How to mantain values if submit fails PHP - php

I have a form that is being validated using PHP server side scripting. The form's action is also to itself. To show multiple errors, I store it in an array and used foreach to loop through the values and echo it for dislay. I thought i'm finished using PHP validation but I noticed that everytime I submit the form, even though it still contains some errors, all of the values that were inputted resets. Why? When I used javascript, the values remains so whats with PHP?
This is how I set my errors and insert data to db.
if(isset($_POST["submit"]))
{
$lname = $_POST["lname"];
$errors = array();
if(strlen($lname) == 0)
{
$errors[] = "Last name is required";
}
if (!empty(errors))
{
//display errors using foreach loop
}
else
{
//connect and insert data to mysql db
}
}
?>

You perhaps write your HTML form without setting up default values.
<input type="text" name="firstname" />
In the case of an error you should set up a default value. Here an example. Do not forget to sanitize your POST vars before outputting.
<input type="text" name="firstname" value="<?php echo $_POST['firstname']; ?>" />

in your form's input element code use it like this
<input type="text" name="username" value="<?php if(isset($_POST['username']) echo $_POST['username'];?>">
using isset to check if value exist or not and than echoing it

Related

keep input values on submit with errors and clear them when the submit is successful (php)

I have a simple form (it´s in a modal and reloads in the same page) which I want to remember the input values if it was filled incorrectly and forget them if it was inserted correctly in the db.
My code remembers the value in both cases and I don't want that.
Thanks for your help.
HTML:
<form>
<input type="text" i" name="user" value = "<?php echo isset($_POST['user']) ? $_POST['user'] : '';?>/">
<button type="submit" name="register-btn">Crear cuenta</button>
</form>
PHP:
<?php
if(isset($_POST['register-btn'])){
include_once('includes/connect_db.php');
$usern = mysqli_real_escape_string($conn, $_POST['user']);
if(empty($usern)){
exit();
echo "error"
}
else{
$varsql = "INSERT INTO registers (user) VALUES ('$usern');";
}
}
?>
Good to see the query value is escaped, Also important to escape before rendering htmlentities($_POST['user])
But to answer your question, You can execute the PHP before the form is rendered, and $_POST['user] = null when the process was successful. Three for the post value will be clear before the form is rendered again.

Button Value Not Changing when Text Field Isn't Empty

Having a small issue getting my submit button to change the value when the php variable isn't empty. So the way I have it set up is that when the button "GO" is pressed it will set the value of the text field to "1234567890" meaning it is no longer empty. Now when the page first loads the text field will be empty and the button should say "GO" once pressed the text field value will change and the button should now say "REFRESH" however it stays saying "GO" can anyone see where I'm going wrong here?
Thanks.
PHP
<?
if (!$HostKey){
$HostBtn = 'GO';
}
else{
$HostBtn = 'REFRESH';
}
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
if (isset($_POST['Go'])) {
$HostKey = "123567890";
}
}
}
?>
HTML
<form action="Home.php" method="post">
<p><strong>HOST:</strong>
<input name="Host" id="Host" type="text" value="<? echo $HostKey; ?>" maxlength="10" disabled>
<input name="Go" id="Go" type="submit" value="<? echo $HostBtn; ?>"></p>
</form>
Your issue is a logical one, as $HostKey isn't defined prior to you doing your check - so $HostKey is null. You define it after that. Then, because the variable is effectively null, when you apply the not-operator !, that condition is always true.
You should refactor your code to check if the form was submitted, and define the value of the variable based on that instead. All you need in PHP would be
$HostBtn = isset($_POST['Go']) ? 'REFRESH' : 'GO';
$HostKey = isset($_POST['Go']) ? 123567890 : '';
This would also fix the "Undefined variable..." notices you should have been getting if you enabled error-reporting,
error_reporting(E_ALL);
ini_set("display_errors", 1);
Which should be enabled while in development. In a live environment, you shouldn't display the actual errors though - but its fine to do that under development.
You should try something like :
<?
if (isset($_POST['Go'])) {
$HostKey = "123567890";
$HostBtn= "REFRESH";
}else{
$HostKey="";
$HostBtn="GO";
}
?>

Need type="text" input fields to display previously-POST'd entries if errors occur

