Develop Reference

How to avoid nul value in URL in android?

I have a strange issue about using Retrofit2 in my android project. I got the issue about the server error since the request is something like that.
https://www.example.com/api/v1/skills?q=Good%00
Since the invalid value "%00" is not acceptable in our server, so it showed error on my activity.
API service
#GET("skills")
Observable<SearchItem> getSkills(#Query("q") String keyword);
In my fragment, I just get the text using following simple statement.
String keyword = editText.getText().toString()
api.getSkills(keyword);
What I want to know is the following:
Is it possible to have a word can be converted to "%00" ?
How to avoid this "Good%00" before I send to getSkills function?

To enable compile time checks on nullity add #NonNull annotation,
#GET("skills")
Observable<SearchItem> getSkills(#NonNull #Query("q") String keyword);
Another way is to change each "%00" in your string, using .replace()

Replace the string with "" if it contains %00
if (text.toString().contains("%00")){
text = text.replace("%00", "");
}
and then call getSkills(text) with updated value

Try this
Use trim()
The java string trim() method eliminates leading and trailing spaces. The unicode value of space character is '\u0020'. The trim() method in java string checks this unicode value before and after the string, if it exists then removes the spaces and returns the omitted string.
SAMPLE CODE
String keyword = editText.getText().toString().trim();
api.getSkills(keyword)
or your can use
replace()
The java string replace() method returns a string replacing all the old char or CharSequence to new char or CharSequence.
SAMPLE CODE
url =url.replace("%00", "");
or You can use URLEncoder
Utility class for HTML form encoding. This class contains static methods for converting a String to the application/x-www-form-urlencoded MIME format. For more information about HTML form encoding
String encodedurl = URLEncoder.encode(yourURL,"UTF-8");

There is two way to solve this
1. you restrict the entry write below code
android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
you can add more thing which you want
String keyword = editText.getText().toString().replace("%00", "");
api.getSkills(keyword);

Escape sequence string processing in android

I have a string variable in my android ap which contains some url http://www.google.com. The string variable is declared as shown below:
String html="";
But there is an error saying that I have to put (semicolon); after the a href=". But actually I want the entire line content ie
(href="http://www.google.com") in the string variable. How can it be achieved?? Thanks in advance.

doubles quotes needs to escaped.
try below one
String html="<a href='http://www.google.com'></a>";
else
String html= "";

You need to escape every " with \ if you want to put it in one String.
For example:
String html="";
If you don't do this, compiler assumes, that you end this String after href= and tries to understand what http://www.google.com means in Java ;)

Using a question mark as a String in android

Im trying to use a question mark as a variable for a string.
I've tried...
strings.xml
<string name="questionMark">\?</string>
.class
String questionMark;
questionMark = getResources().getString(R.string.questionMark);
String delim4 = (questionMark);
This causes a fource close regex error.
and
String delim4 = (\?);
This gets an error Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \ )
and also
I've tried putting 2 backslashes in front of it
String delim4 =(\\?)
System.out.println("delim "+ delim4);
But that just escapes the second slash and sometimes force closes as well.
the output for that was
delim \?
Can any tell me how to put in the question mark as the string. I'm using it as variable to spit a string. The String Im splitting can not be changed.
plz help
Edit added split code
if (FinishedUrl.contains(questionMark)){
String delim3 = (".com/");
String[] parts3 = FinishedUrl.split(delim3);
String JUNK3= parts3[0];
String fIdStpOne = parts3[1];
String fIdStpTwo = fIdStpOne.replaceAll("=#!/","");
String delim4 = (questionMark);
String[] parts4 = fIdStpTwo.split(delim4);
String fIdStpThree= parts3[0];
String JUNK4 = parts3[1];
FId = fIdStpThree;
}

