I'm working with a CSV file that in places, has multiple commas and pound signs. My question is about how to remove the multiple commas and the pound signs, while leaving a single comma between fields.
The part of this task I am on is, using only java and no external libraries to sort through the csv file sort the array by price. I am to input a number as an input parameter and return that number of rows, ordered by price.
What I have currently is around 1000 lines of data that looks like this:
18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,,£307018.48,
I need to remove the double commas and the pound sign, but for the life of me haven't been able to get it to work.
This is the line I am using for the regex.
String currentLine = line.replaceAll("[,{2}|£]", "");
This outputs a line which looks like this:
100086 Norway Maple WayMadelleGeorgeotmgeorgeotrr#hao13.com417175.60
A larger chunk of the code looks like this and by no means is it nearly finished:
public String[] getTopProperties(int n){
String[] properties = new String[n];
String file = "data.csv";
String line = "";
String splitBy = ",";
try (BufferedReader br = new BufferedReader(new FileReader(file))) {
while ((line = br.readLine()) != null) {
String currentLine = line.replaceAll("[,{2}|£]", "");
System.out.println("Current line is: " + currentLine);
String[] user = currentLine.split(splitBy);
}
} catch (IOException e) {
e.printStackTrace();
}
return properties;
}
Issue is it's now removed all the commas and where the price and double commas used to be, they now connect.
Could use some help finding some regex that keeps a single comma between each field, as well as removing the pound sign.
You could simplify this by parsing the CSV file into a 2D array and ignoring the empty column which results from the double comma. Then parsing the currency column is a snap: just ignore the first character.
In your regex .replaceAll("[,{2}|£]", ""); the square-brackets creates a character class, so this means "replace any characters ,, {, 2, }, |, or £ with nothing".
What you really want is to replace the sequence ,,£ with a single comma, which would be .replaceAll(",,£", ",")
In java script this would be...
var line="18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,,£307018.48,";
console.log(' original line: ' + line);
console.log('replacement line: ' + line.replace(/,,£/, ","));
update
Converting this to Java as a stand-alone test program to demonstrate that this does work, I get the following:
public class so50419207
{
public static void main(String... args)
{
String input = "18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,,£307018.48,";
String replaced = input.replace(",,£", ",");
System.out.println("original string: " + input);
System.out.println("replaced string: " + replaced);
}
}
Running this...
$ javac so50419207.java ; java so50419207
original string: 18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,,£307018.48,
replaced string: 18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,307018.48,
Tried the regex (,,)(£)? and tested it in ideone :
Please find the code below:
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
final String regex = "(,,)(£)?";
final String string = "18,,5 Ramsey Lane,,See,Amerighi,,samerighih#trellian.com,,£307018.48,,\n"
+ "18,,5 Ramsey Lane,,See,Amerighi,,samerighih#trellian.com,,£307018.48,,\n"
+ "18,5 Ramsey Lane,,See,Amerighi,,samerighih#trellian.com,,£307018.48,,\n"
+ "18,,5 Ramsey Lane,,See,Amerighi,,samerighih#trellian.com,,£307018.48,,";
final String subst = ",";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
// The substituted value will be contained in the result variable
final String result = matcher.replaceAll(subst);
System.out.println("Substitution result: " + result);
}
}
Output:
Substitution result: 18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,307018.48,
18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,307018.48,
18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,307018.48,
18,5 Ramsey Lane,See,Amerighi,samerighih#trellian.com,307018.48,
I have a parameter of key-value like this:
sign="aaaabbbb="
And I want to get the parameter name sign and the value "aaaabbb="(with quote signs)
I thought I could split the string with = to get the first elem of the array which is the parameter name and do a String.replaceAll() to remove the sign= to get the value. Anyway here is my sample code:
public class TestStringReplace {
public static void main(String[] argvs){
String s = "sign=\"aaaabbbb=\"";
String[] ss = s.split("=");
String value = s.replaceAll("\\[^=]+=","");
//EDIT: s.replaceAll("[^=]+=","") will not do the job either.