I've probably not explained what I"m trying to do in the title very well, so here goes:
I've got a HTML entry form, in a .php file. That entry form when submitted files a POST operation. That POST operation has code to check field input, similar to this:
<?php
...
if ($_POST["submitted"] == 1) {
//"submitted" is a hidden field with value '1' in the form
$isvalid = 1; // Inits to 1, changes to 0 if something fails checks.
$field1 = $_POST["field1"];
$field2 = $_POST["field2"];
...
/*
Checks for validation on each field go here,
but are not relevant to this question here.
*/
}
if ($isvalid == 1) {
// Send email
} else { ?>
<!-- Print form, and error msg. -->
...
<input name="field1" type="text" id="field1" size="32" class="stylingclass">
...
So the above is example code, but here's the real question: How can I get any of the text input fields in my form, when submitted and an error occurred, to automatically contain the previous values that were entered, so that the user can correct the entries accordingly, based on whatever constraints I've set?
You can set the value parameter using a basic ternary operator:
<input name="field1" type="text" id="field1" value="<?php echo isset($_POST["field1"]) ? $_POST["field1"] : ''; ?>" size="32" class="stylingclass">
Note that the ternary operator is used here to prevent an error if there is no index of the specified field in the $_POST array
Just add value="<?=$_POST['field1']?>" (assuming PHP 5.4, use <?php echo $_POST['field1'] ?> otherwise)

Retaining values in forms fields when validation of data fails

I am having problems figuring out how to retain users data when the validation fails. I am somewhat new to PHP so I might be making some huge mistakes in my logic.
Currently if the validation fails all the fields are wiped clean and $_Post data is also gone.
Here is some code assuming the user enters an invalid email I want the Name field to be retained. This code is not working.
<?php
if($_POST['doSubmit'] == 'Submit') {
$usr_name = $data['Name'];
$usr_email = $data['Email'];
if (isEmail($usr_email)==FALSE){
$err = "Email is invalid.");
header("Location: index.php?msg=$err");
exit();
}
//do whatever with data
}
if (isset($_GET['msg'])) {
$msg = mysql_real_escape_string($_GET['msg']);
echo "<div class=\"msg\">$msg</div><hr />";
}
if (isset ($_POST['Name'])){
$reusername = $_POST['Name'];}
else{$reusername = "NOTHING";}//to test
?>
<form action="index.php" method="post" >
<input name="UserName" type="text" size="30" value="<?echo $reusername;?>">
<input name="Email" type="text" size="30">
<input name="doSubmit" type="submit" value="submit">
</form>
}
You can use AJAX to submit your form data to your PHP script and have it return JSON data that specifies whether the validation was successful or not. That way, your fields won't be wiped clean.
Another way is to send back the recorded parameters to the posting page, and in the posting page, populate the fields using PHP.
However, I think the first solution is better.
UPDATE
The edit makes your code clearer and so I noticed something. Your input field is called UserName in the HTML, but you are referring to Name in PHP. That's probably why it's not working. Is your field always being filled with the value NOTHING? Make sure the name of the input field and the subscript you are using in $_POST are the same.
Also, there's no need to redirect to another page (using header) if you have an error. Maintain an $errors array or variable to print error messages in the same page. But like I mentioned before, it's probably better to use the JSON approach since then you can separate your view layer (the html) from the PHP (controller layer). So you'd put your HTML in one file, and your PHP in another file.
EDIT:
Vivin had commented that my assumption regarding the header was incorrect and he was right in that. Further more it looks like what the OP is doing is essentially what i layed out below albeit in a less structured fashion. Further Vivin - caught what is likely the actual problem here - the html name and the array key $_POST do not match.
Its wiped clean because you are using header to redirect to another page. Typicaly you would have a single page that validates the data and if ok does something with it and returns a success view of some sort, or that returns an error view directly showing the form again. By using header youre actually redirecting the browser to another page (ie. starting up an entirely new request).
For example:
// myform.php
if(strtolower($_SERVER['REQUEST_METHOD']) == 'get')
{
ob_start();
include('form.inc.php'); // we load the actual view - the html/php file
$content = ob_get_clean();
print $content; // we print the contents of the view to the browser
exit;
}
elseif(strtolower($_SERVER['REQUEST_METHOD']) == 'post')
{
$form = santize($_POST); // clean up the input... htmlentities, date format filters, etc..
if($data = is_valid($form))
{
process_data($data); // this would insert it in the db, or email it, etc..
}
else
{
$errors = get_errors(); // this would get our error messages associated with each form field indexed by the same key as $form
ob_start();
include('form.inc.php'); // we load the actual view - the html/php file
$content = ob_get_clean();
print $content; // we print the contents of the view to the browser
exit;
}
}
so this assumes that your form.inc.php always has the output of error messages coded into it - it just doesnt display them. So in this file you might see something like:
<fieldset>
<label for="item_1">
<?php echo isset($error['item_1']) ? $error['item_1'] : null; ?>
Item 1: <input id="item_1" value="<?php echo $form['item_1'] ?>" />
</label>
</fieldset>
Could do something similar to if failed then value=$_POST['value']
But vivin's answer is best. I don't know much about AJAX and wouldn't be able to manage that.
Ok, firstly header("Location: index.php?msg=$err"); is not really required. It's best practice not to redirect like this on error, but display errors on the same page. Also, redirecting like this means you lose all of the post data in the form so you can never print it back into the inputs.
What you need to do is this:
<input name="Email" type="text" size="30" value="<?php print (!$err && $usr_email ? htmlentities($usr_email, ENT_QUOTES) : '') ?>">
Here I'm checking whether any errors exist, then whether the $usr_email variable is set. If both these conditions are matched the post data is printed in the value attribute of the field.
The reason I'm using the function htmlentities() is because otherwise a user can inject malicious code into the page.
You appear to be processing the post on the same page as your form. This is an OK way to do things and it means you're nearly there. All you have to do is redirect if your validation is successful but not if it fails. Like this
<?php
if( isset( $_POST['number'] ) ) {
$number = $_POST['number'];
// validate
if( $number < 10 ) {
// process it and then;
header('Location: success_page.php');
} else {
$err = 'Your number is too big';
}
} else {
$number = '';
$err = '';
}
?>
<form method="POST">
Enter a number less than 10<br/>
<?php echo $err ?><br/>
<input name="number" value="<?php echo $number ?>"><br/>
<input type="submit">
</form>

