I got some function:
private fun selectHometown() = File("data/towns.txt")
.readText()
.split("\n")
.shuffled()
.first()
And if I try to get or print some string with the 2 values obtained from this function, the first value disappears. For example:
println("${selectHometown() ${selectHometown() }")
Will only print one city name, while I expect two. I guess the problem is related to string concatenation in Kotlin. Of course, I can get the desired result in a different way, but I'm wondering why this one doesn't work.
Windows way of terminating a line is to use "\r\n" so use it as delimiter :
private fun selectHometown() = File("data/towns.txt")
.readText()
.split("\r\n")
.shuffled()
.first()
println("${selectHometown()} ${selectHometown()}")
I tried:
String test = "racecary";
StringBuilder stringBuilder = new StringBuilder(test);
System.out.println(stringBuilder.reverse()+" -------------");
if (stringBuilder.reverse().toString().equalsIgnoreCase(test)) {
System.out.println("Pass");
}else {
System.out.println("Fail");
}
It always prints pass, even when I mispelled racecar, but passes when I use == inplace of .equalsIgnoreCase.
Am I doing something wrong?
I have already gone through following but didnt get the answer.
JAVA .equalsIgnoreCase not working
JAVA .equalsIgnoreCase not working
StringBuilder.reverse() modifies the builder in place. The test printout is causing the string to be reversed twice, leaving it unmodified. Get rid of the printout and the code will work as expected.
Say I have a URL
http://example.com/query?q=
and I have a query entered by the user such as:
random word £500 bank $
I want the result to be a properly encoded URL:
http://example.com/query?q=random%20word%20%A3500%20bank%20%24
What's the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right.
URLEncoder is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);
When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString() as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8".
Note that spaces in query parameters are represented by +, not %20, which is legitimately valid. The %20 is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?), not in query string (the part after ?).
Also note that there are three encode() methods. One without Charset as second argument and another with String as second argument which throws a checked exception. The one without Charset argument is deprecated. Never use it and always specify the Charset argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.
All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
See also:
What every web developer must know about URL encoding
I would not use URLEncoder. Besides being incorrectly named (URLEncoder has nothing to do with URLs), inefficient (it uses a StringBuffer instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.
Instead I would use URIBuilder or Spring's org.springframework.web.util.UriUtils.encodeQuery or Commons Apache HttpClient.
The reason being you have to escape the query parameters name (ie BalusC's answer q) differently than the parameter value.
The only downside to the above (that I found out painfully) is that URL's are not a true subset of URI's.
Sample code:
import org.apache.http.client.utils.URIBuilder;
URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();
// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24
You need to first create a URI like:
String urlStr = "http://www.example.com/CEREC® Materials & Accessories/IPS Empress® CAD.pdf"
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
Then convert that URI to an ASCII string:
urlStr = uri.toASCIIString();
Now your URL string is completely encoded. First we did simple URL encoding and then we converted it to an ASCII string to make sure no character outside US-ASCII remained in the string. This is exactly how browsers do it.
Guava 15 has now added a set of straightforward URL escapers.
The code
URL url = new URL("http://example.com/query?q=random word £500 bank $");
URI uri = new URI(url.getProtocol(), url.getUserInfo(), IDN.toASCII(url.getHost()), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String correctEncodedURL = uri.toASCIIString();
System.out.println(correctEncodedURL);
Prints
http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$
What is happening here?
1. Split URL into structural parts. Use java.net.URL for it.
2. Encode each structural part properly!
3. Use IDN.toASCII(putDomainNameHere) to Punycode encode the hostname!
4. Use java.net.URI.toASCIIString() to percent-encode, NFC encoded Unicode - (better would be NFKC!). For more information, see: How to encode properly this URL
In some cases it is advisable to check if the URL is already encoded. Also replace '+' encoded spaces with '%20' encoded spaces.
Here are some examples that will also work properly
{
"in" : "http://نامهای.com/",
"out" : "http://xn--mgba3gch31f.com/"
},{
"in" : "http://www.example.com/‥/foo",
"out" : "http://www.example.com/%E2%80%A5/foo"
},{
"in" : "http://search.barnesandnoble.com/booksearch/first book.pdf",
"out" : "http://search.barnesandnoble.com/booksearch/first%20book.pdf"
}, {
"in" : "http://example.com/query?q=random word £500 bank $",
"out" : "http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$"
}
The solution passes around 100 of the test cases provided by Web Platform Tests.
Using Spring's UriComponentsBuilder:
UriComponentsBuilder
.fromUriString(url)
.build()
.encode()
.toUri()
The Apache HttpComponents library provides a neat option for building and encoding query parameters.
With HttpComponents 4.x use:
URLEncodedUtils
For HttpClient 3.x use:
EncodingUtil
Here's a method you can use in your code to convert a URL string and map of parameters to a valid encoded URL string containing the query parameters.
String addQueryStringToUrlString(String url, final Map<Object, Object> parameters) throws UnsupportedEncodingException {
if (parameters == null) {
return url;
}
for (Map.Entry<Object, Object> parameter : parameters.entrySet()) {
final String encodedKey = URLEncoder.encode(parameter.getKey().toString(), "UTF-8");
final String encodedValue = URLEncoder.encode(parameter.getValue().toString(), "UTF-8");
if (!url.contains("?")) {
url += "?" + encodedKey + "=" + encodedValue;
} else {
url += "&" + encodedKey + "=" + encodedValue;
}
}
return url;
}
In Android, I would use this code:
Uri myUI = Uri.parse("http://example.com/query").buildUpon().appendQueryParameter("q", "random word A3500 bank 24").build();
Where Uri is a android.net.Uri
In my case I just needed to pass the whole URL and encode only the value of each parameters.