System.out.println(ss[0]);
System.out.println(value);
}
}
but the output shows this:
sign
sign="aaaabbbb="
Why \\[^=]+= not matching sign= and replace it with empty string here?Quite a newbie of Java regex, need some help.
Thanks in advance.
In Java you can use the following:
String str = "sign=\"aaaabbbb=\"";
String var1 = str.substring(0, str.indexOf('='));
String var2 = str.substring(str.indexOf('=')+1);
System.out.println("var1="+var1+", var2="+var2);
The above would have the following output:
var1=sign, var2="aaaabbbb="
Try the following regex ^\\w+= with replaceAll() instead of your regex:
public class TestStringReplace {
public static void main(String[] argvs){
String s = "sign=\"aaaabbbb=\"";
String[] ss = s.split("=");
String value = s.replaceAll("^\\w+=","");
System.out.println(ss[0]);
System.out.println(value);
}
}
This will remove the sign=.
You can see the DEMO here.
Note that with your "\\[^=]+=" regex you were trying to match the character [ literally in the beginning of your regex.
And it explains why you got sign="aaaabbbb=" as a result with replaceAll() which didn't replace anything because there's no match.
You're probably better off with an actual Pattern and back-references here.
For instance:
String[] test = {
"sign=\"aaaabbbb=\"",
// assuming a HTTP GET-styled parameter list
"blah?sign=\"aaaabbbb=\"",
"foo?sign=\"aaaabbbb=\"&blah=\"hodor\""
};
// | group 1: literal "sign"
// | | literal key-value delimiter and double quote
// | | | group 2: any character reluctantly quantified
// | | | | literal ending double quote
// | | | | | look-ahead for either "&" or end
// | | | | |
Pattern p = Pattern.compile("(sign)=\"(.+?)\"(?=$|&)");
Matcher m = null;
for (String s: test) {
m = p.matcher(s);
while (m.find()) {
System.out.printf(
"Found key: \"%s\" and value: \"%s\"%n", m.group(1), m.group(2)
);
}
}
Output
Found key: "sign" and value: "aaaabbbb="
Found key: "sign" and value: "aaaabbbb="
Found key: "sign" and value: "aaaabbbb="
Notes
I'm assuming a HTTP GET styled parameter list, but maybe you don't need to actually check for a next parameter key-value pair delimiter (i.e. &) - in which case you can remove the & part
I'm also assuming you want the "s out of your value back-reference, which kind of makes the following & check useless
Your current pattern for the replaceAll invocation will match as follows:
// | literal "[" (double-escaped)
// ||literal "^" or "=" (in character class)
// || | ... greedily quantified (1+ occurrences)
// || || literal "="
"\\[^=]+="
Finally, if you really, really want to use String#replaceAll for this, here's a slightly different pattern than the one above:
for (String s: test) {
System.out.println(
s.replaceAll(
".*(sign)=\"(.+?)\"(?=$|&).*",
"Found key: \"$1\" and value: \"$2\""
)
);
}
It still uses back-references and will produce the same result, albeit in a uglier way: you can't reuse the $1 and $2 group values, since you're creating a new String replacing the original one.
Last possible solution, using String#'split. This is the ugliest as it won't work well with a list of parameters:
for (String s: test) {
System.out.println(
// | negative look-behind for start of input
// | | literal "="
// | | | literal "
// | | |
Arrays.toString(s.split("(?<!^)=\""))
);
}
Output
[sign, aaaabbbb]
[blah?sign, aaaabbbb] --> yuck
[foo?sign, aaaabbbb, &blah, hodor"] --> yuck again
The double slash is a mistake, because it is escaping the [ to a literal [, which will never match.
Instead, do this:
String name = s.replaceAll("=.*", "");
String value = s.replaceAll(".*?=", "");
I am trying to replace certain values from a string using java regex
for example the string looks like
:20:1234
6789
:28G::xyz
|20:3456
1234
|29C:pqr
:20|9876
I want to replace tag 20 value (may be multi line value) for second occurrence
|20:3456
1234
with new value(may be multi line value) 6789 so the final replacement string i am expecting is
:20:1234
6789
:28G::xyz
|20:6789
|29C:pqr
:20|9876
Try this regex:
String str = ":20:1234\n 6789\n:28G::xyz\n|20:3456\n 1234\n|29C:pqr\n:20|9876 \n|20:3456\n :20:1234\n";
str = str.replaceAll("(\\|20:)[\\s\\S]*?(?=[|:])","$1" + "6789\n");
Here it is checking until it reaches to anything other than | or :, so that it doesn't pick all.
