public class JavaIntern{
public static void main(String []args){
String str = "JavaIntern"; //hardcoded, but not the problem
char[] s = str.toCharArray();
String result = new String (s,0,1); //this is where the dilemma begins
System.out.println(result);
String result1 = new String (s,0,2);
System.out.println(result1);
String result2 = new String (s,0,3);
System.out.println(result2);
String result3 = new String (s,0,4);
System.out.println(result3);
String result4 = new String (s,0,5);
System.out.println(result4);
String result5 = new String (s,0,6);
System.out.println(result5);
String result6 = new String (s,0,7);
System.out.println(result6);
String result7 = new String (s,0,8);
System.out.println(result7);
String result8 = new String (s,0,9);
System.out.println(result8);
String result9 = new String (s,0,10);
System.out.println(result9); //and this is where it ends... how can I get rid of this?
}
}
//but still get this:
J
Ja
Jav
Java
JavaI
JavaIn
JavaInt
JavaInte
JavaInter
JavaIntern
I guess you want to improve the code and also don't depend on the length of the string.
What about something like this?
public class JavaIntern{
public static void main(String []args){
String str = "JavaIntern"; //hardcoded, but not the problem
String substring = "";
for (char ch: str.toCharArray()) {
substring += ch;
System.out.println(substring);
}
}
}
This will also print:
J
Ja
Jav
Java
JavaI
JavaIn
JavaInt
JavaInte
JavaInter
JavaIntern
The loop gets one character of the string at a time and concatenates it to the substring before printing it.
Im assuming you want to be able to print out one letter more each time.
To do this we use a for loop, and this way it is fairly simple.
public class MainClass {
public static void main(String[] args) {
String str = "JavaIntern";
for (int i = 1; i <= str.length(); i++) {
System.out.println(str.substring(0, i));
}
}
}
We set i to 0 in the loop, keep iterating while i less than or equal to the length of the string, and each time we iterate, add one to i.
We use the substring method to split the string from the first letter, to i.
Beginner here. Sorry for the vague title but the code should put my question into perspective.
public static void main(String [] args)
{
String sentence = "hi. how are you! i'm just dandy.";
String tokenSent;
tokenSent = sentenceCapitalizer(sentence);
System.out.println(tokenSent);
}
public static String sentenceCapitalizer(String theSentence)
{
StringTokenizer strTokenizer = new StringTokenizer(theSentence, ".!", true);
String token = null;
String totalToken = "";
String ch = "";
while(strTokenizer.hasMoreTokens())
{
token = strTokenizer.nextToken().trim();
token = token.replace(token.charAt(0), Character.toUpperCase(token.charAt(0)));
StringBuilder str = new StringBuilder(token);
str.append(" ");
totalToken += str;
}
return totalToken;
}
OUTPUT AS IS: Hi . How are you ! I'm just dandy .
I was able to capitalize the first letter of each sentence but I'm wanting the output to keep the same format as the original String. The problem is that it puts a space before and after the ending punctuation. Is there any way to fix this problem using only a StringBuilder and/or StringTokenizer? Thank you for your time.
You can do this.
String delim = ".!";
StringTokenizer strTokenizer = new StringTokenizer(theSentence, delim, true);
Then,
if(delim.contains(token))
str.append(" ");
When you declare new StringTokenizer(theSentence, ".!", true) ,true enables to return delimiter characters also as next token. Since you appending a space to each token, it is normal to get that output.
You can check if the token is a delimiter character, if so, you add the space.
if(token.length() == 1 &&
delims.indexOf(token.charAt(0)) != -1)
str.append(" "); //we just hit the delim char, add space
I want to decode a Base64 encoded string, then store it in my database. If the input is not Base64 encoded, I need to throw an error.
How can I check if a string is Base64 encoded?
You can use the following regular expression to check if a string constitutes a valid base64 encoding:
^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)?$
In base64 encoding, the character set is [A-Z, a-z, 0-9, and + /]. If the rest length is less than 4, the string is padded with '=' characters.
^([A-Za-z0-9+/]{4})* means the string starts with 0 or more base64 groups.
([A-Za-z0-9+/]{4}|[A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)$ means the string ends in one of three forms: [A-Za-z0-9+/]{4}, [A-Za-z0-9+/]{3}= or [A-Za-z0-9+/]{2}==.
If you are using Java, you can actually use commons-codec library
import org.apache.commons.codec.binary.Base64;
String stringToBeChecked = "...";
boolean isBase64 = Base64.isArrayByteBase64(stringToBeChecked.getBytes());
[UPDATE 1] Deprecation Notice
Use instead
Base64.isBase64(value);
/**
* Tests a given byte array to see if it contains only valid characters within the Base64 alphabet. Currently the
* method treats whitespace as valid.