I didn't find common code to do that, so (!!) so I created this small method to do the job:
public static String encodeUrl(String url) throws Exception {
if (url == null || !url.contains("?")) {
return url;
}
List<String> list = new ArrayList<>();
String rootUrl = url.split("\\?")[0] + "?";
String paramsUrl = url.replace(rootUrl, "");
List<String> paramsUrlList = Arrays.asList(paramsUrl.split("&"));
for (String param : paramsUrlList) {
if (param.contains("=")) {
String key = param.split("=")[0];
String value = param.replace(key + "=", "");
list.add(key + "=" + URLEncoder.encode(value, "UTF-8"));
}
else {
list.add(param);
}
}
return rootUrl + StringUtils.join(list, "&");
}
public static String decodeUrl(String url) throws Exception {
return URLDecoder.decode(url, "UTF-8");
}
It uses Apache Commons' org.apache.commons.lang3.StringUtils.
Use this：
URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());
or this:
URLEncoder.encode(query, "UTF-8");
You can use the following code.
String encodedUrl1 = UriUtils.encodeQuery(query, "UTF-8"); // No change
String encodedUrl2 = URLEncoder.encode(query, "UTF-8"); // Changed
String encodedUrl3 = URLEncoder.encode(query, StandardCharsets.UTF_8.displayName()); // Changed
System.out.println("url1 " + encodedUrl1 + "\n" + "url2=" + encodedUrl2 + "\n" + "url3=" + encodedUrl3);
I am working in a groovy/grails set up and am having some trouble trying to execute a switch statement on a String value.
Basically, I am looping through the attribute names in a webservice response to see if they match pre-defined mappings that are configured on a per user basis. If they have established a mapping on that field, I pull the value out of the response and use it elsewhere.
The code looks something like this:
switch(attributeName)
{
case {attributeName} :
log.info("Currently switching on value... ${attributeName}")
case user.getFirstNameMapping():
model.user.userInfo.firstName = attributeValue
break
case user.getLastNameMapping():
model.user.userInfo.lastName = attributeValue
break
case user.getAuthenticationKeyMapping():
model.authenticationValue = attributeValue
break
case user.getEmailMapping():
model.email = attributeValue.toLowerCase()
break
}
The value being switched on (attributeName) is of type String, and the getter methods for the user instance also return type String.
Based on my research and understanding of the Groovy language, switching on an Object such as a String should end up using String.equals() to make the comparison. The result, however, is that it is matching on the user.getFirstNameMapping() case every time, and repeatedly overwriting the value in the model; therefore, the last value that comes back in the response is what ends up saved, and none of the other values are saved.
What's interesting is that if I use an if/else structure and do something like this:
if(attributeName.equals(user.getFirstNameMapping())
{
...
}
It works fine every time. I've verified through logging that it's not something silly like extra whitespace or a capitalization issue. I've also tried changing things around to run the switch by default and explicitly compare the attributeName in the case like this:
switch(true)
{
case {attributeName} :
log.info("Currently switching on value... ${attributeName}")
case {user.getFirstNameMapping().equals(attributeName)}:
model.user.userInfo.firstName = attributeValue
break
case {user.getLastNameMapping().equals(attributeName)}:
model.user.userInfo.lastName = attributeValue
break
case {user.getAuthenticationKeyMapping().equals(attributeName)}:
model.authenticationValue = attributeValue
break
case {user.getEmailMapping().equals(attributeName)}:
model.email = attributeValue.toLowerCase()
break
}
And it still fails to meet my expectations in the exact same way. So, I'm wondering why this is the behavior when the switch statement should simply be using .equals() to compare the strings, and when I explicitly compare them in an if/else using .equals(), it works as expected.
The issue is in your switch case.
Have a look here :-
case {attributeName} :
log.info("Currently switching on value... ${attributeName}")
case user.getFirstNameMapping():
model.user.userInfo.firstName = attributeValue
break
As you can see your these two cases will run every time because the switch condition is :-
switch(attributeName)
So the first one will get match and will run until it encounters break; which is at after case 2 i.e. case user.getFirstNameMapping(): so i would suggest you to print the value of {attributeName} before the swtich starts.
Hope that will help you.
Thanks
I don't know exactly what's your issue, but the case statement works just fine, even with methods. See my example
String something = "Foo"
class User {
String firstName
String lastName
}
User u = new User(firstName: 'Something', lastName:'Foo')
switch(something) {
case u.getFirstName():
println "firstName: ${u.firstName}"
break;
case u.getLastName():
println "lastName: ${u.lastName}"
break;
default:
println "nothing..."
}
This code will print lastName as expected.
I am currently working on a project that dynamically displays DB content into table.
To edit the table contents i am want to use the dynamically created "string"+id value.
Is there any way to retrieve the appended int value from the whole string in javaScript?
Any suggestions would be appreciative...
Thanks!!!
If you know that the string part is only going to consist of letters or non-numeric characters, you could use a regular expression:
var str = "something123"
var id = str.replace(/^[^\d]+/i, "");
If it can consist of numbers as well, then things get complicated unless you can ensure that string always ends with a non-numeric character. In which case, you can do something like this:
var str = "something123"
var id = str.match(/\d+$/) ? str.match(/\d+$/)[0] : "";
(''+string.match(/\d+/) || '')
Explanation: match all digits in the variable string, and make a string from it (''+).
If there is no match, it would return null, but thanks to || '', it will always be a string.
You might try using the regex:
/\d+$/
to retrieve the appended number