dim pp as string
pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256,
Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156,
Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356,
Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142,
Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100,
Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142,
Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing)
pp.Replace("}", Nothing)
pp.Replace("X=", Nothing)
pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus 9 2135
I think you must do
pp = pp.Replace("{", Nothing)
pp = pp.Replace("}", Nothing)
pp = pp.Replace("X=", Nothing)
pp = pp.Replace("Y=", Nothing)
function replace (string) as string
lobrys
"Crirus" <Cr****@datagroup.ro> a écrit dans le message de
news:ez*************@tk2msftngp13.phx.gbl... dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing) pp.Replace("}", Nothing) pp.Replace("X=", Nothing) pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus
Crirus,
I Thought
Dim Crirus as String = pp.Replace(....................
(String member)
or just
Dim Crirus as String = Replace(
(String Function)
I hope this helps a little bit?
Cor
Sorry, solved...I have to assign it back to pp
"Crirus" <Cr****@datagroup.ro> wrote in message
news:ez*************@tk2msftngp13.phx.gbl... dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing) pp.Replace("}", Nothing) pp.Replace("X=", Nothing) pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus
* "Cor" <no*@non.com> scripsit: Dim Crirus as String = pp.Replace(.................... (String member) or just Dim Crirus as String = Replace( (String Function)
Member of 'Microsoft.VisualBasic.Strings'.
;-)
--
Herfried K. Wagner
MVP · VB Classic, VB.NET
<http://www.mvps.org/dotnet>
Improve your quoting style:
<http://learn.to/quote>
<http://www.plig.net/nnq/nquote.html>
Crirus,
In addition to your & the other comments about need to assign the result
back to a variable.
With four replaces in a row as you have, I would consider using
StringBuilder.Replace instead of String.Replace.
Something like:
Imports System.Text dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
Dim sb As New StringBuilder(pp)
sb.Replace("{", Nothing)
sb.Replace("}", Nothing)
sb.Replace("X=", Nothing)
sb.Replace("Y=", Nothing)
pp = sb.ToString()
Especially if I was using string concatenation to build pp in the first
place. As I would use StringBuild.Append to build the string, then use the
above Replace statements to clean it.
Dim sb As New StringBuilder(pp)
sb.Append(pt1.ToString())
sb.Append(pt2.ToString())
sb.Append(pt3.ToString())
...
sb.Replace("{", Nothing)
...
Hope this helps
Jay
"Crirus" <Cr****@datagroup.ro> wrote in message
news:ez*************@tk2msftngp13.phx.gbl... dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing) pp.Replace("}", Nothing) pp.Replace("X=", Nothing) pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus
Is faster that way? Any other advantages?
"Jay B. Harlow [MVP - Outlook]" <Ja********@email.msn.com> wrote in message
news:%2****************@TK2MSFTNGP11.phx.gbl... Crirus, In addition to your & the other comments about need to assign the result back to a variable.
With four replaces in a row as you have, I would consider using StringBuilder.Replace instead of String.Replace.
Something like:
Imports System.Text
dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
Dim sb As New StringBuilder(pp) sb.Replace("{", Nothing) sb.Replace("}", Nothing) sb.Replace("X=", Nothing) sb.Replace("Y=", Nothing)
pp = sb.ToString()
Especially if I was using string concatenation to build pp in the first place. As I would use StringBuild.Append to build the string, then use the above Replace statements to clean it.
Dim sb As New StringBuilder(pp)
sb.Append(pt1.ToString()) sb.Append(pt2.ToString()) sb.Append(pt3.ToString())
... sb.Replace("{", Nothing) ...
Hope this helps Jay
"Crirus" <Cr****@datagroup.ro> wrote in message news:ez*************@tk2msftngp13.phx.gbl... dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing) pp.Replace("}", Nothing) pp.Replace("X=", Nothing) pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus
Hi Crirus,
To give my opinion to add to the message from Jay B. because I fully agree
with him.
I don't think there is a big advantages in this example.
But the advantage is to get used to the Stringbuilder because the building
of strings is always been in every computer language a time spending part
and has never been really good implemented.
Now we have a better way, so lets use it than we take it the next time
automaticly.
I was glad with this message from Jay B.
I will try to use it next time when I give an example.
Just a thought.
Cor
Crirus,
It really depends on how you are building the string.
For four simple replacements. I would expect it to be a wash (the same
time).
However if you have a series of string concatenations before the
replacements, then yes I would expect the StringBuilder to be faster.