This should work (tested):
str.replaceAll("(\\|" + "20" + ":)[^|:]*\n","$1" + "6789" + "\n");
I want to replace all occurrences of +- with - from a string called myStr in java.
I also want to replace all occurrences of -- with + from myStr.
The following two lines of code are not accomplishing this in java:
myStr.replaceAll("\\+-", "-");
myStr.replaceAll("\\--", "+");
Can anyone show me how to alter these two lines of code to accomplish the desired replacements?
I usually try to avoid regular expressions, but am not sure how to do this operation without them.
You're throwing out the return value of the function. You probably want to use:
myStr = myStr.replaceAll("\\+-", "-").replaceAll("--", "+");
update with additional info from comment:
Be sure to keep the return value of replaceAll.
myStr = myStr.replaceAll("\\+-", "-");
and later
myStr = myStr.replaceAll("--", "+");
public static String escapePlusMinus(String myStr) {
Pattern pattern = Pattern.compile("[+-]-");
Matcher matcher = pattern.matcher(myStr);
StringBuffer result = new StringBuffer();
while (matcher.find()) {
if (matcher.group(0).equals("+-")) {
matcher.appendReplacement(result, "-");
}
else {
matcher.appendReplacement(result, "+");
}
}
matcher.appendTail(result);
return result.toString();
}
escapePlusMinus("+-, --, +-, ---+---") ⇒ "-, +, -, +--+"
The last token is matched as:
"--" ⇒ "+"
"-" skipped, since there are not another "-" following it.
"+-" ⇒ "-"
"--" ⇒ "+"
I want to add Two java JSON String manually , so for this i need to remove "}" and replace it with comma "," of first JSON String and remove the first "{" of the second JSON String .
This is my program
import java.util.Map;
import org.codehaus.jackson.type.TypeReference;
public class Hi {
private static JsonHelper jsonHelper = JsonHelper.getInstance();
public static void main(String[] args) throws Exception {
Map<String, Tracker> allCusts = null;
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String B = "{\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String totalString = A + B;
if (null != totalString) {
allCusts = (Map<String, Tracker>) jsonHelper.toObject(
totalString, new TypeReference<Map<String, Tracker>>() {
});
}
System.out.println(allCusts);
}
}
When adding two Strings A + B
I want to remove the last character of "}" in A and replace it with "," and remove the FIrst character of "{" in B .
SO this should it look like .
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1},";
String B = "\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
I have tried
String Astr = A.replace(A.substring(A.length()-1), ",");
String Bstr = B.replaceFirst("{", "");
String totalString = Astr + Bstr ;
With this i was getting
Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
please suggest .
{ is a control character for Regular Expressions, and since replaceFirst takes a string representation of a Regular Expression as its first argument, you need to escape the { so it's not treated as a control character:
String Bstr = B.replaceFirst("\\{", "");
I would say that using the replace methods is really overkill here since you're just trying to chop a character off of either end of a string. This should work just as well:
String totalString = A.substring(0, A.length()-1) + "," + B.substring(1);
Of course, regex doesn't look like a very good tool for this. But the following seem to work:
String str = "{..{...}..}}";
str = str.replaceFirst("\\{", "");
str = str.replaceFirst("}$", ",");
System.out.println(str);
Output:
..{...}..},
Some issues in your first two statements. Add 0 as start index in substring method and leave with that. Put \\ as escape char in matching pattern and ut a , in second statement as replacement value.
String Astr = A.substring(0, A.length()-1);//truncate the ending `}`
String Bstr = B.replaceFirst("\\{", ",");//replaces first '{` with a ','
String totalString = Astr + Bstr ;
Please note: There are better ways, but I am just trying to correct your statements.