Birthday validation

[demeritrious]

Good evening,

 

I am working on a birthday validation using Jquery datepicker. I am trying to make the birthday section validate whether the birthday input is in a date format and I want to prevent users under 18 from join. Thank you

 

 

accountsettings.php

 
<input type="text" name="Birthday" id="Birthday" maxlength="50" value="
<?php
if($form->value("Birthday") == ""){
   echo $session->userinfo['Birthday'];
}else{
   echo $form->value("Birthday");
}
?>">
 

process.php

 function register( $subuser, $subconfirmuser, $subpass, $subconfirmpass, $subemail, $subconfirmemail, $subFirstname, $subLastname, $subGender, $subBirthday, $subRegion, $subCountry, $subCity_State, $subQuestion, $subAnswer, $subConfirmAnswer){
      global $database, $form, $mailer;  //The database, form and mailer object
 
   
 
 

 /* Birthday error checking */
 $field = "Birthday"; //Use field name for Birthday
 if(!$subBirthday || strlen($subBirthday = trim($subBirthday)) == 0){
 $form->setError($field, "* Birthday not entered");
 }
           else{
         /* I've attempted different formulas but it did not work here */
         $date = date('yyyy-mm-dd');
         if( $subBirthday > $date){
            $form->setError($field, "* You are too young!");
         } 
}

 

[fastsol]

This is not what you think it is for the date() value.

$date = date('yyyy-mm-dd');

You want this instead

$date = date('Y-m-d');

Your way would have produced a date string like this 1999199919991999-0101-3131

[demeritrious]

So should I do a formula that involves?

 $date = date('-18Y-m-d');

Edited July 12, 2014 by demeritrious

[fastsol]

Assuming your date is being posted in a format like 2013-12-31 you could use this.

         if( $subBirthday > date("Y-m-d", strototime("-18 years", date()))){
            $form->setError($field, "* You are too young!");
         }

[Miggy64]

I find working with DateTime() easier:

<?php
$now = new DateTime();
$birthday = new DateTime('1964-08-28');

$years = $now->diff($birthday, true);

echo $now->format('Y-m-d') . '<br>';


echo $years->y; // Number of years since birthday

//var_dump($years);

[demeritrious]

I'm receiving this error:

 

Fatal error: Call to undefined function strototime()

[demeritrious]

I could not figure out how to manipulate it to give me an error confirmation that says the data entered is not in the correct format or if the user is under 18

 

 

I find working with DateTime() easier:

<?php
$now = new DateTime();
$birthday = new DateTime('1964-08-28');

$years = $now->diff($birthday, true);

echo $now->format('Y-m-d') . '<br>';


echo $years->y; // Number of years since birthday

//var_dump($years);

[demeritrious]

By the way this worked for me for Invalid Email validation..maybe this might help someone:

 

 

      /* Email error checking */
      $field = "email";  //Use field name for email
      if(!$subemail || strlen($subemail = trim($subemail)) == 0){
         $form->setError($field, "* Please enter email address!");
      }
      else{
         /* Check if valid email address */
         $regex = "^[_+a-z0-9-]+(\.[_+a-z0-9-]+)*"
                 ."@[a-z0-9-]+(\.[a-z0-9-]{1,})*"
                 ."\.([a-z]{2,}){1}$";
         if(!eregi($regex,$subemail)){
            $form->setError($field, "* Email invalid");
         }
         $subemail = stripslashes($subemail);
      }

[fastsol]

I'm receiving this error:

 

Fatal error: Call to undefined function strototime()

Sorry I misspelled the function, honestly you should know that

It's strtotime()

[demeritrious]

AHHHHH I didn't even notice!!!...Thank you very much for your help! It works the way it is but is there a way to make sure it is in the Y-m-d format in case someone types in random data?

 

This is what worked for me.

 

 /* Birthday error checking */
 $field = "Birthday"; //Use field name for Birthday
 
 if(!$subBirthday || strlen($subBirthday = trim($subBirthday)) == 0){
 $form->setError($field, " * Please enter birthday");
 }
 
else if( $subBirthday > date("Y-m-d", strtotime("-18 years"))){
            $form->setError($field, "* You are too young!");
}