=0A7=
7=0A77=0A134=0A134=0A5=0A5=0A=0AThe second query would return all of the va=
lues to the right of the decimal point:=0A=0A1=0A2=0A1=0A2=0A3=0A1=0A2=0A=
=0A=0ANow, I have been using the following information (although very Greek=
) to try to solve this problem:=0A=0Ahttp://www.postgresql.org/docs/current=
/static/functions-string.html=0A=0AAnd I have been playing around with the =
syntax of the following:=0A=0A=A0=A0=A0 =A0=A0=A0 =A0=A0=A0 substring('112.=
5' from '%#"___.#"_' for '#')=0A=0Abut the aforementioned is not quite work=
ing out... can someone please=A0show me a string function=A0that will produ=
ce the desired results?=0A=0AThanks!=0A~n=0A=0A=0A=0A      
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<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:times new roman, new york, times, serif;=
font-size:12pt"><DIV>Hello, </DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>I have a dile=
ma and I was hoping someone here may offer guidance or assistance. I bet th=
is is a very simple question for someone out there but I am having problems=
 coming up with a solution. Here it is...</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>=
suppose I have a field with the following values:</DIV>=0A<DIV>&nbsp;</DIV>=
=0A<DIV>77.1</DIV>=0A<DIV>77.2</DIV>=0A<DIV>134.1</DIV>=0A<DIV>134.2</DIV>=
=0A<DIV>134.3</DIV>=0A<DIV>5.1</DIV>=0A<DIV>5.2</DIV>=0A<DIV>&nbsp;</DIV>=
=0A<DIV>I need two seperate SELECT queries. One would return the following =
values (everything left of the decimal point)</DIV>=0A<DIV>&nbsp;</DIV>=0A<=
DIV>77</DIV>=0A<DIV>77</DIV>=0A<DIV>134</DIV>=0A<DIV>134</DIV>=0A<DIV>5</DI=
V>=0A<DIV>5</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>The second query would return =
all of the values to the right of the decimal point:</DIV>=0A<DIV>&nbsp;</D=
IV>=0A<DIV>1</DIV>=0A<DIV>2</DIV>=0A<DIV>1<BR>2</DIV>=0A<DIV>3</DIV>=0A<DIV=
>1</DIV>=0A<DIV>2</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>Now,=
 I have been using the following information (although very Greek) to try t=
o solve this problem:</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV><A href=3D"http://ww=
w.postgresql.org/docs/current/static/functions-string.html">http://www.post=
gresql.org/docs/current/static/functions-string.html</A></DIV>=0A<DIV>&nbsp=
;</DIV>=0A<DIV>And I have been playing around with the syntax of the follow=
ing:</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;=
 &nbsp;&nbsp;&nbsp; substring('112.5' from '%#"___.#"_' for '#')</DIV>=0A<D=
IV>&nbsp;</DIV>=0A<DIV>but the aforementioned is not quite working out... c=
an someone please&nbsp;show me a string function&nbsp;that will produce the=
 desired results?</DIV>=0A<DIV>&nbsp;</DIV>=0A<DIV>Thanks!</DIV>=0A<DIV>~n<=
/DIV>=0A<DIV>&nbsp;</DIV></div><br>=0A=0A      </body></html>
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