Hi all,
I have a problem constructing a regular expression using .net.
I have a string, separated with comma, and I want to group the string
together but, I failed to group a numeric character with decimal values.
Example string : 1, 2.3, "two"," three"
So, I want to group this string into 4 groups (1), (2.3), (two) and (three)
The best regular expression that I have so far is:
(?:^|\s*\,\s*)( (?:"(?<SubStrin g>(?:""|[^"])*)")+)|((?<Sub String>(\d))+)
But this regex will return (1), (2), (3), (two) and (three).
So, what is the right regular expression to do this? Please help.
Thanks. 8 3376
Will String.Split(", ") do ?
<Kalpesh/>
No. Consider this string:
string s = "1, 2.3, \"ab,\"\"c\" , \"e.ff,;$\"" ;
I want to split it into (1), (2.3), (ab,""c) and (e.ff,;$).
Got my point? Please help.
"Kalpesh Shah" <ka*****@disc.m icrosoft.com> wrote in message
news:OV******** ******@TK2MSFTN GP10.phx.gbl... Will String.Split(", ") do ?
<Kalpesh/>
Hello
Try this expression
(?:^|\s*\,\s*)( ?:(?:"(?<SubStr ing>(?:""|[^"])*)"\s*)|(?:\s* (?<SubString>(? :\
s*[^\s,]+)*)\s*))
Best regards
Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message
news:uG******** ******@tk2msftn gp13.phx.gbl... Hi all,
I have a problem constructing a regular expression using .net.
I have a string, separated with comma, and I want to group the string together but, I failed to group a numeric character with decimal values.
Example string : 1, 2.3, "two"," three"
So, I want to group this string into 4 groups (1), (2.3), (two) and
(three) The best regular expression that I have so far is: (?:^|\s*\,\s*)( (?:"(?<SubStrin g>(?:""|[^"])*)")+)|((?<Sub String>(\d))+)
But this regex will return (1), (2), (3), (two) and (three).
So, what is the right regular expression to do this? Please help.
Thanks.
Thanx a lot Metainy!
It works. But that regex also matches invalid decimal values. Like, it still
match 1.23aa value. And same case also happened to the double-quote
character, of which I wanted it to start and end with double-quote, with no
trailing character except a comma or no char at all. Got my point?
Just to get it clear:
- 1.23, "abc"qwe" = valid
- 1.23x, "abc"qwe" = invalid
- 1.23, "abc"qwe"xx = invalid
And one more thing is, any good (but free) resource of regex tutorial? ebook
or website.
Still hoping for assistance here.
Thank you.
Hello Ahmad
Try this one
^(?:(?:(?:^|,)\ s*)(?:(?:"(?<Su bString>(""|[^"])*)")|(?<SubStr ing>\d+(?:\.\d+
)?))(?=\s*(?:$| ,))\s*)+$
Best regards,
Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message
news:Ou******** ******@TK2MSFTN GP10.phx.gbl... Thanx a lot Metainy!
It works. But that regex also matches invalid decimal values. Like, it
still match 1.23aa value. And same case also happened to the double-quote character, of which I wanted it to start and end with double-quote, with
no trailing character except a comma or no char at all. Got my point?
Just to get it clear: - 1.23, "abc"qwe" = valid - 1.23x, "abc"qwe" = invalid - 1.23, "abc"qwe"xx = invalid
And one more thing is, any good (but free) resource of regex tutorial?
ebook or website.
Still hoping for assistance here.
Thank you.
Hi ElMetainy,
That one does work, but I also need the double-quote character to be in
between the double quote.
Llike my previous post:
1.23, "abc"qwe" = valid (and by using MatchCollection on <SubString>, this
will return [1.23] and [abc"qwe])
1.23, "abc"qwe"xx = invalid
1.23xx, "abc"qwe" = invalid
Can you help me...just a little bit more? :) You almost got it right.
p/s: Sorry, I still got no time to learn regex. But I really need a quick
solution right now.
"Sherif ElMetainy" <el************ *@wayout.net.NO SPAM> wrote in message
news:%2******** **********@TK2M SFTNGP09.phx.gb l... Hello Ahmad
Try this one
^(?:(?:(?:^|,)\ s*)(?:(?:"(?<Su bString>(""|[^"])*)")|(?<SubStr ing>\d+(?:\.\d+ )?))(?=\s*(?:$| ,))\s*)+$
Best regards, Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message news:Ou******** ******@TK2MSFTN GP10.phx.gbl... Thanx a lot Metainy!
It works. But that regex also matches invalid decimal values. Like, it still match 1.23aa value. And same case also happened to the double-quote character, of which I wanted it to start and end with double-quote, with no trailing character except a comma or no char at all. Got my point?
Just to get it clear: - 1.23, "abc"qwe" = valid - 1.23x, "abc"qwe" = invalid - 1.23, "abc"qwe"xx = invalid
And one more thing is, any good (but free) resource of regex tutorial? ebook or website.
Still hoping for assistance here.
Thank you.
Hello
This can be too complicated
How do I treat the double quote and comma. A ',' between double quotes is
considered a part of the string and a double quote between double quotes is
also considered a part of the string
Imagine this
1.23,"aa,ddd",1 23 this should match [1.23], [aa,ddd] and [123]
1.23,"abc"qwe," ee",124 should match [1.23], [abc"qwe,"ee] and [124] or be
considered invalid??
