We Keep Coding

How can I submit form and stay on the same page after require?

I have a code which its simplified version could look like this :
file1.php
$array = array();
$array = new randomObject(1);
$array = new randomObject(2);
require('file2.php');
file2.php
<form method="post" action="?">
<?php
foreach ($array as $a) {
<p><?php echo $a->getAValue();
<textarea rows="5" cols="70" name="textbox[]">
</textarea>
</p>
<?php } ?>
<input id="isTrue"> //true or false
<input type="submit" >
</form>
The user is supposed to write answers in the textarea and click on submit then his answers are compared to the randomObject values. Then it shows if it's true or false next to each textarea

You are looking for something that the fron-tend will handle for you and an AJAX call is exactly what you need.
First of all, name your form
<form id="myForm" method="post" action="?">
<?php
foreach ($array as $a) {
<p><?php echo $a->getAValue();
<textarea rows="5" cols="70" name="textbox[]">
</textarea>
</p>
<?php } ?>
<input id="isTrue"> //true or false
<input id="submitButton" type="submit" >
</form>
Now you have proper id's both on the submit button and on the form itself.
<script>
let submitB = document.querySelector("#submitButton");
submit.addEventListener("click", function(e) {
e.preventDefault();
});
</script>
From now on you just have to write a proper ajax call to the url you wanted to access and you will be set to go.
If you need help with that let me know and I will throw something your way.

I am guessing you want to retain the values entered by the user, since they go away if you submit the form. (Page reloads)
This can be done by altering the input fields. If a value was submited pass that value to each corresponding input field.
Something like that:
<textarea rows="5" cols="70" name="textbox[]" <?php if(isset(value[SOMETHING])){?> value="<?php echo value[SOMETHING]; ?>" <?php } ?> >
This is just an example of how it would work. Make sure you adapt it to your code!

When clicked on button, go to the next question by ID

I'm building an online survey with statements in php. I retrieve my 2 statements per question via a database.
It should display question by question. So when you click at a statement, it saves the answer and inserts it into a table in the database with the value 0 or 1.
<?php
// I check if it's clicked, then it gets a '0' or a '1'.
if(isset($_GET['manipulate'])){
echo $_GET['manipulate'];
}
foreach($statements as $statement){ // "SELECT * FROM statement;" = $statements
$datas_key[] = ["order"=>$statement["order"], "statement1"=> $statement["statement1"], "statement2"=> $statement["statement2"], "questionnaire_id"=> $statement["questionnaire_id"]];
}
?>
Now I have the 2 options which each a statement that can be clicked like this:
<h1 id="newQ"> <?php $i = 0; ?></h1>
<form method="GET">
<h1>Question <?php echo $i + 1; ?> <br>
<button class="manipulate" name="manipulate" value="0"><?php echo $datas_key[$i]["statement1"]; ?></button>
<button class="manipulate" name="manipulate" value="1"><?php echo $datas_key[$i]["statement2"]; ?></button>
</form>
How do i write code in js or php to make sure that when I click, the id goes +1 AND see the next question with the next two statements ??

Create form inside for loop

I'm using codeignitor and am very new to it so sorry in advance if the question is senseless,but i'm stuck with certain requirement while coding.I have a for loop as below:
<?php foreach($messages as $req):?>
//This loop will execute depending on number of rows and is working fine.
<?php echo form_open('message/addFrom_masterlist','id="myform"'); ?>
//form is having input fields.
<?php echo form_close();?>
\\this acts as a submit button to my form which submits the form using javascript.
<input type="button" name="button" id="b1" class="btn btn-primary" onclick="myFunction1()" value="Submit"/>
<?php endforeach; ?>
//Below is javascript code for from submit.
<script>
function myFunction1() {
document.getElementById("myform").submit();
}
the problem is i want the id name for form to be unique since each time a button is clicked same form is being submitted.I don't want to use the submit button inside the form.Please someone help me

use this code
<?php foreach($messages as $req):?>
<?php $count = 1; ?>
//This loop will execute depending on number of rows and is working fine.
<?php echo form_open('message/addFrom_masterlist','id="myform$count"'); ?>
//form is having input fields.
<?php echo form_close();?>
\\this acts as a submit button to my form which submits the form using javascript.
<input type="button" name="button" id="b1" class="btn btn-primary" onclick="myFunction<?php echo $count; ?>()" value="Submit"/>
<?php $count++; ?>
<?php endforeach; ?>
//Below is javascript code for from submit.
<?php
$arrayCount = count($messages);
if(!empty($arrayCount)){
for($i=1; $i<= $arrayCount; $i++){
?>
<script>
function myFunction<?php echo $arrayCount; ?>() {
document.getElementById("myform<?php echo $arrayCount; ?>").submit();
}
</script>
<?php
}
}
?>
Hope this will help you!!
Note: But the your concept like this is not good.