*
* #param arrayOctet
* byte array to test
* #return {#code true} if all bytes are valid characters in the Base64 alphabet or if the byte array is empty;
* {#code false}, otherwise
* #deprecated 1.5 Use {#link #isBase64(byte[])}, will be removed in 2.0.
*/
#Deprecated
public static boolean isArrayByteBase64(final byte[] arrayOctet) {
return isBase64(arrayOctet);
}
Well you can:
Check that the length is a multiple of 4 characters
Check that every character is in the set A-Z, a-z, 0-9, +, / except for padding at the end which is 0, 1 or 2 '=' characters
If you're expecting that it will be base64, then you can probably just use whatever library is available on your platform to try to decode it to a byte array, throwing an exception if it's not valid base 64. That depends on your platform, of course.
As of Java 8, you can simply use java.util.Base64 to try and decode the string:
String someString = "...";
Base64.Decoder decoder = Base64.getDecoder();
try {
decoder.decode(someString);
} catch(IllegalArgumentException iae) {
// That string wasn't valid.
}
Try like this for PHP5
//where $json is some data that can be base64 encoded
$json=some_data;
//this will check whether data is base64 encoded or not
if (base64_decode($json, true) == true)
{
echo "base64 encoded";
}
else
{
echo "not base64 encoded";
}
Use this for PHP7
//$string parameter can be base64 encoded or not
function is_base64_encoded($string){
//this will check if $string is base64 encoded and return true, if it is.
if (base64_decode($string, true) !== false){
return true;
}else{
return false;
}
}
var base64Rejex = /^(?:[A-Z0-9+\/]{4})*(?:[A-Z0-9+\/]{2}==|[A-Z0-9+\/]{3}=|[A-Z0-9+\/]{4})$/i;
var isBase64Valid = base64Rejex.test(base64Data); // base64Data is the base64 string
if (isBase64Valid) {
// true if base64 formate
console.log('It is base64');
} else {
// false if not in base64 formate
console.log('it is not in base64');
}
Try this:
public void checkForEncode(String string) {
String pattern = "^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{4}|[A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)$";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(string);
if (m.find()) {
System.out.println("true");
} else {
System.out.println("false");
}
}
It is impossible to check if a string is base64 encoded or not. It is only possible to validate if that string is of a base64 encoded string format, which would mean that it could be a string produced by base64 encoding (to check that, string could be validated against a regexp or a library could be used, many other answers to this question provide good ways to check this, so I won't go into details).
For example, string flow is a valid base64 encoded string. But it is impossible to know if it is just a simple string, an English word flow, or is it base 64 encoded string ~Z0
There are many variants of Base64, so consider just determining if your string resembles the varient you expect to handle. As such, you may need to adjust the regex below with respect to the index and padding characters (i.e. +, /, =).
class String
def resembles_base64?
self.length % 4 == 0 && self =~ /^[A-Za-z0-9+\/=]+\Z/
end
end
Usage:
raise 'the string does not resemble Base64' unless my_string.resembles_base64?
Check to see IF the string's length is a multiple of 4. Aftwerwards use this regex to make sure all characters in the string are base64 characters.
\A[a-zA-Z\d\/+]+={,2}\z
If the library you use adds a newline as a way of observing the 76 max chars per line rule, replace them with empty strings.
/^([A-Za-z0-9+\/]{4})*([A-Za-z0-9+\/]{4}|[A-Za-z0-9+\/]{3}=|[A-Za-z0-9+\/]{2}==)$/
this regular expression helped me identify the base64 in my application in rails, I only had one problem, it is that it recognizes the string "errorDescripcion", I generate an error, to solve it just validate the length of a string.
For Flutter, I tested couple of the above comments and translated that into dart function as follows
static bool isBase64(dynamic value) {
if (value.runtimeType == String){
final RegExp rx = RegExp(r'^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)?$',
multiLine: true,
unicode: true,
);
final bool isBase64Valid = rx.hasMatch(value);
if (isBase64Valid == true) {return true;}
else {return false;}
}
else {return false;}
}
In Java below code worked for me:
public static boolean isBase64Encoded(String s) {
String pattern = "^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)?$";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(s);
return m.find();
}
This works in Python:
import base64
def IsBase64(str):
try:
base64.b64decode(str)
return True
except Exception as e:
return False
if IsBase64("ABC"):
print("ABC is Base64-encoded and its result after decoding is: " + str(base64.b64decode("ABC")).replace("b'", "").replace("'", ""))
else:
print("ABC is NOT Base64-encoded.")
if IsBase64("QUJD"):
print("QUJD is Base64-encoded and its result after decoding is: " + str(base64.b64decode("QUJD")).replace("b'", "").replace("'", ""))
else:
print("QUJD is NOT Base64-encoded.")
