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PHP - Keep Value Of Selectbox If There Is An Error On The Form

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Keep Value Of Selectbox If There Is An Error On The Form	View Content

I have a select box on a form that is populated with the following function:

function getStates(){
  // OPEN CONNECTION TO MYSQL ON LOCALHOST
  $dbhost = 'localhost' ;
  $username = 'root' ;
  $password = '' ;
  $conn = mysqli_connect("$dbhost", "$username", "$password");
  if (!$conn) {
    die('Could not connect: ' . mysqli_error()); 
  }
  mysqli_select_db($conn, "mydb"); 
  $retval = mysqli_query($conn, "SELECT * FROM tblstates") or die("Error: " . mysqli_error($conn));
  echo "<option value=''>Select one...</option>";
  while($row = mysqli_fetch_array($retval, MYSQLI_ASSOC)) {
    $rec = $row['name'];
    echo "<option value = " . $row['name']. ">";
    echo $row['name'];
    echo "</option>";
  }
}

On the form I have several text boxes and the following select box:

<div class="formfieldgroup">
  <label class="label">Birth State/Country: <span class="required">&#149;</span></label>
  <select class="fatherbirthstate" name="varfatherbirthstate" tabindex="9"><?=getStates()?></select>
  <span class="form-error2"><?= $varfatherbirthstateErr ?></span>
</div>

TESTING: I intentionally leave a text box blank so I will get an error and I make sure to make a selection from the select box.

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How is it possible, in PHP, to display an error message next to a form input text field if a user is attempting to submit a form with empty fields? Moreover, how is it possible to remove the same error message when the user fills in the input field with the required data and/or refreshes the page?

I have 2 files that I am working with: application.php and process.php.

application.php mainly has the HTML of the form.
process.php has the PHP for processing and validating the form.

I am very new to learning PHP (I started learning it last week) and I have been searching for hours for a clear answer. I have already tried different methods for generating the error message including: using empty(), !isset, $_POST["name"] = "", etc, and have tried using session_start();, $row, echo, print, and other variables to try and display error message on the page, and I have tried using unset();, and = null, to try and remove the error message once the input field has been filled but all to no avail. Either the method I try only half works, or nothing works, and I cannot pinpoint which is the part that is not working.

I only have 2 files to work with, and as an example of what I want to do is:

1. If the first name field is empty when the user clicks submit, an error message should appear next to the input.
2. If the user fills in the input and clicks submit, the form should be processed and added to the database.
3. When the user returns to the application to page to fill in another application form, the error message should not appear.

Is this possible with PHP?

Form Error...

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I have been trying to figure this out for hours and I can't see what's wrong. I'm getting this fatal error:

Fatal error: Call to undefined function form() in /home/content/29/6879529/html/contact.php on line 5

Here's my form code:
Code: [Select]
<html>
<form name="contact" method="POST" action="contact.php" onsubmit="return validateContactForm()">
<input class="textbox" name="fname" type="text" />
<input type="submit" name="submit" value="Send Message" />
</form>
</html>

Here's my php code, which is at the top of the same page the form is on:
Code: [Select]
<?php

if (isset($_POST['submit'])) { // If the form has been submitted.

$fname = form($_POST['fname']); //This is line 5
echo $fname;

}

?>

Any ideas on what could be causing this error? I've used this format MANY times in the past and I've never had a problem until now. I'm baffled!

Error After Php Form

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Hey ,

I've made a website with 2 menubuttons(lotto , euromillions).
When I click on lotto -> I must choose a value of a dropdownlist and click on OK.
Now when I click on "OK" after ive picked a value -> it says that the pagename is empty and gives an error.

To fill the pagename I've added an ELSE function, but it does fix only 1 of the 2 forms which is not the point.

Maybe I can fix this by using menuarray to make an IF-ELSE code for the 2 menu.

Can someone give me an idea to solve this ?

Index page -> to write the pagename which selects the included file.
    if (array_key_exists($_GET["page"], $arrMenuItems)){
        $strPage = $_GET["page"];
    }else{
        //$strPage = "lotto";
        //$strPage = "euromillions";
    }

MenuArray -> links to pages
$arrMenuItems = array(
     "lotto" => array(
            'menuitem'=>'lotto' ,
            'page_id' => '?page=lotto',
            'page_name' =>'lotto',
            'page_content'=> 'lotto/index.php',
            'page_css'=>'lotto.css',

           ),
     "euromillions" => array(
            'menuitem'=>'euromillions' ,
            'page_id' => '?page=euromillions',
            'page_name' =>'euromillions',
            'page_content'=> 'euromillions/index.php',
            'page_css' => 'euromillions.css',
           ),
);

Error With Form To Email

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Validation Form Error

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Hello..

I have two php form and one has categories and their subjects. In there, users can select category and relevant subjects to that category. Likewise users can select upto 3 categories and their subjects in first form. when those select and after submitting those should handle in my second page.

But I have made so frustrated here when Im going to handle those in my second page.

Please Can anybody help me to do this.

I use this, but......

Code: [Select]
if ( ( isset($_POST['submitted2'])) && ( isset($_SESSION['category'])) && (isset( $_SESSION['subjects'])) && sizeof( $_SESSION['category']) == 1) {

//$_SESSION['category'] = $_POST['category'];

print_r ( $_SESSION['category']);

echo 'good';

} elseif ( ( isset($_POST['submitted2'])) && ( isset($_SESSION['category'])) &&
(isset( $_POST['main_category'] ))) {

print_r ( $_SESSION );

echo 'very good';

}
Thanks in advance.