Is there a function to append padding digits to the left of a digit.I want to form a 3 digit number, so when the value is '3' i need to make it '003'
e.g in php we have
$m_ccode=str_pad($m_casecode,3,"0",str_pad_left);
same i want to convert in asp.net c#. How can i do it, i have numeric value stored in string 's'
String s = DropDownList3.SelectedItem.Value;
string variant1 = "3".PadLeft(3, '0'); // if you have a string
string variant2 = 3.ToString("000"); // if you have a number
In your case you need to unbox your int if the numeric value is an int
String s = ((int)DropDownList3.SelectedItem.Value).ToString("000");
Try the following:
String s = DropDownList3.SelectedItem.Value.PadLeft(3, '0');
Related
I have strings which are in a format below:
"p100"
"p231"
"p000"
.
.
.
these strings are button names and the last 3 characters are referring to a cell of a 3D array,and I need to convert those strings to integers to get the cell address, I use this:
string str = clickedButton.Name.ToString();
int xi ;
int xj;
int xz;
xi = Convert.ToInt32(str[1]);
xj = Convert.ToInt32(str[2]);
xz = Convert.ToInt32(str[3]);
please note that I don't use "str[0]" because it is "p".
but when I compile my code , the value of xi,xj,xz are the Ascci values of the string characters.
how should I convert string to int so that it won't happen?
Do the following:
int xi = (int)Char.GetNumericValue(str[1]);
Also refer to this
Why not just format the string to remove p from the name directly?
string str = clickedButton.Name.ToString();
str = str.Replace("p","");
int a = Convert.ToInt32(str);
int onesplace = a%10;
int secondplace = (a/10)%10;
int thirdplace = a/100;
The 'places' start from right to left. Ones place is the right most digit
If you want the individual values for the cell, you can then convert this string into an integer and use the modulus function
EDIT: updated code for it
Your code is going after the character at the position.
What you want is to get the string at the position.
xi = Convert.ToInt32(str.Substring(1,1));
xj = Convert.ToInt32(str.Substring(2,1));
xk = Convert.ToInt32(str.Substring(3,1));
Am reading from an excel sheet column and need to save into an sql table.
this is what the fiedl looks like in excel;'33349836', but I need it to be saved this way in the database '0033349836', because that field needs to be 10characters.
You can use String.PadLeft Method (Int32, Char) overload.
Returns a new string that right-aligns the characters in this instance
by padding them on the left with a specified Unicode character, for a
specified total length.
string s = "33349836";
string newstring = s.PadLeft(10, '0');
Remember, 0033349836 will be a string representation of your numeric values. Don't keep this kind of data in a numeric column type. Keep it in a some character type of column like nvarchar.
There are a couple of ways to do this using C#.
One such way is this:
static void Main(string[] args)
{
string s = "33349836";
int width = 10;
char padding = '0';
string s1 = s.PadLeft(width, padding);
Console.WriteLine(s);
Console.WriteLine(s1);
}
That code will output these values:
33349836
0033349836
You can use PadLeft() function to add zeros to your string.
Try This:
var str = "33349836";
str = str.PadLeft(10,'0');
I want to have numbers with a fixed digit count.
example: 00001, 00198, 48484
I can do like this:
string value;
if (number < 10)
{
value = "0000" + number.ToString();
}
else if (number < 100)
{
value = "000" + number.ToString();
}
else if (number < 1000)
{
...
}
But this is a bit odd. Is there any built in function for my purpose?
Yes, there is:
string value = String.Format("{0:D5}", number);
According to the MS reference: http://msdn.microsoft.com/en-us/library/dd260048.aspx
You can pad an integer with leading zeros by using the "D" standard
numeric format string together with a precision specifier. You can pad
both integer and floating-point numbers with leading zeros by using a
custom numeric format string.
So:
To display the integer as a decimal value, call its ToString(String)
method, and pass the string "Dn" as the value of the format parameter,
where n represents the minimum length of the string.
Code:
string value = number.ToString("D5");
.NET fiddle: http://dotnetfiddle.net/0U9A6N
You should use the ToString() method with custom formating - see the docs. In particular the 0 specifier.
Replaces the zero with the corresponding digit if one is present; otherwise, zero appears in the result string.
eg,
value = number.Tostring("00000");
string value = number.ToString("00000");
You can do it this way :
number.ToString("00000")
If you wish to return 5 digits numbers, you should use the PadLeft() function;
int Value = 101;
char pad = '0';
String sValue = Value.ToString();
sValue = sValue.s.PadLeft(5, char)
In this case, you don't have to test whether to add 1, 2 or 3 zeros, it'll automatically add the number of zeros needed to make it 5 digits number.
int input_number = Convert.ToInt32(txtinput.Text);
string number_value = input_number.ToString("00000");
I hope that it will solve your problem. It worked well for me in my previous project.
Test this code in your development. It should be worked properly without doubt.
Same as #Jojo's answer, but using C# 6's interpolated strings:
var value = $"{number:00000}";
Apart from String.Format, You can also use String.PadLeft
value = number.ToString().PadLeft(5, '0');
I my application due to some reason I have two numbers in 5 digits.
The following code give you brief idea.
string s = "00001"; // Initially stored somewhere.
//Operation start
string id = DateTime.Now.ToString("yy") + DateTime.Now.AddYears(-1).ToString("yy") + s;
//Operation end
//Increment the value of s by 1. i.e 00001 to 00002
This can be done easily by convert the value of s to int and increment it by 1 but after all that I have to also store the incremented value of s in 5 digit so it will be "00002".
This think give me a pain...
use
string s = "00001";
int number = Convert.ToInt32(s);
number += 1;
string str = number.ToString("D5");
to get atleast 5 digits.
The "D" (or decimal) format specifier
If required, the number is padded with zeros to its left to produce the number of digits given by the precision specifier. If no
precision specifier is specified, the default is the minimum value
required to represent the integer without leading zeros.
This seems to work for me.
string s = "00001";
int i = Int32.Parse(s);
i++;
s = i.ToString("D" + s.Length);
So I think you want to know how to convert an int to a 5 digit string.
You can do this:
int i = 1;
string s = i.ToString("D5");
//s = "00001"
There are plenty of format examples here.
Use String.Format() to achieve this:
string str = String.Format({0:#####}, s);
Look here.
This works using the PadLeft function:
int i = 1; // Initially stored somewhere.
//Operation start
string id = DateTime.Now.ToString("yy") + DateTime.Now.AddYears(-1).ToString("yy") + i.ToString().PadLeft(5, '0');
//Operation end
I am adding a new record to XML file, first I'm querying all existing items and storing the count in an int
int number = query.count()
and then increment the number by 1.
number = number + 1;
Now I want to format this value in a string having N00000000 format
and the number will occupy the last positions.
Pseudo code:
//declare the format string
sting format = "N00000000"
//calculate the length of number string
int length =number.ToString().Length();
// delete as many characters from right to left as the length of number string
???
// finally concatenate both strings with + operator
???
String output = "N" + String.Format ("00000000", length)
Alternatively if you change your formatstring to "'N'00000000" you can even use:
String output = String.Format (formatString, length)
Which means you can fully specify your output by changing your formatstring without having to change any code.
int i = 123;
string n = "N" + i.ToString().PadLeft(8, '0');
var result = number.ToString("N{0:0000000}");
HTH
You can use the built in ToString overload that takes a custom numeric format string:
string result = "N" + number.ToString("00000000");
Here is a another one ...
result = String.Format("N{0:00000000}",number);