I want to identify the elements in letters and elements in numbers in String array.Is there any other way to do it?
String a[]={"aaaa","111111","bbbbbb"};
for(int i=0;i<a.length;i++)
{
post your code for this FOR LOOP
}
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2 Answers
They say that you have a problem... so you choose regular expressions to solve it, now you have two problems. :-)
However, if speed is not that much of an issue, you could attempt it, assuming you have good unit tests in place. Something along the lines of:
public static void testRegularExpressionForElement() {
String[] a = new String[] {"test1", "13", "blah", "1234.44"};
Pattern pattern = Pattern.compile("[-+]?[0-9]*\\.?[0-9]+");
for (String element : a) {
if (pattern.matcher(element).matches()) {
System.out.println(element + " is a number");
} else {
System.out.println(element + " is not a number");
}
}
}
What's nice about the above, is that you can adapt the expression to match exactly what you want.
Another approach would be is to use Integer.parseInt() and catching the exceptions, but this is bad programming as you're using exceptions for logic.
1 Comment
Jaco Van Niekerk
The above will work as is, it's been in Java since 1.5 (I think?). If you really insists on the old style, add the line "String element = a[i];" directly after the loop declaration. However, I really think you should be using the improved for loop.
Traverse the array and write a function to check isNumeric or not.
for (a1 : a){
boolean isNumbr= isNumeric(a1);
}
...
public static boolean isNumeric(String str)
{
try
{
double d = Double.parseDouble(str);
}
catch(NumberFormatException nfe)
{
return false;
}
return true;
}
answered Aug 20, 2012 at 8:29
Subhrajyoti Majumder
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2 Comments
Jaco Van Niekerk
This will work Quoi, but the "catch" for program flow is something I don't like. I agree that there is no other way for the above. Functional java with "Either" as a return type would be a nice alternative.
Subhrajyoti Majumder
@JacoVanNiekerk - Even in that case Pattern matching would be alternative.