php if statement from html option value error

Question

I am trying to replicate the following:

php if statement html option value and am following the directions as suggested:

Give name to your select.

<select name="selectedValue">
    <option value="Newest">Newest</option>
    <option value="Best Sellers">Best Sellers</option>
    <option value="Alphabetical">Alphabetical</option>
</select>

In php

Example:

switch($_POST['selectedValue']) {
    case 'Newest':
        // do Something for Newest
    break;

    case 'Best Sellers':
        // do Something for Best seller
    break;

    case 'Alphabetical':
        // do Something for Alphabetical
    break;

    default:
      // Something went wrong or form has been tampered.
}

However, I am getting the following error:

Notice: Undefined index: selectedValue in C:\xampp\trial.php on line 7

Please help

the error occurs as soon as I load the page and even when I select an option, nothing changes. — Prasinus Albus
– Prasinus Albus, Commented Mar 6, 2018 at 5:38
$_POST is only filled in when you submit the form. It won't have any values in it when you first load the page. You need to check if the form was submitted before trying to use the parameters. — Barmar
– Barmar, Commented Mar 6, 2018 at 5:39

Pavan Sikarwar · Accepted Answer · 2018-03-06 05:57:03Z

2

if you are sending form data then you have to set method post in your form tag or if you are using ajax send it by type post and in your php file put code inside if condition after form has been posted like isset($_POST['selectedValue'])

<form method="post" action="url to your php file">
   <select name="selectedValue">
      <option value="Newest">Newest</option>
      <option value="Best Sellers">Best Sellers</option>
      <option value="Alphabetical">Alphabetical</option>
   </select>
   <button type="submit">Submit</button>
</form>

edited Mar 6, 2018 at 5:57

answered Mar 6, 2018 at 5:42

Pavan Sikarwar

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Comments

Andrei Todorut · Accepted Answer · 2018-03-06 07:10:44Z

0

That's happens because you don't always have the key selectedValue in your POST array.

if(isset($_POST['selectedValue'])){
   // your code
}

answered Mar 6, 2018 at 7:10

Andrei Todorut

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Prasinus Albus · Accepted Answer · 2018-03-07 13:10:55Z

0

My problem was solved after adding a submit button within the form tag of HTML

<form  action="#" method="post">
    <select name="selectedValue">
        <option value="Newest">Newest</option>
        <option value="Best Sellers">Best Sellers</option>
        <option value="Alphabetical">Alphabetical</option
    </select>
    <input type="submit" style="float:left; margin-left: 2%;" name="submit" value="Search"/>
</form>

edited Mar 7, 2018 at 13:10

user8280225

answered Mar 6, 2018 at 5:56

Prasinus Albus

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