How to Send Free SMS From Online?

As you all know guyz, the era that we are living is an era of internet, so everyone is hanging on their PC all the time. Everyday we are sending some SMS to our friends in order to inform something to them. So, if there is some method that can send SMS for free, everyone gonna use that method and loved it. Today, in this tutorial we are gonna discuss about that method. And also if you wanna send the message to your friends without displaying your number and name, this tutorial is for you! sometime we are in the mode of cracking jokes and wanna fool your friends or sending some threatening message(only for entertainment purpose otherwise it will be illegal) you can use this method.

All you have do is go to http://www.txtlocal.co.uk and there you need to create one account. after creating account. (you need to fill up some of your information and txtlocal will send you a verification email into your email and verify your account [watch the video for detail]) you have to hover your mouse over send menu and then click on send text SMS, click on that option and you need to provide the number that you want to send the SMS and you can select your default display name and you can also schedule your SMS which will looks like this:

May be after sending 10 SMS your credit will be finished! but don't be sad you can send more and more SMS but you need to create another account with another email account! that's it. you may not be that much happy to hear this (sending only 10 SMS) but don't forget you are able to send these 10 SMS for free and also there is one line "To get something, you must lose something". it means you have to spend some time in creating account in txtlocal.co.uk in order to send free SMS.

Finally, if you guyz think, this is awesome tutorial and help you in some extend please share this tutorials with your all friend so that they can know about this method of sending SMS!

for more details watch the video:

Watch the video

How to Create Login Attempts Limit Form in Java Using NetBeans IDE and MySQL?[With Source Code]

watch the video for the whole procedure and designing part:

write the following code inside the class:

    Connection con = null;
    PreparedStatement pst = null;
    ResultSet rs = null;
    int attempt =1;

Write the following code inside the listening event of button:

       
        String sql = "select * from mysqllimitnew where username = ? and password=?";
        if(attempt<4){
        try{
            con = DriverManager.getConnection("jdbc:mysql://localhost/mysqllimitnew", "root", "");
            pst = con.prepareStatement(sql);
            pst.setString(1, txtuser.getText());
            pst.setString(2, txtpass.getText());
            rs=pst.executeQuery();
            if(rs.next()){
            JOptionPane.showMessageDialog(null,"username and password matched"+attempt);
            }else
            {
            JOptionPane.showMessageDialog(null, "error"+attempt);}
            
            
        }
        catch(Exception ex){
        JOptionPane.showMessageDialog(null, ex);}
        }
        else if(attempt!=4){
            JOptionPane.showMessageDialog(null,"try once again"+attempt);
        }
        else{
        JOptionPane.showMessageDialog(null,"exceed"+attempt);
        txtuser.setEnabled(false);
        txtpass.setEditable(false);
        }
        if(attempt==5){
            this.dispose();
        }
        attempt++;
       
 

watch the video:

How to Create Login Attempts Limit Form in Java Using NetBeans IDE?[With Source Code]

watch the video for the designing part!
write the following code inside the click event of the button:

Declare following variable inside the class but outside the click event of the button as:

         int attemtp=1;

        String user = txtuser.getText();
        String pass = txtpass.getText();
        if(attempt <4 && user.equals("TSN") && pass.equals("TSN")){
        JOptionPane.showMessageDialog(null, "accessed!");
        }
        else if(attempt!=4){
            JOptionPane.showMessageDialog(null, "denied"+attempt);
        }
        else
        {
        JOptionPane.showMessageDialog(null, "attempt exceed!"+attempt);
        txtuser.setEditable(false);
        txtpass.setEnabled(false);
        }
        if(attempt==5){
        this.dispose();
        }
        attempt++;
    }

Watch the video:

How to Show jTable Selected Row into jTextField in java?[With Source Code]

Before doing this, you need to watch this video!

watch the designing part in the video section given below:

Connection code:
    Connection con = null;
    PreparedStatement pst = null;
    ResultSet rs = null;

First create one function showTableData() and write the following code inside this function!

       
    try{
         con = DriverManager.getConnection("jdbc:mysql://localhost/combobox","root","");
         String sql = "SELECT * FROM combobox";
         pst = con.prepareStatement(sql);
         rs=pst.executeQuery();
         jTable2.setModel(DbUtils.resultSetToTableModel(rs));
 }
     catch(Exception ex){
         JOptionPane.showMessageDialog(null, ex);
         
