I was trying to have the value as 456,987,214.
But For me it is coming like without comma.
Here is my code, Did i mistake anything
const string price = "^\\d{3},\\d{3},\\d{3}$";
string pricelist = query.price.ToString();
string price1 = "";
if (System.Text.RegularExpressions.Regex.IsMatch(pricelist.ToString(), price))
{
price1 = query.price.ToString();
}
I noticed it, it is number what you intended to format why not using angular built-in formatting feature like: {{price | number}}
Online Demo
If you want to format this with C# then you can format it using .Net built-in formatters like:
double value = 1234567890;
Console.WriteLine(value.ToString("#,#", CultureInfo.InvariantCulture));
// outputs 1,234,567,890
Try this simple Method:
Just pass your value like this ,
double value = double.Parse(query.price.ToString());
string price1= value.ToString("#,#", CultureInfo.InvariantCulture);
Output will be like 123,456,789
Try this simple method: Just pass your value like this, then you will get the output as you want.
double value = double.Parse(query.price.ToString());
string price1= value.ToString("#,#", CultureInfo.InvariantCulture);
Output will be like 123,456,789
Related
I have a list of filenames in a struct array, example:
4x1 struct array with fields:
name
date
bytes
isdir
datenum
where files.name
ans =
ts.01094000.crest.csv
ans =
ts.01100600.crest.csv
etc.
I have another list of numbers (say, 1094000). And I want to find the corresponding file name from the struct.
Please note, that 1094000 doesn't have preceding 0. Often there might be other numbers. So I want to search for '1094000' and find that name.
I know I can do it using Regex. But I have never used that before. And finding it difficult to write for numbers instead of text using strfind. Any suggestion or another method is welcome.
What I have tried:
regexp(files.name,'ts.(\d*)1094000.crest.csv','match');
I think the regular expression you'd want is more like
filenames = {'ts.01100600.crest.csv','ts.01094000.crest.csv'};
matches = regexp(filenames, ['ts\.0*' num2str(1094000) '\.crest\.csv']);
matches = ~cellfun('isempty', matches);
filenames(matches)
For a solution with strfind...
Pre-16b:
match = ~cellfun('isempty', strfind({files.name}, num2str(1094000)),'UniformOutput',true)
files(match)
16b+:
match = contains({files.name}, string(1094000))
files(match)
However, the strfind way might have issues if the number you are looking for exists in unexpected places such as looking for 10 in ["01000" "00101"].
If your filenames match the pattern ts.NUMBER.crest.csv, then in 16b+ you could do:
str = {files.name};
str = extractBetween(str,4,'.');
str = strip(str,'left','0');
matches = str == string(1094000);
files(matches)
I have a string like this: 001,"John Marvin","doctor", "full time"
I want to delete everything after (001) with substr, but, the length of (001) is not always 3 so I can not put something like thie:
string chain = "001,\"John Marvin\",\"doctor\", \"full time\"";
std::string partial = chain.substr(0,3);
How can I proceed in this case?
You could find the index of the first comma and use that to determine where to cut off the string.
Something like:
string chain = "001,\"John Marvin\",\"doctor\", \"full time\"";
int cutoff = chain.find(',');
string newString = chain.substr(0, cutoff);
Tested here.
I have the following string:
{'output',{'variable','VGRG_Pos_Var1/Parameters/D_foo'},'date',734704.60904050921}
I would like to verify the format of the string that the word 'variable' is the second word and i would like to retrive the string after the last '/' in the 3rd string (In this example 'D_foo').
how could i verify this and retrive the sting i search?
I tried the following:
regexp(str,'{''\w+'',{''variable'',''([(a-z)|(A-Z)|/|_])+')
without success
REMARK
The string to analysis is not splited after the komma, it is only due to length of the string.
EDIT
my string is:
'{''output'',{''variable'',''VGRG_Pos_Var1/Parameters/D_foo''},''date'',734704.60904050921}';
and not a cell, which could be understood. I added the sybol ' at the start and end of the string to symbolizied that it is a string.
I realise that you mention using regexp in the question, but I'm not sure if this is a requirement? If other solutions are acceptable you could try this:
str='{''output'',{''variable'',''VGRG_Pos_Var1/Parameters/D_foo''},''date'',734704.60904050921}';
parts1=textscan( str, '%s','delimiter',{',','{','}'},'MultipleDelimsAsOne',1);
parts2=textscan( parts1{1}{3}, '%s','delimiter',{'/',''''},'MultipleDelimsAsOne',1);
string=parts2{1}{end}
match=strcmp(parts1{1}{2},'variable')
To answer the first part of your question, you can write this:
str = {'output',{'variable','VGRG_Pos_Var1/Parameters/D_foo'},'date',734704.60904050921};
temp = str(2); %this holds the cell containing the two strings
if cmpstr(temp{1}(1), 'variable')
%do stuff
end
For the second part you can do this:
str = {'output',{'variable','VGRG_Pos_Var1/Parameters/D_foo'},'date',734704.60904050921};
temp = str(2); %like before, this contains the cell
temp = temp{1}(2); %this picks out the second string in the cell
temp = char(temp); %turns the item from a cell to a string
res = strsplit(temp, '/'); %splits the string where '/' are found, res is an array of strings
string = res(3); %assuming there will always be just 2 '/'s.
