PHP function argument not processed to output

Question

    <?php

Function runSearch($name)
{
If(isset($_POST['submit']))
    {
    $name = $_POST['name'];
    echo "Results for " .$name;
    }
}

?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
   Search String: <input type="text" name="name"><br>
   <input type="submit" name="submit" value="Submit"><br>
</form>

This code is suppose to display what is entered into the Search String text box. When I don't use a function it works fine. But as soon as I place the code into the function runSearch there is no output. I'm new to php can an argument be sent to a php function and then displayed on the screen?

Atheist · Accepted Answer · 2021-04-17 05:03:50Z

3

you need to call your function, otherwise nothing will happen. Also you need to removed the $name-parameter:

<?php

function runSearch()
{
    if(isset($_POST['submit']))
    {
        $name = $_POST['name'];
        echo "Results for " .$name;
    }
}
runSearch();
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
   Search String: <input type="text" name="name"><br>
   <input type="submit" name="submit" value="Submit"><br>
</form>

answered Apr 17, 2021 at 5:03

Atheist

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