I Have some text file. theses texts contain a string like this(a part of text):
<abbr class="word p1"">dd</abbr>
<img src"D:\Images\1.png">
<abbr class="word p1">dd</abbr>
<img src"D:\ticket\t\1.png">
In each text file,(D:\Images\1.png) png name is different but it is always numbers(from 1 to 114)for example(1,2,3,10,...)
I want to replace this text D:\Images\[number].png with a specific text for expample:
string newtext=Replace("D:\Images\[number].png","Something");
How can i do this?
thanks.
Use a regular expression:
string newtext = Regex.Replace(text, #"(D:\\Images\\)\d+(.png)","$1Something$2");
It will replace the full match, including D:\Images\ and .png, so $1 and $2 puts back what's caught by the parentheses, so that Somthing only replaces the digits.
Use regular expressions that are represented mostly be the Regex class. See these links:
http://www.codeproject.com/Articles/93804/Using-Regular-Expressions-in-C-NET
http://msdn.microsoft.com/en-us/library/ms228595%28v=vs.80%29.aspx
Related
I had to split string data based on Comma.
This is the excel data:-
Please find the excel data
string strCurrentLine="\"Himalayan Salt Body Scrub with Lychee Essential Oil from Majestic Pure, All Natural Scrub to Exfoliate & Moisturize Skin, 12 oz\",SKU_27,\"Tombow Dual Brush Pen Art Markers, Portrait, 6-Pack\",SKU_27,My Shopify Store 1,Valid,NonInventory".
Regex CSVParser = new Regex(",(?=(?:[^\"]\"[^\"]\")(?![^\"]\"))");
string[] lstColumnValues = CSVParser.Split(strCurrentLine);
I have attached the image.The problem is I used the Regex to split the string with comma but i need the ouptut just like SKU_27 because string[0] and string2 contains the forward and backward slash.I need the output string1 and remove the forward and backward slash.
The file seems to be a CVA file. For CVA to be properly formatted, it will use quotes "" to wrap strings that contains comma, such as
id, name, date
1,"Some text, that includes comma", 2020/01/01
Simply split the string by comma, you will get the 2nd column with double quote.
I'm not sure whether you are asking how to remove the double-quotes from lstColumnValues[0] and lstColumnValues[2], or add them to lstColumnValues[1].
To remove the double-quotes, just use Replace:
string myString = lstColumnValues[0].Replace("\"", "");
If you need to add them:
string myString = $"\"{lstColumnValues[1]}\"";
I'm trying to split a string (separated with the HTML break tag), without deleting the break tag. I think it's pretty messy to add a break as string after splitting, so is there any function/possibility to keep the separator while "splitting"?
Example:
<HTML><BODY><p>some text<br/>some more text</p></BODY></HTML>
Expected result:
<HTML><BODY><p>some text<br/>
some more text</p></BODY></HTML>
As far as I know SPLIT removes the separator from the results and it doesn't seem like you can change that.
But you could create your own separator by first replacing your <br/> tag with <br/> plus an arbitrary string that is highly unlikely to ever appear in your HTML source, and then split the HTML using this arbitrary string as a separator instead.
types:
begin of t_result,
segment(2000) type c,
end of t_result.
DATA:
source type string,
separator type string,
brtag type string,
repl type string,
result_tab type standard table of t_result,
result_row TYPE t_result.
brtag = '<br/>'.
separator = '|***SEP***|'.
concatenate brtag separator into repl.
source = '<HTML><BODY><p>some text<br/>some more text</p></BODY></HTML>'.
replace all occurrences of brtag in source with repl.
split source at separator into table result_tab.
LOOP AT result_tab INTO result_row.
WRITE:
result_row-segment.
ENDLOOP.
Output of that example report:
<HTML><BODY><p>some text<br/>
some more text</p></BODY></HTML>
The caveat of this solution is that your custom separator, if not chosen with some care, might appear in your HTML source on its own. I therefore would choose an arbitrary string with a special character or two that would be encoded in HTML (like umlauts) and therefore not appear in your source.
