I have a query that looks like this:
INSERT INTO table VALUES ('47677;2019;2019;10T-1001-10010AS;A05;International;TieLineKoman-KosovoB;L_KOM-KOSB;2018;NULL;NULL;;NULL;Tieline;NULL;10XAL-KESH-----J;0;3')
that is produced by parsing a csv file.
The query is not in a valid form, I have to replace all semicolons with the string ',' (comma inside single quotes). What I want to get is:
('47677','2019','2019','10T-1001-10010AS','A05','International','TieLineKoman-KosovoB','L_KOM-KOSB','2018','NULL','NULL','','NULL','Tieline','NULL','10XAL-KESH-----J','0','3')
I have tried to do this in many different ways, but I end up with backshlashes added in my string. This is what I get:
"INSERT INTO AllocatedEICDetail VALUES ('47677\\',\\'2019\\',\\'2019\\',\\'10T-1001-10010AS\\',\\'A05\\',\\'International\\',\\'TieLineKoman-KosovoB\\',\\'L_KOM-KOSB\\',\\'2018\\',\\'NULL\\',\\'NULL\\',\\'\\',\\'NULL\\',\\'Tieline\\',\\'NULL\\',\\'10XAL-KESH-----J\\',\\'0\\',\\'3')"
Any ideas how to do this properly without having the backslashes added?
Thank you!
//the string you have
const string = '47677;2019;2019;10T-1001-10010AS;A05;International;TieLineKoman-KosovoB;L_KOM-KOSB;2018;NULL;NULL;;NULL;Tieline;NULL;10XAL-KESH-----J;0;3';
//the string you need:
const targetString = string.replace(/\;/g,',');
You specify a small regex between the forward slashes in replace which is a simple ';', give it a 'g' flag for global which will replace all instances, and in the second argument supply what you need it replaced with.
Related
I have varibale which contains raw data as shown below.
I want to replace the comma inside double quotes to nothing/blank.
I used replaceAll(',',''), but all other commas are also getting replaced.
So need regex function to identify pattern like "123,456,775" and then replace here comma into blank.
var = '
2/5/2023,25,"717,990","18,132,406"
2/4/2023,27,"725,674","19,403,116"
2/3/2023,35,"728,501","25,578,008"
1/31/2023,37,"716,580","26,358,186"
2/1/2023,37,"720,466","26,494,010"
1/30/2023,37,"715,685","26,517,878"
2/2/2023,37,"723,545","26,603,765" '
Tried replaceAll, but did not work
If you just want to replace "," with "", you have to escape the quotes this will do:
var.replaceAll(/\",\"/, /\"\"/)
If you want to replace commas inside the number strings, "725,674" with "725674" you will have to use a regex and capture groups, like this:
var.replaceAll(/(\"\d+),(\d+\")/, /$1$2/)
It will change for three groupings, like "18,132,406", you will have to use three capture groups.
I am trying to clean text strings containing any ' or ' (which includes an ; but if i add it here you will see just ' again. Because the the ANSI is also encoded by stackoverflow. The string content contains ' and when it does there is an error.
when i insert the string to my database i get this error:
psycopg2.ProgrammingError: syntax error at or near "s"
LINE 1: ...tment and has commenced a search for mr. whitnell's
the original string looks like this:
...a search for mr. whitnell's...
To remove the ' and ' ; I use:
stripped_content = stringcontent.replace("'","")
stripped_content = stringcontent.replace("' ;","")
any advice is welcome, best regards
When you try to replace("' ;","") it literally searching for "' ;" occurrences in string. You need to convert "' ;" to its character equivalent. Try this:
s = "That's how we 'roll"
r = s.replace(chr(int('''[2:])), "")
and with this chr(int('''[2:])) you'll get ' character.
Output:
Thats how we roll
Note
If you try to run this s.replace(chr(int('''[2:])), "") without saving your result in variable then your original string would not be affected.
I am trying to add changes data in a csv file:
This is the sample data:
DATE status code value value2
"2016-01-26","Subscription All","119432660","1315529431362550","0.0080099833517888"
"2016-01-26","Subscription All","119432664","5836995058433524","0.033825584764444"
"2016-01-26","Subscription All","119432664","8287300074499777","0.076913377834744"
"2016-01-26","Subscription All","119432664","14870697739968326","0.0074188355187426"
My code used to format the data:
CSVReader reader = new CSVReader(new FileReader(new File(fileToChange)), CSVParser.DEFAULT_SEPARATOR, CSVParser.NULL_CHARACTER, CSVParser.NULL_CHARACTER, 1)
info "Read all rows at once"
List<String[]> allRows = reader.readAll();
CSVWriter writer = new CSVWriter(new FileWriter(fileToChange), CSVWriter.DEFAULT_SEPARATOR, CSVWriter.NO_QUOTE_CHARACTER)
writer.writeAll(allRows)
writer.close()
The output i get is this, with extra quote added instead of removing it.
