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somiadiscon

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Posts posted by somiadiscon

  1. Remove the quotes embedding the number you are comparing to :)This would cause the script think it is an actual numberr instead of a string. :)
    thanx man i will try and see ok
    Remove the quotes embedding the number you are comparing to :)This would cause the script think it is an actual numberr instead of a string. :blink:
    thanx man i will try and see ok :blink:
  2. I am using 5.1.4 php with a 1.3.x apache server i am not using mySQL normaly below php file show end-user to age is older or younger but this time nothing in my browser so why is that any one can help me to figure this out thanx :) <?php if ($_POST) { $age = $_POST['age']; if ($age < "14") { echo "You are younger than 14"; } else { echo "You are older than 14"; } } else { echo "<h4>Age Checker Software!</h4>"; echo "<form action=age.php method=post>Your Age:<input type=text name=age><br /><br />"; echo "<input type=submit value=Go!></form>"; } ?>

  3. could you please tell me there is an error says line five error i want to make a function that will display some text but i couldn't :) Parse error: parse error, unexpected T_STRING in c:\PHP\uploadtemp\Copy.php on line 12<?php function about() { // this declares a function and gives it a name of about, the () is used for arguments but we will not b needing those for this. echo("Hello, My name is John Doe and I am X years old. My Life's goal is to be a goalkeeper... bye now!"); // echo("");means that PHP should display this text. in this case some random message } this is a closing bracket teling PHP to end the function. about(); // to display all of that text now all we need to do is put the name of the function: about(); Don't forget the semi-colon afterabout();otherwise you'll get a bunch of errors ?>

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