Just use the replace command. replace <br/> with <br/>CR_LF
The CR_LF refers to the carriage return linefeed character.
In more complex cases you can use regex expressions in abap.
class ZTEST_SO definition public create public .
public section.
methods t1.
ENDCLASS.
CLASS ZTEST_SO IMPLEMENTATION.
METHOD T1.
data: my_break type string,
my_string type string
value '<HTML><BODY><p>some text<br/>some more text</p></BODY></HTML>'.
my_break = '<br/>' && CL_ABAP_CHAR_UTILITIES=>CR_LF.
replace all occurrences of '<br/>' in my_string with my_break in character mode.
"check my_string in the debugger :)
"<HTML><BODY><p>some text<br/>
"some more text</p></BODY></HTML>
ENDMETHOD.
ENDCLASS.
I want to print item.rate but only with having integer values in it.Currently it is printing the rate and also the currency in character.I want to hide the chars.
<td> ${item.rate} </td>
You should be able to use Freemarker's replace function. See https://freemarker.apache.org/docs/ref_builtins_string.html#ref_builtin_replace for reference.
I'd recommend using a regular expression to find everything that's not a number. That should look something like:
${item.rate}?replace('\D', '', 'r')
Reference for Freemarker's regular expression syntax is here: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
I'm trying to read a string in a specific format
RealSociedad
this is one example of string and what I want to extract is the name of the team.
I've tried something like this,
houseteam = sscanf(str, '%s');
but it does not work, why?
You can use regexprep like you did in your post above to do this for you. Even though your post says to use sscanf and from the comments in your post, you'd like to see this done using regexprep. You would have to do this using two nested regexprep calls, and you can retrieve the team name (i.e. RealSociedad) like so, given that str is in the format that you have provided:
str = 'RealSociedad';
houseteam = regexprep(regexprep(str, '^<a(.*)">', ''), '</a>$', '')
This looks very intimidating, but let's break this up. First, look at this statement:
regexprep(str, '^<a(.*)">', '')
How regexprep works is you specify the string you want to analyze, the pattern you are searching for, then what you want to replace this pattern with. The pattern we are looking for is:
^<a(.*)">
This says you are looking for patterns where the beginning of the string starts with a a<. After this, the (.*)"> is performing a greedy evaluation. This is saying that we want to find the longest sequence of characters until we reach the characters of ">. As such, what the regular expression will match is the following string:
<ahref="/teams/spain/real-sociedad-de-futbol/2028/">
We then replace this with a blank string. As such, the output of the first regexprep call will be this:
RealSociedad</a>
We want to get rid of the </a> string, and so we would make another regexprep call where we look for the </a> at the end of the string, then replace this with the blank string yet again. The pattern you are looking for is thus:
</a>$
The dollar sign ($) symbolizes that this pattern should appear at the end of the string. If we find such a pattern, we will replace it with the blank string. Therefore, what we get in the end is:
RealSociedad
Found a solution. So, %s stops when it finds a space.
str = regexprep(str, '<', ' <');
str = regexprep(str, '>', '> ');
houseteam = sscanf(str, '%*s %s %*s');
This will create a space between my desired string.
How would I extract in Lua a text between a pattern. For example
s="this is a test string. <!2014-05-03 23:12:08!> something more"
I would need only the date/time as result: 2014-05-03 23:12:08
print(string.gsub(s, "%<!.-%!>")) doesn't work
I would need all the text WITHOUT the date/time like: "this is a
test string. something more"
The pattern "<!.-!>" works, but you need to use string.match to get the date/time part:
print(string.match(s, "<!(.-)!>"))
Note that you don't need to escape ! or < in a pattern. Of course escaping them is not an error.
To get the string without the date/time part, replace it with an empty string:
local result = string.gsub(s, "<!.-!>", "")
print(result)
You can also expand the pattern .- to validate the format of date/time more:
result = string.gsub(s, "<!%d%d%d%d%-%d%d%-%d%d%s+%d%d:%d%d:%d%d!>", "")