The other major advantage, each time you call string.Replace you are
creating a new temporary string object, if you do 4 replacements you will
have 4 temporary string objects.
With the StringBuilder you will have 1 temporary stringbuilder object, so
you are saving 3 objects that the GC needs to worry about. Would I worry
about just the 3 objects, probably not. However if I decided I needed 10
replacements or 20 replacements I think you will find that the StringBuilder
quickly becomes the better deal...
Hope this helps
Jay
"Crirus" <Cr****@datagroup.ro> wrote in message
news:eZ**************@TK2MSFTNGP11.phx.gbl... Is faster that way? Any other advantages?
"Jay B. Harlow [MVP - Outlook]" <Ja********@email.msn.com> wrote in
message news:%2****************@TK2MSFTNGP11.phx.gbl... Crirus, In addition to your & the other comments about need to assign the result back to a variable.
With four replaces in a row as you have, I would consider using StringBuilder.Replace instead of String.Replace.
Something like:
Imports System.Text
dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
Dim sb As New StringBuilder(pp) sb.Replace("{", Nothing) sb.Replace("}", Nothing) sb.Replace("X=", Nothing) sb.Replace("Y=", Nothing)
pp = sb.ToString()
Especially if I was using string concatenation to build pp in the first place. As I would use StringBuild.Append to build the string, then use
the above Replace statements to clean it.
Dim sb As New StringBuilder(pp)
sb.Append(pt1.ToString()) sb.Append(pt2.ToString()) sb.Append(pt3.ToString())
... sb.Replace("{", Nothing) ...
Hope this helps Jay
"Crirus" <Cr****@datagroup.ro> wrote in message news:ez*************@tk2msftngp13.phx.gbl... dim pp as string pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285, Y=156}{X=356, Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200, Y=177.6142}{X=177.6142, Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150, Y=100}{X=177.6142, Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
pp.Replace("{", Nothing) pp.Replace("}", Nothing) pp.Replace("X=", Nothing) pp.Replace("Y=", Nothing)
after this, pp is exactly the same as before??!!
Crirus
Good to see this for future needs...thanks
"Jay B. Harlow [MVP - Outlook]" <Ja********@email.msn.com> wrote in message
news:eW**************@tk2msftngp13.phx.gbl... Crirus, It really depends on how you are building the string.
For four simple replacements. I would expect it to be a wash (the same time).
However if you have a series of string concatenations before the replacements, then yes I would expect the StringBuilder to be faster.
The other major advantage, each time you call string.Replace you are creating a new temporary string object, if you do 4 replacements you will have 4 temporary string objects.
With the StringBuilder you will have 1 temporary stringbuilder object, so you are saving 3 objects that the GC needs to worry about. Would I worry about just the 3 objects, probably not. However if I decided I needed 10 replacements or 20 replacements I think you will find that the
StringBuilder quickly becomes the better deal...
Hope this helps Jay
"Crirus" <Cr****@datagroup.ro> wrote in message news:eZ**************@TK2MSFTNGP11.phx.gbl... Is faster that way? Any other advantages?
"Jay B. Harlow [MVP - Outlook]" <Ja********@email.msn.com> wrote in message news:%2****************@TK2MSFTNGP11.phx.gbl... Crirus, In addition to your & the other comments about need to assign the
result back to a variable.
With four replaces in a row as you have, I would consider using StringBuilder.Replace instead of String.Replace.
Something like:
Imports System.Text
> dim pp as string > pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, > Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, > Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285,
Y=156}{X=356, > Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200,
Y=177.6142}{X=177.6142, > Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, > Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150,
Y=100}{X=177.6142, > Y=100}{X=200, Y=122.3858}{X=200, Y=150}"
Dim sb As New StringBuilder(pp) sb.Replace("{", Nothing) sb.Replace("}", Nothing) sb.Replace("X=", Nothing) sb.Replace("Y=", Nothing)
pp = sb.ToString()
Especially if I was using string concatenation to build pp in the
first place. As I would use StringBuild.Append to build the string, then use the above Replace statements to clean it.
Dim sb As New StringBuilder(pp)
sb.Append(pt1.ToString()) sb.Append(pt2.ToString()) sb.Append(pt3.ToString())
... sb.Replace("{", Nothing) ...