     }
       
 

Call this function inside the constructor of the class! i.e. in the next line of initComponents();

now create the mouse clicked event of the jtable and write the following code!

       
    try{
            con = DriverManager.getConnection("jdbc:mysql://localhost/combobox","root","");
            int row = jTable2.getSelectedRow();
            String tblclick = (jTable2.getModel().getValueAt(row, 0).toString());
            String sql = "select * from combobox where Rollno ="+tblclick+"";
            pst = con.prepareStatement(sql);
            rs=pst.executeQuery();
            if(rs.next())
            {
                String Roll = String.valueOf(rs.getInt("Rollno"));
                jTextField1.setText(Roll);
                String firstname = rs.getString("fname");
                jTextField2.setText(firstname);
                String lastname = rs.getString("lname");
                jTextField3.setText(lastname);
                String add = rs.getString("address");
                jTextField4.setText(add);
            }
        }
        catch(Exception ex)
        {
            JOptionPane.showMessageDialog(null, ex);
        }
       
 

You need to import two library file mysql connector and rs2xml jar file. The link of that file is provided in the video description section.

Watch the video:

How to Connect jComboBox with Database in java?[With Source Code]

watch the video for the designing part!

write the following code inside the class:

    Connection con = null;
    PreparedStatement pst = null;
    ResultSet rs = null;

Create one function and write the following code inside that function!

       
    try
        {
            con= DriverManager.getConnection("jdbc:mysql://localhost/combobox", "root","");
            String sql = "select * from combobox ";
            pst = con.prepareStatement(sql);
            rs = pst.executeQuery();
            while(rs.next())
            {
                String name = rs.getString("fname");
                jComboBox1.addItem(name);
            }
            
        }catch(Exception ex)
        {
            JOptionPane.showMessageDialog(null, ex);
        }
       
 

call this function inside the constructor of that class i.e. under the initComponents();

 Watch The Video:

How to Create a Login Form in Java with MySQL Database using NetBeans IDE?[With Source Code]

see the designing part and library connection part in video:

Source Code:
Connection code inside the class!

    Connection con = null;
    PreparedStatement pst =null;
    ResultSet rs = null;


Code Inside the login button!

       
    String sql="select * from login where username=? and password=?";
        try{
            con = DriverManager.getConnection("jdbc:mysql://localhost/testsql","root","");
            pst=con.prepareStatement(sql);
            pst.setString(1, user.getText());
            pst.setString(2,pass.getText());
            rs=pst.executeQuery();
            if(rs.next())
            {
                JOptionPane.showMessageDialog(null,"username and password matched");
                
            }
            else
            {
                JOptionPane.showMessageDialog(null, "username and password do not matched");
            }}
            catch(SQLException | HeadlessException ex)
                    {
                    JOptionPane.showMessageDialog(null,ex);
                    }
       
 

Watch the video:

How to Create Login Form in ASP.NET using SQL Server Database[With Source Code]

See the Designing part in Video:

<connectionStrings>
    <add name="dbconnection" connectionString="Data Source=RAN-PC;Initial Catalog=aspnew;Integrated Security=True"/>
  </connectionStrings>

Watch The Database Connection Part in Video.

Code inside the Login Button:

       
     string constr = WebConfigurationManager.ConnectionStrings["dbconnection"].ConnectionString;
        SqlConnection con = new SqlConnection(constr);
        try
        {
            con.Open();
            SqlCommand cmd = new SqlCommand("select count(*) from aspnew where username='" + txtuser.Text + "' and password ='" + txtpass.Text + "' ", con);
            SqlDataAdapter sda = new SqlDataAdapter(cmd);
            DataTable dt = new DataTable();
            sda.Fill(dt);
            cmd.ExecuteNonQuery();
            if(dt.Rows[0][0].ToString()=="1")
            {
                Response.Write("alert('successful in login')"); //write Script tag before writing alert and close script tag as shown in below
                Response.Redirect("~/Default2.aspx");
            }
            else
            {
                Response.Write("alert('error in login')"); //write Script tag before writing alert and close script tag
            }
        }catch(Exception ex)
        {
            Response.Write(ex.Message);
        }

    }
       
 

Write the Script tag as:

Response.Write("<script>alert('successful in login')</script");

Watch The Video: