delimiter

Java parsing string

由 孤人 提交于 2019-12-06 00:31:27
I'm looking to parse the following string in java <some lines here> Key1:thingIWantToKnow Key2:otherThing Key3:bla Key4:bla Key5:bla <(possibly) more lines here> All lines end with a newline (\n) character. I'm looking to store the value pair once I find the key's I'm care about. If a Map is what you want: Map<String, String> keyValueMap = new HashMap<String,String>(); String[] lines = input.split("\n"); if (lines == null) { //Compensate for strange JDK semantics lines = new String[] { input }; } for (String line : lines) { if (!line.contains(":")) { //Skip lines that don't contain key-value

PHP preg_split with two delimiters unless a delimiter is within quotes

由 不打扰是莪最后的温柔 提交于 2019-12-06 00:28:26
Further on from my previous question about preg_split which was answers super fast, thanks to nick; I would really like to extend the scenario to no split the string when a delimiter is within quotes. For example: If I have the string foo = bar AND bar=foo OR foobar="foo bar" , I'd wish to split the sting on every space or = character but include the = character in the returned array (which works great currently), but I don't want to split the string either of the delimiters are within quotes. I've got this so far: <!doctype html> <?php $string = 'foo = bar AND bar=foo'; $array = preg_split('/

Cut string after first occurrence of a character

由 邮差的信 提交于 2019-12-05 23:20:54
问题 I have strings like 'keepme:cutme' or 'string-without-separator' which should become respectively 'keepme' and 'string-without-separator'. Can this be done in PostgreSQL? I tried: select substring('first:last' from '.+:') But this leaves the : in and won't work if there is no : in the string. 回答1: Use split_part(): SELECT split_part('first:last', ':', 1) AS first_part Returns the whole string if the delimiter is not there. And it's simple to get the 2nd or 3rd part etc. Substantially faster

Changing the scanf() delimiter

由 巧了我就是萌 提交于 2019-12-05 22:33:00
My objective is to change the delimiter of scanf to " \n ". I tried using scanf("%[^\n]s",sen); and works fine for single inputs. But when i put the same line inside a for loop for multiple sentences it gives me garbage values. Does anyone know why? Here's my code: char sen[20]; for (i=0;i<2;i++) { scanf("%[^\n]s",sen); printf("%s\n",sen); } Consider this (C99) code: #include <stdio.h> int main(void) { char buffer[256]; while (scanf("%255[^\n]", buffer) == 1) printf("Found <<%s>>\n", buffer); int c; if ((c = getchar()) != EOF) printf("Failed on character %d (%c)\n", c, c); return(0); } When I

JS RegEx to match all characters (including newline) between two strings but without the two strings?

由 最后都变了- 提交于 2019-12-05 21:33:40
Example Text: <div id="not-wanted"> no no no </div> <div id="wanted">I want only this text </div> no no no no no <div id="not-wanted"> no no no </div> <div id="wanted">no no no no</div> <div id="wanted"> no no </div> Should deliver: I want only this text Or better: I want only this text Unfortunately, my solution catches the 2 delimitation strings also: $('#put').append(/<div id="wanted">[^<>]*<\/div>/.exec(strg)[0]); ==> <div id="wanted">I want only this text </div> Online example http://regex101.com/r/rF7jR9 Question What regular expression for Java Script can deliver the characters between

Replacing delimiter characters in file path

由 余生颓废 提交于 2019-12-05 14:24:02
I'm developing a C# web application in VS 2008. I let the user select an input file and then I store the file path in a string variable. However, it stores this path as "C:\\folder\\..." . So my question is how do I convert this file path into single "\"? Thank you guys for all your helps! Please forgive me as I am a newbie to ASP.NET development. This is more of my code in context. First I want to see if the directory exists. I guess I don't have to check this if I check if the file exists. But this should still work right? And currently my "path" string variable is not showing up the way I

D3 remove comma delimiters for thousands

由 ◇◆丶佛笑我妖孽 提交于 2019-12-05 09:39:33
问题 I have a .json with 3 columns where one of them is 'Year'. The column contains only years. No dates!. When I am graphing it on the 'x' axis the years come out with a comma delimiter for thousands. So in the .json the date is this format :"Year":1990 and on the 'x' axis it comes out like that 1,990 I have been trying to figure out how to parse the year but I have no luck so far. I have tried the following: var parseDate = d3.time.format("%y").parse var x = d3.time.scale() .range([0, width]); /

Error near 'DELIMITER $$'

由 萝らか妹 提交于 2019-12-05 02:30:47
问题 when I change Delimeter from mysql console or MySQL Workbench I do not get any error, but when I embed the same code in ruby on rails I get error mysql> DELIMITER $$ mysql> gives no error. but ActiveRecord::Base.connection.execute(%Q{ DELIMITER $$ }) gives: ActiveRecord::StatementInvalid: Mysql2::Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'DELIMITER $$' at line 1: 回答1: DELIMITER is actually a

Java CSV parser with string separator (multi-character)

由 耗尽温柔 提交于 2019-12-05 01:08:43
Is there any Java open source library that supports multi-character (i.e., String with length > 1) separators (delimiters) for CSV? By definition, CSV = Comma-Separated Values data with a single character (',') as the delimiter. However, many other single-character alternatives exist (e.g., tab), making CSV to stand for "Character-Separated Values" data (essentially, DSV: Delimiter-Separated Values data). Main Java open source libraries for CSV (e.g., OpenCSV ) support virtually any character as the delimiter, but not string (multi-character) delimiters. So, for data separated with strings

Replacing one character with another in a string

由 房东的猫 提交于 2019-12-04 22:32:37
I have a data like below: A:B:C:D and I want to replace the C with data (say, Z ) so that it may look like A:B:Z:D How can I do it? =SUBSTITUTE(A1,"C","Z") Although I wasn't clear on whether you wanted G or Z , you mentioned G but your example output shows Z . jrad If you have A:B:C:D in cell A1, then this works: =CONCATENATE(MID(A1, 1, SEARCH(":", MID(A1, SEARCH(":", A1) + 1, LEN(A1) - SEARCH(":", A1) + 1)) + SEARCH(":", A1)), "Z", MID(MID(MID(A1, SEARCH(":", A1) + 1, LEN(A1) - SEARCH(":", A1) + 1), SEARCH(":", MID(A1, SEARCH(":", A1) + 1, LEN(A1) - SEARCH(":", A1) + 1)) + 1, LEN(MID(A1,
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