PHP Self-referencing script - php

Question

I am trying to embed a self-referencing PHP script inside an HTML form with following code:
Undefined index: conv
<form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post">
<input type = "number" id = "temp2" name = "temperature2" placeholder = "28">
<label for = "temp2"> degrees </label>
<select>
<option name = "conv" value = "f"> Fahrenheit </option>
<option name = "conv" value = "c"> Celsius </option>
</select>
<input type = "submit" value = "equals">
<?php
$type = $_POST["conv"];
$tmp = $_POST["temperature2"];
if ($type == "f") {
$newTmp = (9/5 * $tmp) + 32;
echo $newTmp . " degrees Celsius.";
}
elseif ($type == "c") {
$newTmp = (5 * ($tmp - 32)) / 9;
echo $newTmp . " degrees Fahrenheit.";
}
?>
</form>
And I am getting this messages:
Notice: Undefined index: conv
Notice: Undefined index: temperature2
Everything worked fine when the PHP script was in another file.
Anyone knows what am I doing wrong?

Answer 1

You must verify that you was send the page and $_POST exist. And correct the select element
<form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post">
<input type = "number" id = "temp2" name = "temperature2" placeholder = "28">
<label for = "temp2"> degrees </label>
<select name = "conv">
<option value = "f"> Fahrenheit </option>
<option value = "c"> Celsius </option>
</select>
<input type = "submit" value = "equals">
<?php
if(isset($_POST["temperature2"])) {
$type = $_POST["conv"];
$tmp = $_POST["temperature2"];
if ($type == "f") {
$newTmp = (9/5 * $tmp) + 32;
echo $newTmp . " degrees Celsius.";
}
elseif ($type == "c") {
$newTmp = (5 * ($tmp - 32)) / 9;
echo $newTmp . " degrees Fahrenheit.";
}
}
?>
</form>

Answer 2

The variable ($type = $_POST["conv"];) is not set until the form is processed. Do
if (!empty($_POST["conv"])) {
$type = $_POST["conv"];
}

Answer 3

Here is my answer...
First, it is better to verify, is it submitted or not ??, if submit button is invoked then code proceed rest. Else you get error. Moreover result and variable will not be shown until you click the submit button.
<form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post">
<input type = "number" id = "temp2" name = "temperature2" placeholder = "28">
<label for = "temp2"> degrees </label>
<select name = "conv">
<option value = "f"> Fahrenheit </option>
<option value = "c"> Celsius </option>
</select>
<input type = "submit" name="submit" value = "equals">
<?php
if(isset($_POST["submit"])) {
$type = $_POST["conv"];
$tmp = $_POST["temperature2"];
if ($type == "f") {
$newTmp = (9/5 * $tmp) + 32;
echo $newTmp . " degrees Celsius.";
}
elseif ($type == "c") {
$newTmp = (5 * ($tmp - 32)) / 9;
echo $newTmp . " degrees Fahrenheit.";
}
}
?>
</form>

Answer 4

Your PHP code will run every time you load the page, not only when someone presses submit. This means it looks out there for $_POST['conv'] and $_POST['temperature2'] but doesn't find anything because the form hasn't been posted.
You need to name your submit button and then surround all your PHP processing with an if like this:
<input type = "submit" name="mysubmit" value = "equals">
<?php
if (#$_POST['mysubmit']) {
$type = $_POST["conv"];
$tmp = $_POST["temperature2"];
if ($type == "f") {
$newTmp = (9/5 * $tmp) + 32;
echo $newTmp . " degrees Celsius.";
}
elseif ($type == "c") {
$newTmp = (5 * ($tmp - 32)) / 9;
echo $newTmp . " degrees Fahrenheit.";
}
}
?>
Now it will only look at that PHP code when someone has actually submitted something. Putting the # before the #$_POST['mysubmit'] makes it so you don't get the same error that you were getting before on this new array key.