special-characters

I am trying to validate a string, that should contain letters numbers and special characters &-._ only. For that I tried with a regular expression. var pattern = /[a-zA-Z0-9&_\.-]/ var qry = 'abc&*'; if(qry.match(pattern)) { alert('valid'); } else{ alert('invalid'); } While using the above code, the string abc&* is valid. But my requirement is to show this invalid. ie Whenever a character other than a letter, a number or special characters &-._ comes, the string should evaluate as invalid. How can I do that with a regex? Add them to the allowed characters, but you'll need to escape some of

I have configured the apache httpClient like so: HttpProtocolParams.setContentCharset(httpParameters, "UTF-8"); HttpProtocolParams.setHttpElementCharset(httpParameters, "UTF-8"); I also include the http header "Content-Type: application/json; charset=UTF-8" for all http post and put requests. I am trying to send http post/put requests with a json body that contains special characters (ie. chinese characters via the Google Pinyin keyboard, symbols, etc.) The characters appear as gibberish in the logs but I think this is because DDMS does not support UTF-8, as descibed in this issue. The problem

EDIT: %.% operator is now deprecated. Use %>% from magrittr. ORIGINAL QUESTION What does this %.% operator do?? I've seen it used a lot with the dplyr package, but can't seem to find any supporting documentation on what it is or how it works. It seems to chain commands together, but that's as far as I can tell...While I'm at it, can anyone explain what the gambit of those special operators that hang around with the % sign do and when is technically the right time to use them to code better? Carlos Cinelli I think Hadley would be the best person to explain to you, but I will give it a shot. %.%

This question already has an answer here: String replace method is not replacing characters 5 answers I'm trying to replace all special characters with a "%", like: "123.456/789" -> "123%465%798" my regular expression is: [^a-zA-Z0-9]+ In online tools* it works perfecly, but in java s.replaceAll("[^a-zA-Z0-9]+", "%"); strings remain untouched. *I tried: http://www.regexplanet.com/ http://regex101.com/ and others Strings are immutable. You forgot to reassign new String to the s variable :) s = s.replaceAll("[^a-zA-Z0-9]+", "%"); // ^ this creates a new String replaceAll() like all methods in

I'm trying to get the following to work & it's not, help me please: curl -s 'https://cryptofresh.com/api/asset/markets?asset=MKR' | jq .OPEN.BTC The variable in question includes a period, I tried just about everything to escape the period && also tried surrounding it in quotes; no go ; this is the variable I'm looking to pull from (I ran jq without any filters, and truncated the output here to show what I need. Thanks in advance future problem solver! curl -s 'https://cryptofresh.com/api/asset/markets?asset=MKR' | jq .... "OPEN.BTC": { "volume24": 0.932166, "price": 0.09995, "updated": "2016