php and regex, how to strip the dollar sign and percentage sign if a user inputs it?

Question

Basically what this needs to do is only accept a price value, that can only be 40 or 39.95 or 100.09 or 4 etc, if the user enters anything other than a digit into the field it comes back with an error.

My question: how can I change it so that if the user inputs a dollar sign in the input field that it just gets stripped out rather than returning an error in that particular case?

if (ereg_replace('/^\d+(\.\d{2})?$/', $_POST['itemAmt'])) {
    echo '<h1>The original cost of the item: $' . $_POST['itemAmt'] . ' </h1>';
} else {
    echo '<h1 style="color:red; font-weight:900;">The price value was not numeric, try again :) </h1><br>';
}

All the ereg_* functions are deprecated. Check the docs http://php.net/ereg_replace — Mike B, Jan 06 '13 at 20:19
Never change data, make validators, that validate user's input, and return return if user inputed wrong data... — Glavić, Jan 06 '13 at 20:20

score 6 · Answer 1 · answered Jan 06 '13 at 20:19

6

$itemAmt = str_replace(array('$', '%'), '', $_POST['itemAmt']);

answered Jan 06 '13 at 20:19

zerkms

249,484
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score 2 · Answer 2 · answered Jan 06 '13 at 20:21

2

preg_replace('#[^0-9\.]+#','',$_POST['itemAmt']);

answered Jan 06 '13 at 20:21

MDEV

10,730
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49

score 0 · Answer 3 · answered Jan 06 '13 at 20:32

0

possible way

if ( preg_match( '/^(\d+(?:\.\d+)?)[\$%]?$/', $_POST['itemAmt'] ) ) {
    $validprice = preg_replace( '/^(\d+(?:\.\d+)?)[\$%]?$/', '\1', $_POST['itemAmt'] );
} else {
    // invalid input
}

answered Jan 06 '13 at 20:32

bukart

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40

php and regex, how to strip the dollar sign and percentage sign if a user inputs it?

3 Answers3