How to remove all characters before a specific character in Python?
Use re.sub. Just match all the chars upto I then replace the matched chars with I. re.sub(r’^.*?I’, ‘I’, stri)
Use re.sub. Just match all the chars upto I then replace the matched chars with I. re.sub(r’^.*?I’, ‘I’, stri)
Use LEFT combined with CHARINDEX: UPDATE MyTable SET MyText = LEFT(MyText, CHARINDEX(‘;’, MyText) – 1) WHERE CHARINDEX(‘;’, MyText) > 0 Note that the WHERE clause skips updating rows in which there is no semicolon. Here is some code to verify the SQL above works: declare @MyTable table ([id] int primary key clustered, MyText varchar(100)) insert … Read more
Use REPLACE: SELECT REPLACE(t.column, ‘est1’, ‘rest1’) FROM MY_TABLE t If you want to update the values in the table, use: UPDATE MY_TABLE t SET column = REPLACE(t.column, ‘est1’, ‘rest1’)
Either use location or window.location instead of document.location as the latter is a non-standard. window.location.hash=”#food”; This will replace the URL’s hash with the value you set for it. Reference
Try this: website=$(sed ‘s”https://stackoverflow.com/”\\/|g’ <<< $website) Bash actually supports this sort of replacement natively: ${parameter/pattern/string} — replace the first match of pattern with string. ${parameter//pattern/string} — replace all matches of pattern with string. Therefore you can do: website=${website////\\/} Explanation: website=${website // / / \\/} ^ ^ ^ ^ | | | | | | | … Read more
Try using String.replace() or String.replaceAll() instead. String my_new_str = my_str.replace(“&”, “&”); (Both replace all occurrences; replaceAll allows use of regex.)
Using RedGate Profiler using the following code class Program { static string data = “abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz”; static Dictionary<string, string> values; static void Main(string[] args) { Console.WriteLine(“Data length: ” + data.Length); values = new Dictionary<string, string>() { { “ab”, “aa” }, { “jk”, “jj” }, { “lm”, “ll” }, { “yz”, “zz” }, { “ef”, “ff” }, … Read more
You could extend the String object with your own function that does what you need (useful if there’s ever missing functionality): String.prototype.replaceArray = function(find, replace) { var replaceString = this; for (var i = 0; i < find.length; i++) { replaceString = replaceString.replace(find[i], replace[i]); } return replaceString; }; For global replace you could use regex: … Read more
And when I debug this the logic does fall into the sentence.replace. Yes, and then you discard the return value. Strings in Java are immutable – when you call replace, it doesn’t change the contents of the existing string – it returns a new string with the modifications. So you want: sentence = sentence.replace(“and”, ” … Read more
Try this: find /home/user/ -type f | xargs sed -i ‘s/a\.example\.com/b.example.com/g’ In case you want to ignore dot directories find . \( ! -regex ‘.*/\..*’ \) -type f | xargs sed -i ‘s/a\.example\.com/b.example.com/g’ Edit: escaped dots in search expression