""2016-01-26"",""Subscription All"",""119432660"",""1315529431362550"",""0.0080099833517888""
""2016-01-26"",""Subscription All"",""119432664"",""5836995058433524"",""0.033825584764444""
""2016-01-26"",""Subscription All"",""119432664"",""8287300074499777"",""0.076913377834744""
""2016-01-26"",""Subscription All"",""119432664"",""14870697739968326"",""0.0074188355187426""
I want to remove the quotes.
Please can someone help.
Also, is it possible to change the date format to yyyymmdd instead of yyyy-mm-dd?
allRows.each { String[] theLine ->
String newDate = theLine[0].replaceAll('-', '')
String newline = theLine.eachWithIndex { String s, int i -> return i > 0 ? s : newDate}
writer.writeLine(newline)
}
Thanks
When you instantiated your CSVReader you told it to treat no characters as quotes, therefore it read the existing quotes as data and did not remove them.
When you told CSVWriter not to add any quotes it honored your request. However, the input data contained quote characters, and the convention for including quotes inside a string in CSV is to double the quotes. Thus the
string value
ABC"DEF
gets coded in CSV as
"ABC""DEF"
So the result you see is the combination of not removing the quotes on input (you told it not to) and then doubling the quotes on output.
To solve this change the input option from NULL_CHARACTER to DEFAULT_QUOTE_CHARACTER. However be aware that if any of your data actually contains embedded quotes or commas the resulting output will not be valid CSV.
Also I think this might be a valid bug report against OpenCSV. I believe that OpenCSV needs to inform you if it is about to generate invalid CSV when you told it to omit quotes, probably via a runtime exception. Although I suppose they might argue that you chose to work without a net and should accept whatever you get. Personally I go for the "principle of least surprise", which IMHO would be not to double quotes when the output is unquoted.
Because quotation in your CSVReader is set to CSVParser.NULL_CHARACTER " is treated as normal character which is part of read token. This causes your array to contain data in form:
["2016-01-26", "Subscription All", "119432660", "1315529431362550", "0.0080099833517888"]
rather than:
[2016-01-26, Subscription All, 119432660, 1315529431362550, 0.0080099833517888]
So try changing option from CSVParser.NULL_CHARACTER to either
'"'
CSVParser.DEFAULT_QUOTE_CHARACTER (it also stores '"').
CsvToBean csvToBean = new CsvToBeanBuilder(new StringReader(csv))
.withMappingStrategy(strategy)
.withIgnoreLeadingWhiteSpace(true)
.withSeparator(',')
.withIgnoreEmptyLine(true)
.withQuoteChar('\'')
.withQuoteChar('"')
.build();
I have a string to modify as per the requirements.
For example:
The given string is:
str1 varchar = '123,456,789';
I want to show the string as:
'456,789'
Note: The first part (delimited) with comma, I want to remove from string and show the rest of string.
In SQL Server I used STUFF() function.
SELECT STUFF('123,456,789',1,4,'');
Result:
456,789
Question: Is there any string function in PostgreSQL 9.3 version to do the same job?
you can use regular expressions:
select substring('123,456,789' from ',(.*)$');
The comma matches the first comma found in the string. The part inside the brackets (.*) is returned from the function. The symbol $ means the end of the string.
A alternative solution without regular expressions:
select str, substring(str from position(',' in str)+1 for length(str)) from
(select '123,456,789'::text as str) as foo;
You could first turn the string to array and return second and third cell:
select array_to_string((regexp_split_to_array('123,456,789', ','))[2:3], ',')
Or you could use substring-function with regular expressions (pattern matching):
SELECT substring('123,456,789' from '[0-9]+,([0-9]+,[0-9]+)')
[0-9]+ means one or more digits
parentheses tell to return that part from the string
Both solutions work on your specific string.
Your The SQL Server example indicates you just want to remove the first 4 characters, which makes the rest of your question seem misleading because it completely ignores what's in the string. Only the positions matters.
Be that as it may, the simple and cheap way to cut off leading characters is with right():
SELECT right('123,456,789', -4);
SQL Fiddle.
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.