suttercain Posted September 20, 2007 Share Posted September 20, 2007 Hi guys, I have this basic script I wrote. Right now I only have a single form input for demo purposes which will not process if the field is left blank. The thing I cannot get to work is the error to also be displayed if the form is submitted while no value is entered: <?php if(isset($_POST['submit'])){ // If the form was submitted validate_input(); // Check for empty fields if(count($errors) != 0){ // If there are errors, displayForm(); // redisplay the form } else { print_r($_POST); //Include Processing } } else { displayForm(); } function validate_input(){ global $errors; if ($_POST['name'] == "") { //Left blank show error. $errors['name']="Required!"; } } function displayForm() { ?> <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post"> <input type="text" name="name" value="" /> <?php echo $errors['name']; //display error ?> <input type="submit" name="submit" value="Submit" /> </form> <?php } ?> Link to comment https://forums.phpfreaks.com/topic/70055-solved-server-side-validation-cannot-get-the-error-to-display/ Share on other sites More sharing options...
Jessica Posted September 20, 2007 Share Posted September 20, 2007 change it to <?php if(!isset($_POST['name']) || strlen(trim($_POST['name'])){ $errors['name']="Required!"; } ?>[code] [/code] Link to comment https://forums.phpfreaks.com/topic/70055-solved-server-side-validation-cannot-get-the-error-to-display/#findComment-351808 Share on other sites More sharing options...
suttercain Posted September 20, 2007 Author Share Posted September 20, 2007 I got it. I was missing the declaration of setting the $errors to global within the displayForm() function. Thanks. SC Link to comment https://forums.phpfreaks.com/topic/70055-solved-server-side-validation-cannot-get-the-error-to-display/#findComment-351818 Share on other sites More sharing options...
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