how to get form POST value in php [duplicate]

Question

I need how to get form POST value on the same page. Below is my script to try and extract the value as well as the form html:

<form action="" id ='list' name= "list" class="form-inline" method="POST">
    <div class="select">
        <select name="lamp" id ="lamp" onchange="gender(this)" style="background:transparent">
            <option id ='gender' hidden="hidden">Gender</option>
            <option value="0">Boy's</option>
            <option value="1">Girl's</option>
        </select>
    </div>
</form>

My Php code:

if (isset($_POST)) {  
    $pg_type =$_POST['lamp'];
    echo $pg_type;die;
}

I have echo the my variabe it is showing like:

Notice: Undefined index: lamp in C:\xampp\www\htdocs\rentozy\assets\includes\list-header.php on line 20

how does form will send request to your php file until you haven't define action="" ? — Ravi
– Ravi, Commented Jan 13, 2018 at 5:20
I imagine it has something to do with your inline javascript gender(this) sending the wrong value since there is no submit on this form. Probably related to your other question: stackoverflow.com/questions/48219386/… — Rasclatt
– Rasclatt, Commented Jan 13, 2018 at 5:23
As $_POST is a superglobal, I would believe isset($_POST) will always return true, even if it is empty (ie on initial page load). So you may want to check the for the actual field isset($_POST['lamp']) — Sean
– Sean, Commented Jan 13, 2018 at 5:27

ino · Accepted Answer · 2018-01-13 06:04:35Z

1

The form:
1. be consistent - use " or ' but not both for the HTML attributes;
2. set the name of select to gender and remove the first option since it is nonsense in this example;

<form action="" id="list" name="list" class="form-inline" method="POST">
    <div class="select">
        <select name="gender" id="gender" style="background:transparent">
            <option value="0">Boy's</option>
            <option value="1">Girl's</option>
        </select>
    </div>
</form>

PHP:
1. change the if check to the exact key such as gender not to the whole POST array 2. To check what was actually send via POST you can print the whole array:

if (isset($_POST['gender'])) {  
    $pg_type=$_POST['gender'];
    die('Gender value in $pg_type is: ' . $pg_type); // for debugging
}
else {
     // for debugging
     print_r($_POST);
}

answered Jan 13, 2018 at 6:04

ino

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