a String contain 821560888v.I want to validate this String every time a user enter a similar type-java coding?
339901
To validate text input converted to a int try this:\n\ntry {\n    String s = new JTextField().getText(); // The user enters text in the text field\n    int value = Integer.parseInt(s);\n}\ncatch(NumberFormatException ex) { // Invalid input, not a number\n    ex.printStackTrace();\n}\n\nHope this helps
674350
91316  \n \n \n91388    (UNIQUE)      \n \n \n91401  \n \n \n91402  \n \n \n91403  \n \n \n91404    (PO BOX)      \n \n \n91405  \n \n \n91406  \n \n \n91407    (PO BOX)      \n \n \n91408    (PO BOX)      \n \n \n91409    (PO BOX)      \n \n \n91410    (PO BOX)      \n \n \n91411  \n \n \n91412    (PO BOX)      \n \n \n91413    (PO BOX)      \n \n \n91416    (PO BOX)      \n \n \n91423  \n \n \n91426    (PO BOX)      \n \n \n91436  \n \n \n91470    (UNIQUE)      \n \n \n91482    (UNIQUE)      \n \n \n91495    (UNIQUE)      \n \n \n91496    (UNIQUE)      \n \n \n91497    (UNIQUE)      \n \n \n91499    (UNIQUE)
770189
Number will have to be the actual number it is working it out from..and it might not work as a long and definately not an integer.\n\nNumber as String\nOdd as Long (or Integer)\nEven as Long ( or Integer)\n\nIf Not (Number <1 or Number > 10) Then\nNumber = Number / 2\nIf InStr(Number, ".5") then\nOdd = Odd + 1\nElse\nEven = Even + 1\nEnd If\nEnd If\nMe.Print "Odd: " & Odd & " - " & "Even: " & Even
981844
2 +28 -65 +48 +99,002,456,486 +1 -87 -998\n= 99,002,455,415\nNote to other answerers:  If you hold your mouse over the string of characters, it will show everything.
449829
The validation options are:\n\n1.None. \n2.Valid chars validation. \n3.Invalid chars validation. \n4.Letters validation. \n5.Numbers validation. \n6.MaskEdit. \n7.Regular expression validation (E-mail, IP, URL, ZIP, Fecha and Custom). \n8.Using the code\n\nTry the following link for more info.
1168828
1237548*26636686+5567562356-46\n\n32969745048238
717813
In JavaScript you can do it. In Java you cannot, unless perhaps you typeset a char to a byte to an int. But that's not what you mean, is it?\n\nYou need to use at least one method.\n\nEdit:\nOf course, what the people who answered below me don't realize is that everything is in a method ("function" in other languages). Perhaps you want to write your own method to convert strings to numbers... it's still part of a method.\n\nBut why would you want to do that when there are parseInt and valueOf?\nRead down from http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Integer.html#parseInt(java.lang.String)
457195
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
164812
If it does not contain a decimal place I guess:\n5?
436488
I assume we're in C or Java, right???\n\na) Loss of precision\nb) can't assign a string to a char (Java - in C there are no strings)\nc) value out of range\nd) incompatible type\ne) WORKS!!!
