The third parameter of String.prototype.replace() function was never defined as a standard, so most browsers simply do not implement it. The best way is to use regular expression with g (global) flag. var myStr=”this,is,a,test”; var newStr = myStr.replace(/,/g, ‘-‘); console.log( newStr ); // “this-is-a-test” Still have issues? It is important to note, that regular expressions … Read more
This is the idiom I’ve always used to get a modified copy of a string without changing the original: (my $newstring = $oldstring) =~ s/foo/bar/g; In perl 5.14.0 or later, you can use the new /r non-destructive substitution modifier: my $newstring = $oldstring =~ s/foo/bar/gr; NOTE: The above solutions work without g too. They also … Read more
You can use xmlstarlet to edit a XML file in a shell like this : xmlstarlet edit -L -u “/scs-shop/price[@type=”double”]” -v ‘99.66’ file.xml NOTE “/scs-shop/price[@type=”double”]” is a Xpath expression see xmlstarlet ed –help
You could use simple regex: var result = fileAsString.replace(/string to be replaced/g, ‘replacement’); So… var fs = require(‘fs’) fs.readFile(someFile, ‘utf8’, function (err,data) { if (err) { return console.log(err); } var result = data.replace(/string to be replaced/g, ‘replacement’); fs.writeFile(someFile, result, ‘utf8’, function (err) { if (err) return console.log(err); }); });
You should be using DOM to parse HTML, not regular expressions… Edit: Updated code to do simple regex parsing on the href attribute value. Edit #2: Made the loop regressive so it can handle multiple replacements. $content=” <p><a href=”http://www.website.com”>This is a text link</a></p> <a href=”http://sitename.com/#foo”>bah</a> <a href=”#foo”>I wont change</a> “; $dom = new DOMDocument(); $dom->loadHTML($content); … Read more
To my knowledge there is no built-in function to replace empty strings across all columns of a table. You can write a plpgsql function to take care of that. The following function replaces empty strings in all basic character-type columns of a given table with NULL. You can then cast to integer if the remaining … Read more
Although it is not one call, you can nest the replace() calls: SET mycol = replace( replace(mycol, ‘{1}’, ‘myoneval’), ‘{2}’, mytwoval)
I have found that the easiest and best way to remove all escape characters from your JSON string, is to pass the string into Regex.Unescape() method. This method returns a new string with no ecapes, even \n \t etc, are removed. See this MSDN article for more details: Regex.Unescape Method (String) (System.Text.RegularExpressions)
PriceList[0] is a float. PriceList[0][1] is trying to access the first element of a float. Instead, do PriceList[0] = PriceList[1] = …code omitted… = PriceList[6] = PizzaChange or PriceList[0:7] = [PizzaChange]*7
Solution with replace by dictionary: df[‘prod_type’] = df[‘prod_type’].replace({‘respon’:’responsive’, ‘r’:’responsive’}) print (df) prod_type 0 responsive 1 responsive 2 responsive 3 responsive 4 responsive 5 responsive 6 responsive If need set all values in column to some string: df[‘prod_type’] = ‘responsive’