As pointed out by user laalto, ? is a meta-character in regex. You must work around that.
Let's see what's happening here. Firstly, some ground rules:
`?` is not a special character in Java
`?` is a reserved character in regex
This entails:
String test = "?"; // Valid statement in java, but illegal when used as a regex
String test = "\?"; // Illegal use of escape character
Why is the second statement wrong? Because we are trying to escape a character that isn't special (in Java). Okay, we'll get back to this.
Now, for the split(String) method, we need to escape the ? - it being a meta-character in regex. So, we need \? for the regex.
Coming back to the string, how do we get \?? We need to escape the \(backslash) - not the question mark!
Here's the workflow:
String delim4 = "\\?";
This statement gives us \? - it escapes the \(backslash).
String[] parts4 = fIdStpTwo.split(delim4);
This lets us use \? as a regex in the split() method. Since delim4 is being passed as a regex, \? is used as ?. Here, the prefix \ is used to escape ?.
Your observations:
String delim4 = (\?);
This gets an error Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \ )
I covered this above. You are escaping ? at the java level - but it isn't a special character and needs no escaping - hence the error.
String delim4 =(\\?)
System.out.println("delim "+ delim4);
But that just escapes the second slash and sometimes force closes as well. the output for that was
delim \?
This is what we want. It is easier to think of this as a two stage process. The first stage deals with successfully placing a \(backslash) in front of the ?. In the second stage, regex finds that the ? has been prefixed by a \ and uses ? as a literal instead of a meta-character.
And here's how you can place the regex in your res/values/strings.xml:
<string name="questionMark">\\?</string>
By the way, there's another option - not something I use on a regular basis these days - split() works just fine.
You can use StringTokenizer which works with delimiters instead of regex. Afaik, any literal can be used as a delimiter. So, you can use ? directly:
StringTokenizer st = new StringTokenizer(stringToSplit, "?");
while (st.hasMoreTokens()) {
// Use tokens
String token = st.nextToken();
}

Easiest way is to quote or backslash them:
<string name="random">"?"</string>
<string name="random">\?</string>

The final code.
String startDelim = ("\\?");
String realDelim = (startDelim);
String[] parts4 = fIdStpOne.split(realDelim);
String fIdStpTwo= parts4[0];
String JUNK4 = parts4[1];

Normally you'd just put it literally, like
String q = "?";
However, you say you're using it to split a string. split() takes a regular expression and ? is a metacharacter in a regex. To escape it, add a backslash in front. Backslash is a special character in Java string literals so it needs to be escaped, too:
String q = "\\?";

replaceAll function not working while removing special character

I am working with a project where some data are retrieved by JSON parsing. Unfortunately, invalid character '\' escapes. I need to remove them. I tried calling .replace("\\'","\");. This solution is not working. No exceptions are thrown, but the string does not change. Here is my code:
shop_name = c.getString(TAG_SHOP_NAME);
if(shop_name.contains("\\'")==true)
{
//try{
shop_name=shop_name.replaceAll(Pattern.quote("\\'"), "'");
Log.e("vvvvvv","new shop name: "+shop_name);}
//catch(Exception q){Log.e(TAG+" vvvv","EXPTN",q);}
}
send JSON object is: Bimal\'s
required object: Bimal's
Please let me know whether I went wrong somewhere or if there is any other method other than replaceAll.

You need to double escape the backslash as it's an escape character in both strings and regex:
shop_name.replaceAll("\\\\'", "'");
Or without using regex (as it's not needed in this circumstance):
shop_name.replace("\\'", "'");

Escape the meta character " with "\" :
string.replaceAll("\"", "");
Remember to assign it back to the String reference , because it returns a new String object.
You should use replace() instead:
str = str.replace("\"", "");
replaceAll() is used for replacing regular expressions.

PatternSyntaxException: String.replaceAll() android

I want to remove all { } as follow:
String regex = getData.replaceAll("{", "").replaceAll("}", "");
but force close my app with log.
java.util.regex.PatternSyntaxException: Syntax error U_REGEX_RULE_SYNTAX
what have i done wrong ?

You need to escape {:
String regex = getData.replaceAll("\\{", "").replaceAll("\\}", "");

Curly brackets are used to specify repetition in regex's, therefore you will have to escape them.
Furthermore, you should also consider removing all the brackets in one go, instead of called replaceAll(String, String) twice.
String regex = getData.replaceAll("\\{|\\}", "");

For what you want to do you don't need to use a regex!
You can make use of the replace method instead to match specific chars, which increases readability a bit:
String regex = getData.replace("{", "").replace("}", "");
Escaping the \\{ just to be able to use replaceAll works, but doesn't make sense in your case