How to replace a string at a particular position

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Is there a way to replace a portion of a String at a given position in java script. For instance I want to replace 00 in the hours column with 12 in the below string.The substring comes at 13 to 15.

Mar 16, 2010 00:00 AM 

The following is one option:

var myString = "Mar 16, 2010 00:00 AM";

myString = myString.substring(0, 13) + 
           "12" + 
           myString.substring(15, myString.length);

Note that if you are going to use this to manipulate dates, it would be recommended to use some date manipulation methods instead, such as those in DateJS.

Replace a character at a specific index in a String in Java , Creating a Cell at specific position in Excel file using Java � StringBuilder replace () in Java with Examples � HashMap replace(key, value) method in Java with� A slightly faster and more concise approach: for the second part of the string, use myString.substr(15) - this will get the string from position 15 till the end of the string. – user6269864 Mar 23 '18 at 7:00

A regex approach

"Mar 16, 2010 00:00 AM".replace(/(.{13}).{2}/,"$112")
Mar 16, 2010 12:00 AM

Replace character at specific index in a Java String, character. There are several ways to replace a character at a specific index in a String – replace character at the specified position in char array. chars[pos]� Hi All, My requisite is to search for the string "0108"(which is the year and has come in the wrong year format) in a particular column say 4th column in a tab delimited file and then replace it with 2008(the correct year format) in the same position where 0108 was found..The issue is the last

One option would be

>>> var test = "Mar 16, 2010 00:00 AM";
>>> test.replace(test.substring(13,15),"12")

Replace character at specific position in a string in C#, Remove Method. In order to replace character present at specific index in the string, the idea is to remove the character present at a given position in the string and then insert the specified character at the very same position. Your problem is with the replace function. When you call the replace function it replaces ALL instances of the first argument with the second argument. so, if newhand = AKAK9, newhand.replace ("A","Q") will result in newhand = QKQK9. If possible, change the string to a list, then do the following to change the specific index:

if it is always 00: in hours,

you can just replace 00: with 12:

using replace() ,

if not u need find the indexOf the : character ,

and then replace 2 digit before with 12.

Python String – Replace character at Specific Position, Python - Replace character at given index - To replace a character with a given character at a specified index, you can use python string slicing; or convert the� Hi all, need some help. I want to replace the second character in a string, regardless of what character that is. If I try the replace the second character with the (substring 1, 1) all instances of that character gets replaced. I just want to replace the one sitting at position 2 in the string. Ex Good -> Gxod, not Gxxd Thanks in advance

You can direclty use replace() method along with indexOf() of string in Javascript.

Replace a Character at a Specific Index in a String in Java, Here, the idea is to convert the String to char[] and then assign the new char at the given index. Finally, we construct the desired String from that� Method 1: Using String Class There is no predefined method in String Class to replace a specific character in a String, as of now. However this can be achieved indirectly by constructing new String with the 2 different substrings: one from beginning till the specific index – 1, the new character at the specific index and the other from the index + 1 till the end.

C# Replace or remove a char at a given index (pos) in a string, Replace or remove a char at a given index (position) in a string. Strings in c# are immutable objects - it's not possible to replace or remove a char directly. Search for a pattern and replace a space at specific position with a Character in File In file, we have millions of records each of 1000 in length. And at specific position say 800 there is a space, we need to replace it with Character X if the ID in that row starts with 123.

Add, Remove, Replace String In C#, Remove() method removes a given number of characters from a string at a specified position. The position is a 0 index position. That means the 0� If you want to remove a character by its position in the string, just use '' as the replacement string. $str = substr_replace ($str,'',$pos,1); If you have further questions, you should really start your own new SO question. – Matthew Feb 11 '14 at 1:33.

%REPLACE (Replace Character String), %REPLACE returns the character string produced by inserting a replacement string into the source string, starting at the start position and replacing the specified� sublen : Number of characters of string object to be copied into another string object. n : Number of characters to be copied into an another string object. Return value. This function does not return any value. Example 1. First example shows how to replace given string by using position and length as parameters.

Comments
  • A slightly faster and more concise approach: for the second part of the string, use myString.substr(15) - this will get the string from position 15 till the end of the string.
  • I like this approach but if anyone is concerned about the performance here's a test comparison: jsperf.com/substring-replace
  • right, regex are generally slower. That I won't use myself in 2016.
  • Ten times slower, but still extremely fast unless you need to perform a million operations per second.
  • Wouldn't that give "Mar 16, 2010 12:12 AM"? You could change it to test.replace(test.substring(13,16),"12:") I think (similar to haim's method).
  • @Dominic: Good point, but actually it will replace it correctly in this case, because the JavaScript replace() method only replaces the first occurrence. But if the date was "Mar 16 2000", it would not have worked.
  • @Daniel - interesting - seems like an odd implementation of String::replace. Thanks for the correction!
  • Can you elaborate an example?
  • @EuryPérezBeltré I can think about a straight way to do that: 'Mar 16, 2010 00:00 AM'.replace('00:', '12:'), (even not using indexOf())