How to change php code to show errors under each input box

Question

I have the following code :

    if(isset($_POST['submit'])){

        if (! isset($_POST['firstname'])) {
            $error[] = "Please fill out all fields";
        }

        if (! isset($_POST['surname'])) {
            $error[] = "Please fill out all fields";
        }
........

with validation:

if (strlen($_POST['firstname']) < 2){
        $error[] = 'First name cannot be empty';
    }
    if (strlen($_POST['surname']) < 2){
        $error[] = 'Please provide your surname';
    }

......

More checks are made with the database....

This checks for errors and displays them in one go:

if(isset($error)){
                    foreach($error as $error){
                        echo '<p class="error-login">'.$error.'</p>';
                    }
                }

While this is working fine, I would like errors to be shown under each input box where there is an error happening.

I don't want to change the entire code, just want to make the necessary changes to this one, which I am incapable of doing myself.

Is putting them in array the only approach here or is there a simpler way ?

Thanks.

Add a key to errors: $error['firstname'] = '...'; and under each input check for errors under key. — u_mulder
– u_mulder, Commented Apr 6, 2020 at 11:03

u_mulder · Accepted Answer · 2020-04-06 11:10:14Z

1

The approach is - add errors to $error under a certain key, I presume - name of the input field:

if(isset($_POST['submit'])){

    // I use key `all` for errors that don't belong to any field

    if (! isset($_POST['firstname'])) {
        $error['all'] = "Please fill out all fields";
    }

    if (! isset($_POST['surname'])) {
        $error['all'] = "Please fill out all fields";
    }

    if (strlen($_POST['surname']) < 2){
        $error['surname'] = 'Please provide your surname';
    }

In your html markup:

// general errors, not related to inputs
if(isset($error['all'])){
    foreach($error['all'] as $err){
        echo '<p class="error-login">'.$err.'</p>';
    }
}

<input type="text" name="surname" />
<?php
if(isset($error['surname'])){
    foreach($error['surname'] as $err){
        echo '<p class="error-login">'.$err.'</p>';
    }
}

answered Apr 6, 2020 at 11:10

u_mulder

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2 Comments

davidev Over a year ago

Looks like the best approach for that question.

icbxw Over a year ago

Thanks, I appreciate your response however, to my code, I've made every modification as pointed out but it's not working. The form when submitted shows no errors and even when I fill out all fields it does not submit. I thought it might be the database processing as it returns this "$error[] = $e->getMessage();" but I've removed email & password and still the same. Thought it might be input values when returning "<?php if(isset($error)){ echo htmlspecialchars($_POST['firstname'], ENT_QUOTES); } ?>" but I've changed these to "if(isset($error['firstname'])" and still nothing. Any ideas ?

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