在PHP中选择选项

Question

Hi I am trying to get the value for 3 different option select menu's and process them via a php script. The problem I am having is getting the value's to be sent to the server. When I run tests it is showing that the option is not 'set'. Here is my code -

HTML

    <div class='buy'>
            <form method="post">
            <select name="season1" id="purchasequantity" onchange="displaysub(this)">
            <option value="">Quantity</option>
            <option value="1">1</option>
            <option value="2">2</option>
            <option value="3">3</option>
            <option value="4">4</option>
            <option value="5">5</option>
            </select>
            </input type="submit">
            </form>

PHP

    <?php
    if(!isset($_POST['season1'])){
            echo "hello";

            $select1 = $_POST['season1'];
            switch ($select1) {
                    case '1':
                           echo 'this is value1<br/>';
                           break;
                    case '2':
                           echo 'value2<br/>';
                           break;
                    default:
                           # code...
                           break;
            return true;
}

} ?>

Answer 1

You condition is incorrect.

You need to check if the field isset yet your if statement is checking if it is NOT set because of the rouge ! .

Change,

if(!isset($_POST['season1'])) {

To,

if(isset($_POST['season1'])) {

Also

You input for your submit is incorrect HTML,

Change,

</input type="submit">

To,

<input type="submit">

Personally...

I'd change the code so the PHP checks if the submit button has been pressed as that makes for sense.

HTML

<input type="submit" name="submitButton">

PHP

if (isset($_POST['submitButton'])) {
    /** Your code here now. **/

}

Answer 2

   if(!isset($_POST['season1'])){

I made the above piece of code for debugging purposes. The ! is meant to be there.

There are 3 optional select menus that are all submited via one button.

The js function just displays the value selected from the select menu when selected

Answer 3

Maybe if you are using both the codes in same page :

 <div class='buy'>
            <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
            <select name="season1" id="purchasequantity" onchange="displaysub(this)">
            <option value="">Quantity</option>
            <option value="1">1</option>
            <option value="2">2</option>
            <option value="3">3</option>
            <option value="4">4</option>
            <option value="5">5</option>
            </select>
            <input type="submit">
            </form>
            <?php
    if(isset($_POST['season1'])){
            echo "hello";

            $select1 = $_POST['season1'];
            switch ($select1) {
                    case '1':
                           echo 'this is value1<br/>';
                           break;
                    case '2':
                           echo 'value2<br/>';
                           break;
                    default:
                           # code...
                           break;
            return true;
}
}else

echo 'value not set';   ?>