POST BACK in PHP or JAVASCRIPT?

how can I post back the data that are already in the text field?
example:
if I miss one of the required field an error will prompt when i click the submit button.
How can I make an post back data in that form using php or javascript and make the cursor of the mouse directly located to the field that caused an error?
There is no automated ways in PHP to write back the informations of the fields so you just have to echo it back.
Let's say you've got a "username" field ( <input type="text" name="username" /> ) you just need to add this:
value="<?php echo isset($_POST['username']) ? $_POST['username'] : ''; ?>"
or if you like more:
value="<?php if(isset($_POST['username'])) echo $_POST['username']; ?>"
changed "" to ''
This sounds like basic form validation. I would recommend reading some of these tutorials or looking for some pre-built PHP form validation mechanisms.
Form validation using PHP
PHP/CSS Form validation
PHP Form Validation
Some frameworks such as CodeIgniter will do this for you if you use their own libraries. It's worth checking out such a framework as they provide a lot of other benefits. Of course it's not always possible to transfer an existing application but it's still useful to bear in mind for the future.
If I understand this correctly you want to keep whatever data the user has already entered, tell him what he did wrong and preferably focus on the bad field.
If so then here's a very basic example using a form with two fields where both need to be filled in to proceed.
<?php
$field1=$_POST['field1'];
$field2=$_POST['field2'];
$badField="";
if($_POST['form_action']=="submitted") {
//Check incoming data
if(empty($field1)) {
$badField="field1";
echo 'field1 is empty<br>';
}
elseif(empty($field2)) {
$badField="field2";
echo 'field2 is empty<br>';
}
else { //Everything ok - move to next page
header('Location: <next page>');
}
}
echo '<form name="mybo" action="' . $_SERVER['PHP_SELF'] . '" method="POST">
<input type="text" name="field1" value="' . $field1 . '"><br>
<input type="text" name="field2" value="' . $field2 . '"><br>
<input type="submit" name="Submit" value=" Enter ">
<input type="hidden" name="form_action" value="submitted">
</form>';
//Focus on empty field
if(!empty($badField)) {
echo '<SCRIPT language="JavaScript">
document.mybo.' . $badField . '.focus(); </SCRIPT>';
}
?>
I think the Moav's answer is "philosophically" correct however if you want do that you can:
1) pass via GET or POST the text control id;
2) on the server check that error condition;
3) fill an hidden input field with that value on the page returns
4) if error that with JS you can do:
window.onload = init; // init stuff here
function init()
{
checkForError();
}
function checkForError()
{
var h = document.getElementById("error_field");
var v = h.value;
if(v)
document.getElementById(v).focus();
}
However, if you will do that for every error field there will be a post and this is
by a user perspective very boring...so it is better to adopt other approaches...
I would take a different approach:
Validation should be in JS, and as such you never loose data, as you don't submit.
Any wrong data that was submitted and caught on the server is due to someone trying to pass over your JS validation, which means he has criminal thoughts, usually.

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