How to read varbinary of SQL as string in C#?

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I have an web application which reads varbinary from SQL as a byte array in code. Basically it does a ExecuteScalar of the column and returns it as a byte array.

I am now doing a mini version of the application as a Windows application. For this, I am trying to skip the DB connection. I am directly inputting the varbinary value from SQL to a RichTextBox and trying to read it as a byte array. However, I do not get the same byte array. What am I doing wrong and how should I proceed ?

Basically, I have in hand the varbinary value (eg: "B0015379737"). This is a varbinary value i have taken from SQL. Now, I want to simply paste this value in a textbox and read it as byte array. Basically the source is varbinary. I am giving input as a string and I want the output as byte array which matches the source varbinary

Posted 8-Mar-15 23:43pm

Divakar Raj M

Updated 9-Mar-15 0:08am

v2

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3 solutions

Solution 1

Hi,

Depending on which encoding was used, use the following code:

C#

byte[] binaryString = (byte[])reader[1];

// if the original encoding was ASCII
string ascii = Encoding.ASCII.GetString(binaryString);

// if the original encoding was UTF-8
string utf = Encoding.UTF8.GetString(binaryString);

// if the original encoding was UTF-16
string utfs = Encoding.Unicode.GetString(binaryString);

Posted 8-Mar-15 23:56pm

Praneet Nadkar

Comments

Divakar Raj M 9-Mar-15 6:04am

See, I have in hand the varbinary value (eg: "B0015379737"). This is a varbinary value i have taken from SQL. Now, I want to simply paste this value in a textbox and read it as byte array. Basically the source is varbinary. I am giving input as a string and I want the output as byte array which matches the source varbinary

phil.o 9-Mar-15 9:12am

B0015379737 is not a binary value. It the the hexadecimal representation (so, a string) of binary value 0000000000000000000010110000000000010101001101111001011100110111.

Divakar Raj M 9-Mar-15 9:13am

It's a value of datatype varbinary in SQL

phil.o 9-Mar-15 9:47am

No, it is its string hexadecimal representation.

Solution 2

If you start from the hexadecimal representation of the value, and you want to work with it, you have to do it in several steps:

- First, translate this hexadecimal string representation of a byte array into a byte array

- Second, pass this byte array to the GetString() method of the chosen Encoding.

Something like:

C#

using System.Globalization;
using System.Text;

public static byte[] ByteArrayFromHexaString(string hexa) {
   int length = hexa.Length;
   List<byte> result = new List<byte>();
   // Fetch whether there are an odd or even number of chars
   bool isOdd = ((length & 1) == 1);
   if (isOdd) {
      result.Add(byte.Parse(hexa[0], NumberStyles.HexNumber));
   }
   string s;
   for (int i = (isOdd) ? 1 : 0; i < length; i += 2) {
      s = hexa.SubString(i, 2);
      result.Add(byte.Parse(s, NumberStyles.HexNumber));
   }
   return result.ToArray();
}

public static string StringFromByteArray(byte[] bytes, Encoding encoding) {
   return encoding.GetString(bytes);
}

You now have both methods you need to get a string from the hexadecimal representation of its encoded raw value.

Good luck. Hope this helps.

Posted 9-Mar-15 3:12am

phil.o

Comments

Divakar Raj M 10-Mar-15 9:44am

Thanks. You got me thinking in the right path. Though this code didn't work for me, I did some googling and was able to figure out a piece of code which works to my requirement.

Member 10377494 24-Nov-17 1:04am

what is this NumberStyles

phil.o 24-Nov-17 2:06am

NumberStyles Enumeration

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Divakar Raj M · Accepted Answer · 2015-03-10T03:45:00

Solution 3

I got it using this piece of code

C#

string hex = richTextBox1.Text
int NumberChars = hex.Length;
byte[] bytes = new byte[NumberChars / 2];
for (int i = 0; i < NumberChars; i += 2)
bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2).ToString(), 16);
return bytes;

Posted 10-Mar-15 3:45am

Divakar Raj M

Comments

phil.o 10-Mar-15 9:53am

hex.Substring(i, 2).ToString() is ugly :)
SubString method already returns a string. There is no need to transform a string to a string.

Divakar Raj M 11-Mar-15 7:14am

Yup. Thanks