1153308
import java.io.IOException;\n\npublic class PosNeg {\n\npublic static void main(String[] args) {\n\nint [] wNumbers;\nwNumbers = new int[5];\nchar c = ' ';\n\nfor (int i=0;i<5;i++) {\nSystem.out.print("Enter the #"+i+": ");\ntry {\nwNumbers[i]=System.in.read()-48;\nif (wNumbers[i]<0)\nwNumbers[i]=(System.in.read()-48)*-1;\nwhile (System.in.read()!='&#92;n'){\n}\n} catch (IOException e) {\ne.printStackTrace();\n}\nSystem.out.print("The number you entered is ");\nif (Math.abs(wNumbers[i])!=wNumbers[i])\n   System.out.println("negative.");\n else\n   System.out.println("positive.");\n}\n}\n}
684775
SQL has got a convert(datatype(size,precision),value) method\n eg:convert(decimal(18,2),12345.6789) as no_with_10.2_precisoin\n\nu should be able to use it to convert ur string to small int too. but make sure u validate the entry before sending to sql (avoid a typecaste error)
918056
There is probably a better way, or at least a more streamline way, of doing this, but off the top of my head here is what I would do:\n\n1. Create a loop to run through all the characters in your string.\n2. Test each character to see if it falls between ASCII values 47 - 58 (i.e. 0-9). If it does then the character is a number.\n3. If the first test fails then you know that it isn't a number; therefore, you know that it must be a character, so now test to see if it is a character between 'a' and 'd' (i.e. ASCII 97-100).\n\nHere is some sample code (hopefully it can help):\npublic static void main(String[] args){\nString teststr = "d3";\nchar x;\n\nSystem.out.println("Test String:" + teststr);\n\nfor(int i = 0; i < teststr.length(); i++){\nx = teststr.charAt(i);\nif(x >= 47 && x <= 58){\n// Store it however\nSystem.out.println(Integer.parseInt(""+ x));\n}\nelse if(x >= 97 && x <= 100){\nif(x == 'a')\nSystem.out.println("a=" + 1);\nelse if(x == 'b')\nSystem.out.println("b=" + 2);\nelse if(x == 'c')\nSystem.out.println("c=" + 3);\nelse if(x == 'd')\nSystem.out.println("d=" + 4);\n}\n}\n}
565108
You are on the right track, but try the ASCII conversion.\n\nif ASC((mid$(stringname,1,1)) > 64 AND \n          ASC((mid$(stringname,1,1)) < 91 Then\n     'code here, 65 is UpperCase "A" and 90 is upperCase "Z"\nelse\n  if ASC((mid$(stringname,1,1)) > 96 AND \n          ASC((mid$(stringname,1,1)) < 123 Then\n   '97 is lowercase "a" and 122 is lowercase "z"\nEnd if
977141
' Validate an e-mail address\n' Example:\n'   MessageBox.Show(IsValidEmail("mbellinaso@vb2themax.com")) ' True\n'   MessageBox.Show(IsValidEmail("vb2themax.com"))            ' False\n'   MessageBox.Show(IsValidEmail("mbellinaso@vb2themax"))     ' False\n\nVB version:\n\nFunction IsValidEmail(ByVal email As String) As Boolean\n    Return System.Text.RegularExpressions.Regex.IsMatch(email, _\n        "^&#92;w+([-+.]&#92;w+)*@&#92;w+([-.]&#92;w+)*&#92;.&#92;w+([-.]&#92;w+)*$")\nEnd Function\n\nAs C#:\nbool IsValidEmail(string email)\n{\n    return System.Text.RegularExpressions.Regex.IsMatch(email,  "^&#92;w+([-+.]&#92;w+)*@&#92;w+([-.]&#92;w+)*&#92;.&#92;w+([-.]&#92;w+)*$");\n}
278590
You ae confusing yourself!!! 0 - 255 is the range of ASCII characters. For eg. 65 stands for A and 97 stands for a and 1 stands for ☺. So you try 256 it will again display ☺ because there are only 255 characters. To know the value of an ASCII character press Alt + the req no. For eg Alt + 129 yeilds to ü. Go ahead try it yourself in any text editor. So a character variable can store only characters (ie. 0 - 255) NOTE:- Even all numbers are characters. Eg. Alt + 49 yeilds to 1 which is considered as a character. You know what I mean??? Don't you???
177153
There is no flaw as long as you recognize that .99999 repeats forever.  If you were to end it at some point and subtract, then as a previous poster said, you'd actually get 8.99....1.  But since it repeats, there is always another 9 and so it works.\n\nAnd consider this - if .9999999repeating did NOT equal 1, then there would be another number between them.  What would that number be?\n\n~~~~~~~~~~~~~~~~~~~~\nNo, the number in question is not infinitessimal, it doesn't exist.  You can never get to the end of the string of nines to try and squeeze another number in there :)