PHP - check / compare input from textfield

Question

Until now I had a PHP Kontaktform with empty textfields and checked the needed fields with:

$name = check_input($_POST['name'], "Please enter a Name.");

Now the lable "your name" is in it the textfield itself. So the value is not empty anymore. How do I check if there's still only "your name" in the field instead of a real name?

Thank you (sorry php-noob)

@nile - isn't there a "your name" missing, or does it check the standard value by itself? — fab_ulus
– fab_ulus, Commented Jun 18, 2013 at 12:22

aebersold · Accepted Answer · 2013-06-18 12:27:00Z

4

use the HTML5 placeholder tag in the first place.

<input placeholder="Your Name" value="" name="name">

or, if that is not possible, check for that given string:

$name = (strstr('your name', $_POST['name']) ? '' : check_input($_POST['name'], "Please enter a Name."));

edited Jun 18, 2013 at 12:27

answered Jun 18, 2013 at 12:21

aebersold

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1 Comment

aebersold Over a year ago

there are several polyfills for ie8 and lower.

Triclops200 · Accepted Answer · 2013-06-18 12:23:20Z

1

Compare $name to "your name". If they are the same, then you didn't get an actual name.

for example

if ($name == "your name") {
    // do stuff
} else {
    // they have a valid name
}

edited Jun 18, 2013 at 12:23

answered Jun 18, 2013 at 12:21

Triclops200

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1 Comment

fab_ulus Over a year ago

This is exacly, what I was looking for. Thank you. ...but the placeholder solution is nicer I learned. ;)

som · Accepted Answer · 2013-06-18 12:24:09Z

0

You can do like this :

$name = isset( $_POST['name'] ) ? ($_POST['name'] == 'your name' ? '' : $_POST['name']) : '';
$name = check_input($name, "Please enter a Name.");

answered Jun 18, 2013 at 12:24

som

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