I have a list of several phrases in the following format
thisIsAnExampleSentance
hereIsAnotherExampleWithMoreWordsInIt
and I'm trying to end up with
This Is An Example Sentance
Here Is Another Example With More Words In It
Each phrase has the white space condensed and the first letter is forced to lowercase.
Can I use regex to add a space before each A-Z and have the first letter of the phrase be capitalized?
I thought of doing something like
([a-z]+)([A-Z])([a-z]+)([A-Z])([a-z]+) // etc
$1 $2$3 $4$5 // etc
but on 50 records of varying length, my idea is a poor solution. Is there a way to regex in a way that will be more dynamic? Thanks
A Java fragment I use looks like this (now revised):
result = source.replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
result = result.substring(0, 1).toUpperCase() + result.substring(1);
This, by the way, converts the string givenProductUPCSymbol into Given Product UPC Symbol - make sure this is fine with the way you use this type of thing
Finally, a single line version could be:
result = source.substring(0, 1).toUpperCase() + source(1).replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
Also, in an Example similar to one given in the question comments, the string hiMyNameIsBobAndIWantAPuppy will be changed to Hi My Name Is Bob And I Want A Puppy
For the space problem it's easy if your language supports zero-width-look-behind
var result = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "(?<=[a-z])([A-Z])", " $1");
or even if it doesn't support them
var result2 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "([a-z])([A-Z])", "$1 $2");
I'm using C#, but the regexes should be usable in any language that support the replace using the $1...$n .
But for the lower-to-upper case you can't do it directly in Regex. You can get the first character through a regex like: ^[a-z] but you can't convet it.
For example in C# you could do
var result4 = Regex.Replace(result, "^([a-z])", m =>
{
return m.ToString().ToUpperInvariant();
});
using a match evaluator to change the input string.
You could then even fuse the two together
var result4 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "^([a-z])|([a-z])([A-Z])", m =>
{
if (m.Groups[1].Success)
{
return m.ToString().ToUpperInvariant();
}
else
{
return m.Groups[2].ToString() + " " + m.Groups[3].ToString();
}
});
A Perl example with unicode character support:
s/\p{Lu}/ $&/g;
s/^./\U$&/;
I have a huge log file with different types of string rows, and I need to extract data in a "smart" way from these.
Sample snippet:
2011-03-05 node32_three INFO stack trace, at empty string asfa 11120023
--- - MON 23 02 2011 ERROR stack trace NONE
For instance, what is the best way to extract the date from each row, independent of date format?
You could make a regex for different formats like so:
(fmt1)|(fmt2)|....
Where fmt1, fmt2 etc are the individual regexes, for yor example
(20\d\d-[01]\d-[0123]\d)|((?MON|TUE|WED|THU|FRI|SAT|SUN) [0123]\d [01]\d 20\d\d)
Note that to prevent the chance to match arbitrary numbers I restricted year, month and day numbers accordingly. For example, a day number cannot start with 4, neither can a month number start with 2.
This gives the following pseudo code:
// remember that you need to double each backslash when writing the
// pattern in string form
Pattern p = Pattern.compile("..."); // compile once and for all
String s;
for each line
s = current input line;
Matcher m = p.matcher(s);
if (m.find()) {
String d = m.group(); // d is the string that matched
....
}
Each individual date pattern is written in () to make it possible to find out what format we had, like so:
int fmt = 0;
// each (fmt) is a group, numbered starting with 1 from left to right
for (int i = 1; fmt == 0 && i <= total number of different formats; i++)
if (m.group(i) != null) fmt = i;
For this to work, inner (regex) groups must be written (?regex) so that they do not count as capture-groups, look at updated example.
If you use Java, you may want to have a look at Joda time. Also, read this question and related answers. I think Joda DateTimeFormat should give you all the flexibility that you need to parse the various date/time format of your log file.
A quick example:
String dateString = "2011-04-18 10:41:33";
DateTimeFormatter formatter =
DateTimeFormat.forPattern("yyyy-MM-dd HH:mm:ss");
DateTime dateTime = formatter.parseDateTime(dateString);
Just define a String[] for the formats of you date/time, and pass each element to DateTimeFormat to get the corresponding DateTimeFormatter. You can use regex just separate date strings from other stuff in the log lines, and then you can use the various DateTimeFormatters to try and parse them.