Summary: IsBase64("string here") returns true if string here is Base64-encoded, and it returns false if string here was NOT Base64-encoded.
C#
This is performing great:
static readonly Regex _base64RegexPattern = new Regex(BASE64_REGEX_STRING, RegexOptions.Compiled);
private const String BASE64_REGEX_STRING = #"^[a-zA-Z0-9\+/]*={0,3}$";
private static bool IsBase64(this String base64String)
{
var rs = (!string.IsNullOrEmpty(base64String) && !string.IsNullOrWhiteSpace(base64String) && base64String.Length != 0 && base64String.Length % 4 == 0 && !base64String.Contains(" ") && !base64String.Contains("\t") && !base64String.Contains("\r") && !base64String.Contains("\n")) && (base64String.Length % 4 == 0 && _base64RegexPattern.Match(base64String, 0).Success);
return rs;
}
There is no way to distinct string and base64 encoded, except the string in your system has some specific limitation or identification.
This snippet may be useful when you know the length of the original content (e.g. a checksum). It checks that encoded form has the correct length.
public static boolean isValidBase64( final int initialLength, final String string ) {
final int padding ;
final String regexEnd ;
switch( ( initialLength ) % 3 ) {
case 1 :
padding = 2 ;
regexEnd = "==" ;
break ;
case 2 :
padding = 1 ;
regexEnd = "=" ;
break ;
default :
padding = 0 ;
regexEnd = "" ;
}
final int encodedLength = ( ( ( initialLength / 3 ) + ( padding > 0 ? 1 : 0 ) ) * 4 ) ;
final String regex = "[a-zA-Z0-9/\\+]{" + ( encodedLength - padding ) + "}" + regexEnd ;
return Pattern.compile( regex ).matcher( string ).matches() ;
}
If the RegEx does not work and you know the format style of the original string, you can reverse the logic, by regexing for this format.
For example I work with base64 encoded xml files and just check if the file contains valid xml markup. If it does not I can assume, that it's base64 decoded. This is not very dynamic but works fine for my small application.
This works in Python:
def is_base64(string):
if len(string) % 4 == 0 and re.test('^[A-Za-z0-9+\/=]+\Z', string):
return(True)
else:
return(False)
Try this using a previously mentioned regex:
String regex = "^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{4}|[A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)$";
if("TXkgdGVzdCBzdHJpbmc/".matches(regex)){
System.out.println("it's a Base64");
}
...We can also make a simple validation like, if it has spaces it cannot be Base64:
String myString = "Hello World";
if(myString.contains(" ")){
System.out.println("Not B64");
}else{
System.out.println("Could be B64 encoded, since it has no spaces");
}
if when decoding we get a string with ASCII characters, then the string was
not encoded
(RoR) ruby solution:
def encoded?(str)
Base64.decode64(str.downcase).scan(/[^[:ascii:]]/).count.zero?
end
def decoded?(str)
Base64.decode64(str.downcase).scan(/[^[:ascii:]]/).count > 0
end
Function Check_If_Base64(ByVal msgFile As String) As Boolean
Dim I As Long
Dim Buffer As String
Dim Car As String
Check_If_Base64 = True
Buffer = Leggi_File(msgFile)
Buffer = Replace(Buffer, vbCrLf, "")
For I = 1 To Len(Buffer)
Car = Mid(Buffer, I, 1)
If (Car < "A" Or Car > "Z") _
And (Car < "a" Or Car > "z") _
And (Car < "0" Or Car > "9") _
And (Car <> "+" And Car <> "/" And Car <> "=") Then
Check_If_Base64 = False
Exit For
End If
Next I
End Function
Function Leggi_File(PathAndFileName As String) As String
Dim FF As Integer
FF = FreeFile()
Open PathAndFileName For Binary As #FF
Leggi_File = Input(LOF(FF), #FF)
Close #FF
End Function
import java.util.Base64;
public static String encodeBase64(String s) {
return Base64.getEncoder().encodeToString(s.getBytes());
}
public static String decodeBase64(String s) {
try {
if (isBase64(s)) {
return new String(Base64.getDecoder().decode(s));
} else {
return s;
}
} catch (Exception e) {
return s;
}
}
public static boolean isBase64(String s) {
String pattern = "^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{4}|[A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)$";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(s);
return m.find();
}
For Java flavour I actually use the following regex:
"([A-Za-z0-9+]{4})*([A-Za-z0-9+]{3}=|[A-Za-z0-9+]{2}(==){0,2})?"
This also have the == as optional in some cases.
Best!
I try to use this, yes this one it's working
^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)?$
but I added on the condition to check at least the end of the character is =
string.lastIndexOf("=") >= 0