Hope this helps Jay
"Crirus" <Cr****@datagroup.ro> wrote in message news:ez*************@tk2msftngp13.phx.gbl... > dim pp as string > pp="{X=356, Y=256}{X=356, Y=311.2285}{X=311.2285, Y=356}{X=256, > Y=356}{X=200.7715, Y=356}{X=156, Y=311.2285}{X=156, Y=256}{X=156, > Y=200.7715}{X=200.7715, Y=156}{X=256, Y=156}{X=311.2285,
Y=156}{X=356, > Y=200.7715}{X=356, Y=256}{X=200, Y=150}{X=200,
Y=177.6142}{X=177.6142, > Y=200}{X=150, Y=200}{X=122.3858, Y=200}{X=100, Y=177.6142}{X=100, > Y=150}{X=100, Y=122.3858}{X=122.3858, Y=100}{X=150,
Y=100}{X=177.6142, > Y=100}{X=200, Y=122.3858}{X=200, Y=150}" > > pp.Replace("{", Nothing) > pp.Replace("}", Nothing) > pp.Replace("X=", Nothing) > pp.Replace("Y=", Nothing) > > after this, pp is exactly the same as before??!! > > Crirus > > >
This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
by: Qwiati |
last post by:
Hi
My user has put ' with editable place and then press enter so this same way
put text information with ' string into database.
When I try tu user command : UPDATE and try to change text with '...
|
by: Thomas |
last post by:
It looks like the String.replace doesn't work in IE6.1. Anyone else
has the same problem. I am using newest service package of IE and
Win2K.
Thanks
|
by: Hp |
last post by:
Hi All,
Thanks a lot for all your replies.
My requirement is as follows:
I need to read a text file, eliminate certain special characters(like !
, - = + ), and then convert it to lower case and...
|
by: Jay S |
last post by:
Hi,
Can't figure out the following:
<concept>
I want to match a specified string followed by any string that is *not* a
different specified string.
</concept>
<specific example>
|
by: tshad |
last post by:
Can you do a search for more that one string in another string?
Something like:
someString.IndexOf("something1","something2","something3",0)
or would you have to do something like:
if...
|
by: Joebloggs |
last post by:
Hi
I am trying to do an ldap lookup. I can pick up the domain name in the
standard format DOMAIN\USERNAME. The problem is the company I work for
expects the query in the format DOMAIN:USERNAME....
|
by: Lester |
last post by:
I need a regexp function which makes a match when the string contains
<img...AND the img tag above dows NOT contain a certain path
Here is what I have:
<img\s.*(src).+>
This matches if my...
|
by: Peted |
last post by:
Is it possible to concatanate a string and a byte array, into a
"string" variable, send it as a string to an ip socket device and have
the bytes, seen as a sequence of bytes, not char or string ?
...
|
by: DotNetNewbie |
last post by:
Hi,
I want to parse a string, ONLY allowing alphanumeric characters and
also the underscore '_' and dash '-' characters.
Anything else in the string should be removed.
I think my regex is...
|
by: Charles Arthur |
last post by:
How do i turn on java script on a villaon, callus and itel keypad mobile phone
|
by: aa123db |
last post by:
Variable and constants
Use var or let for variables and const fror constants.
Var foo ='bar';
Let foo ='bar';const baz ='bar';
Functions
function $name$ ($parameters$) {
}
...
|
by: ryjfgjl |
last post by:
If we have dozens or hundreds of excel to import into the database, if we use the excel import function provided by database editors such as navicat, it will be extremely tedious and time-consuming...
|
by: ryjfgjl |
last post by:
In our work, we often receive Excel tables with data in the same format. If we want to analyze these data, it can be difficult to analyze them because the data is spread across multiple Excel files...
|
by: BarryA |
last post by:
What are the essential steps and strategies outlined in the Data Structures and Algorithms (DSA) roadmap for aspiring data scientists? How can individuals effectively utilize this roadmap to progress...
|
by: Sonnysonu |
last post by:
This is the data of csv file
1 2 3
1 2 3
1 2 3
1 2 3
2 3
2 3
3
the lengths should be different i have to store the data by column-wise with in the specific length.
suppose the i have to...
|
by: marktang |
last post by:
ONU (Optical Network Unit) is one of the key components for providing high-speed Internet services. Its primary function is to act as an endpoint device located at the user's premises. However,...
|
by: Oralloy |
last post by:
Hello folks,
I am unable to find appropriate documentation on the type promotion of bit-fields when using the generalised comparison operator "<=>".
The problem is that using the GNU compilers,...
|
by: jinu1996 |
last post by:
In today's digital age, having a compelling online presence is paramount for businesses aiming to thrive in a competitive landscape. At the heart of this digital strategy lies an intricately woven...
| |