表单的PHP错误[关闭]

Question

I want to keep my div layout to my site when being directed with my error codes. Thing is I don't want multiple errors to have two separate css/div layouts. Here's what I got so far:

if(!preg_match($email_exp,$email_from)) {
  $error_message .= '<font color="blue">The Email Address you entered does not appear to be valid.<br /></font>';
}

$string_exp = "/^[A-Za-z .'-]+$/";

if(!preg_match($string_exp,$first_name)) {
  $error_message .= 'The First Name you entered does not appear to be valid.<br />';
}

if(!preg_match($string_exp,$last_name)) {
  $error_message .= 'The Last Name you entered does not appear to be valid.<br />';
}

if(strlen($comments) < 2) {
  $error_message .= 'The Comments you entered do not appear to be valid.<br />';

I want them all to open up on a page and have the same layout. If there are more than one errors, I want them to not each have separate code. How do I do that?

Answer 1

So I think you're asking for a single div to show any errors, something like this:

<div id="errors">
    <?php echo (!empty($error_message) ? $error_message : ""); ?>
</div>

But if you want the errors to show up on the same page as the form, you're far better off doing that in Javascript. Put something like this on your page:

<form>
    Sample: <input id="sample" type="text" onKeyUp="validateInputs()">
    <button type="button">Button Text</button>
</form>
<div id="error-div"></div>

<script>
    function validateInputs() {
    var errorMessage = '';
    var sampleText = document.getElementById('sample').value;
    if( !sampleText || sampleText.length < 4 ) {
        errorMessage = 'input "Sample" must be as least 4 chars in length';
    }
    if( errorMessage.length > 0 ) {
        document.getElementById('error-div').innerHTML = errorMessage;
    } else {
        document.getElementById('error-div').innerHTML = 'no errors';   
    }
  }
</script>

Here's a working jsfiddle example.