To take this decision you have to understand the nature of the data (for
example being able to distinguish a contact's first name from his
nickname) which is not possible with regular expressions.
This is why it is difficult to match one double quote between 2 double
quotes.
Here is where the "" resolves the ambiguity
1.23,"abc""qwe, ""ee",124 should match [1.23], [abc"qwe,"ee] and [124]
meaning that 2 consecutive double quotes within double quotes should be
treated as a one double quote which is a part of the string. The "" is
standard in formats like csv.
Best regards,
Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message
news:#n******** ******@TK2MSFTN GP09.phx.gbl... Hi ElMetainy,
That one does work, but I also need the double-quote character to be in between the double quote. Llike my previous post:
1.23, "abc"qwe" = valid (and by using MatchCollection on <SubString>, this will return [1.23] and [abc"qwe]) 1.23, "abc"qwe"xx = invalid 1.23xx, "abc"qwe" = invalid
Can you help me...just a little bit more? :) You almost got it right.
p/s: Sorry, I still got no time to learn regex. But I really need a quick solution right now.
"Sherif ElMetainy" <el************ *@wayout.net.NO SPAM> wrote in message news:%2******** **********@TK2M SFTNGP09.phx.gb l... Hello Ahmad
Try this one
^(?:(?:(?:^|,)\ s*)(?:(?:"(?<Su bString>(""|[^"])*)")|(?<SubStr ing>\d+(?:\.\d+ )?))(?=\s*(?:$| ,))\s*)+$
Best regards, Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message news:Ou******** ******@TK2MSFTN GP10.phx.gbl... Thanx a lot Metainy!
It works. But that regex also matches invalid decimal values. Like, it still match 1.23aa value. And same case also happened to the double-quote character, of which I wanted it to start and end with double-quote,
with no trailing character except a comma or no char at all. Got my point?
Just to get it clear: - 1.23, "abc"qwe" = valid - 1.23x, "abc"qwe" = invalid - 1.23, "abc"qwe"xx = invalid
And one more thing is, any good (but free) resource of regex tutorial? ebook or website.
Still hoping for assistance here.
Thank you.
Hi,
I know that it was too complicated, that's why I'm here.
But, I think I have my way out now. I can use MatchColelction and break the
string apart between the comma, and use another regex to validate every
broken string. :)
Anyway, you've been a great help ElMetainy. Thanks a lot.
Bye.
"Sherif ElMetainy" <el************ *@wayout.net.NO SPAM> wrote in message
news:uy******** *****@TK2MSFTNG P11.phx.gbl... Hello
This can be too complicated
How do I treat the double quote and comma. A ',' between double quotes is considered a part of the string and a double quote between double quotes
is also considered a part of the string Imagine this
1.23,"aa,ddd",1 23 this should match [1.23], [aa,ddd] and [123]
1.23,"abc"qwe," ee",124 should match [1.23], [abc"qwe,"ee] and [124] or be considered invalid?? To take this decision you have to understand the nature of the data (for example being able to distinguish a contact's first name from his nickname) which is not possible with regular expressions.
This is why it is difficult to match one double quote between 2 double quotes.
Here is where the "" resolves the ambiguity 1.23,"abc""qwe, ""ee",124 should match [1.23], [abc"qwe,"ee] and [124] meaning that 2 consecutive double quotes within double quotes should be treated as a one double quote which is a part of the string. The "" is standard in formats like csv.
Best regards, Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message news:#n******** ******@TK2MSFTN GP09.phx.gbl... Hi ElMetainy,
That one does work, but I also need the double-quote character to be in between the double quote. Llike my previous post:
1.23, "abc"qwe" = valid (and by using MatchCollection on <SubString>,
this will return [1.23] and [abc"qwe]) 1.23, "abc"qwe"xx = invalid 1.23xx, "abc"qwe" = invalid
Can you help me...just a little bit more? :) You almost got it right.
p/s: Sorry, I still got no time to learn regex. But I really need a
quick solution right now.
"Sherif ElMetainy" <el************ *@wayout.net.NO SPAM> wrote in message news:%2******** **********@TK2M SFTNGP09.phx.gb l... Hello Ahmad
Try this one
^(?:(?:(?:^|,)\ s*)(?:(?:"(?<Su bString>(""|[^"])*)")|(?<SubStr ing>\d+(?:\.\d+ )?))(?=\s*(?:$| ,))\s*)+$
Best regards, Sherif
"Ahmad A. Rahman" <je************ @yahoo.com> wrote in message news:Ou******** ******@TK2MSFTN GP10.phx.gbl... > Thanx a lot Metainy! > > It works. But that regex also matches invalid decimal values. Like,
it still > match 1.23aa value. And same case also happened to the double-quote > character, of which I wanted it to start and end with double-quote, with no > trailing character except a comma or no char at all. Got my point? > > Just to get it clear: > - 1.23, "abc"qwe" = valid > - 1.23x, "abc"qwe" = invalid > - 1.23, "abc"qwe"xx = invalid > > And one more thing is, any good (but free) resource of regex
tutorial? ebook > or website. > > Still hoping for assistance here. > > Thank you. > > > >
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