Question

The situation is that the user enters a value in the textbox.Now whenever user enters the value, the value is to be compared to a value(field) which has been retrieved form the database.If the value the user has entered is higher than that of the database then they should be displayed with a popup.How do i do that?The language which i am using is PHP.The popup should be automatic

Thanks in advance

Answer 1

Online demo of following example: http://jsbin.com/ezusi3/

You can’t do “popups” with PHP. You’ll need to use a client-side solution, like Javascript. Below is a simple function that you could call in order to do what you are asking:

function checkVals() {
  // Retrieve the user-provided value
  var userVal = Number(document.getElementById("userVal").value);
  // Retrieve the server-provided value
  var dataVal = Number(document.getElementById("dataVal").value);
  // If the user value is too high, alert the user
  if (userVal > dataVal) {
    alert("Your value is too high.");
  }
}

Note that this is a Javascript function, not a PHP function.