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How to insert multiple tabs string in Java?
This example:
getName() + '\t' + '\t' + getLastName(), does not work.
Your example should work; however there's no reason to append each tab character individually. This works, too:
getName() + "\t\t" + getLastName();
The errors you are getting are not related to the tab characters.
Your syntax is screwy. Try this.
String whatever0 = "firstname"+"\t"+"\t"+"lastname";
String whatever1 = "firstname"+"\t"+"\t"+"\t"+"lastname";
System.out.println(whatever0);
System.out.println(whatever1);
You'll see that they give different results.
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Any ideas how to effeciently convert the String "TDMaturityReinvestOnNotSelected" to "TD Maturity Reinvest On Not Selected" using a java function?
Cheers
Shaun
This brilliant answer to RegEx to split camelCase or TitleCase (advanced) should work nicely.
Below is an excerpt from that answer:
final String pattern = "(?<!(^|[A-Z]))(?=[A-Z])|(?<!^)(?=[A-Z][a-z])";
for (String w : "TDMaturityReinvestOnNotSelected".split(pattern))
{
System.out.println(w);
}
And the ouput to show it running:
Edit: You'll need to reassemble the split words with spaces, but that should be trivial to work out.
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I want to convert a hexadecimal string into IP4 string , I googled but didn't get any good way. Could anybody suggest me a simple way.
Thanks .
Please try this method.
String hexValue = "Hex IP";
String ip = "";
for(int i = 0; i < hexValue.length(); i = i + 2) {
ip = ip + Integer.valueOf(hexValue.subString(i, i+2), 16) + ".";
}
System.out.println("Ip = " + ip);
In this post the author gave a method for printing out an int into hexadecimal.
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Given a string "Hello". I only want all the letters after the first letter to be in my new string.
E.g.: "ello"
How could this be done?
Tried pattern matching but cant get it to work :(
Try this:
String s = "Hello";
String newS = s.substring(1); // newS is "ello"
The above will create a new string containing all the characters of the original, except the first one. See the documentation for more details.
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I am looking for a regular expression to remove all #[x] present in a string
Where x can be any string.
For example #[title]
#\[[^\]]*\]?
As in:
String s = "asda #[asdagf] dsgfdg";
System.out.println(s.replaceAll("#\\[[^\\]]*\\]",""));
will print out:
asda dsgfdg
String original = "Hello #[world]!";
String replaced = original.replaceAll("#\\[.*?\\]", "");
// replaced = "Hello !"`enter code here`
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I am currently trying to make use of the java string function someString.replaceAll() to find commonly used words (and, the, by, of, etc) and replace them with " ". Based on the answers to the question at Whitespace Matching Regex - Java, I produced this function call:
data.replaceAll("(?i)\\sthe\\s", " ")
However, it isnt working and I'm really not sure why. Nothing about it looks wrong based on what I've found. Please help me!
Strings are immutable!
data = data.replaceAll("(?i)\\sthe\\s", " ");