hi Im making use of VB 2008, i attempted including the Russian characters and it worked. And as a be counted of actuality I never pronounced the UTF set in my question. under is the code i Used. have a seem. Dim sqlcon As OleDbConnection = New OleDbConnection("information source=C:login.mdb; service=Microsoft.Jet.OLEDB.4.0; person id=Admin") sqlcon.Open() Dim ssql As String = "INSERT INTO login VALUES (@call,@password)" Dim sqlcmd As OleDbCommand = New OleDbCommand(ssql, sqlcon) sqlcmd.Parameters.AddWithValue("@call", TextBox1.text cloth) sqlcmd.Parameters.AddWithValue("@passwor... TextBox2.text cloth) sqlcmd.ExecuteNonQuery() I COPIED THE RUSSIAN CHARACTERS TO TEXTBOX1 and that they artwork.
Comments
SQL INSERT INTO Statement
http://www.w3schools.com/sql/sql_insert.asp
Refer the below link
http://www.ntchosting.com/mysql/insert-data-into-t...
It is the same as SQL syntax, you run below command
INSERT INTO phonebook(phone, firstname, lastname, address) VALUES('+1 123 456 7890', 'John', 'Doe', 'North America')
hi Im making use of VB 2008, i attempted including the Russian characters and it worked. And as a be counted of actuality I never pronounced the UTF set in my question. under is the code i Used. have a seem. Dim sqlcon As OleDbConnection = New OleDbConnection("information source=C:login.mdb; service=Microsoft.Jet.OLEDB.4.0; person id=Admin") sqlcon.Open() Dim ssql As String = "INSERT INTO login VALUES (@call,@password)" Dim sqlcmd As OleDbCommand = New OleDbCommand(ssql, sqlcon) sqlcmd.Parameters.AddWithValue("@call", TextBox1.text cloth) sqlcmd.Parameters.AddWithValue("@passwor... TextBox2.text cloth) sqlcmd.ExecuteNonQuery() I COPIED THE RUSSIAN CHARACTERS TO TEXTBOX1 and that they artwork.
INSERT INTO table_name (values,values) VALUES (data,data);
For eg:
$name = $_post['name'];
$address = $_post['address'];
$phone = $_post['phone'];
$result= mysql_query("insert into table_name (id,name,address,phone) values (' ', '$name', '$address', '$phone').mysql_error()");
if($result)
{
echo "<script>alert('successfully inserted');
}
else
{
echo "failed to insert";
}