php - select option value not passed to post

Im pretty stuck at this code, I really can't see why it should not work, i don't know if some javascript code im running beforehand is interfering?
Only showing relevant part of the code
The first section with javascript updates page when selecting another dropdown, and is placed before the code that im struggling with:
`
<script type="text/JavaScript">
function changeDropDown(){
var elLocation = document.getElementById('form_location_id');
var location = elLocation.options[elLocation.selectedIndex].value;
document.getElementById("form1").action = "choose.php?id=" + location;
document.getElementById("form1").submit();
}
</script>
<form id="form1" name="form1" method="post">
<select size="1" name="form_location_id" id="form_location_id" onchange='changeDropDown(this);'>
<?php
if ($chosen_id == "1") { ?>
<option value = "1" selected><?php echo "dropdown text" ?></option>
<? } else { ?>
<option value = "1"><?php echo "dropdown text" ?></option>
<?php } ?>
</select>
</form>
<form method="post" action="update.php">
<select size="1" id="choice" name="value">
<?php
while($row = mysqli_fetch_array($query)) {
$id = $row['id'];
$number = $row['number'];
>?
<option value = "<?php echo ($id) ?>"><?php echo "ID=" . ($id) . " - #" . ($number) . ""?></option>
<?php
}
mysqli_close($db_conn);
?>
</select>
<input name="submit" type="submit" value="Submit">
</form>
update.php:
<?php
if (isset($_POST['submit'])) {
$chosen_id = $_POST['id'];
}
?>
`
I've only posted the code handling the select option and the post part...
Why is the $chosen_id variable always 0 ?
The while loop works, and fill's the variable, as this is tested with echo command inside the option line
Any help is much appreciated...

$_POST['id'] and <select size="1" id="choice" name="value">
Use $_POST['value']

You are trying to print the wrong key
if you trying to get the value of form_location_id
$chosen_id = $_POST['form_location_id'];
And if you trying to get the value of choice
$chosen_id = $_POST['value'];
This is why when you post a form to php it use html name attribute as key to assign the value to $_POST Array

I'd change Update.php to
<?php
if (isset($_POST['value'])) {
$chosen_id = $_POST['value'];
}
?>

You need to use the form_location_id to get the required value. You are using the wrong key to access the data. You need to use the name of the input. In this case, the input is the select and the name of that is form_location_id. So, you need to do this.
$value = $_POST['form_location_id'];
Try it out and do let me know if it worked out for you.

Thanks for everyone posting idea's - i've actually got it working, the code was actually working, only error was a misplaced tag, which was placed inside a tag, when placed outside this tag it works ;)

Problems with simple Javascript form validation

I'm working on putting together a multi-page set of forms that are interlinked by a session and use Javascript for form validation whenever each section is submitted. My first form is working fantastically. My second one, not so much. It's a dynamically created form that depends on the previous form for the number of drop down boxes (between 1 and 20) that are populated from a database. I will spare you all the query code to create the array that fills in my dropdowns and get straight to the form validation script and then the form itself.
The validation script from the head without the <script> tags:
function validateForm()
{
var count = <?php echo $_SESSION['credits']; ?>;
var i = 1;
for (i = 1; i <= count; i++)
{
var j = i;
for (j++; j <= count; j++)
{
var a = document.forms["addcredits"]["creditdropdown_"+i].value;
var b = document.forms["addcredits"]["creditdropdown_"+j].value;
if ((a == b) && (a != 0))
{
alert('Cannot select the same writer more than once.');
return false;
}
}
}
var foundOne = false;
var h = 1;
while (h <= count && !foundOne)
{
if (document.forms["addcredits"]["creditdropdown_"+h].value != 0)
{
foundOne = true;
}
h++;
}
if (!foundOne)
{
alert('You must select at least one writer to credit.');
return false;
}
}
The form code:
<form id="addcredits" class="appintro" method="post" action="recordcheck.php" onsubmit="return validateForm()">
<div class="form_description">
<h2>Song Title Submission</h2>
<p>Step 2: Select writing credits for song title</p>
</div>
<ul>
<?php
for ($i = 1; $i <= $_SESSION['credits']; $i++)
{ ?>
<li id="li_<?php echo $i; ?>">
<label class="description" for="creditdropdown_<?php echo $i; ?>">Song Credit #<?php echo $i; ?>:</label>
<select "creditdropdown_<?php echo $i; ?>">
<option value="0">Select a writer</option>
<?php foreach ($writers as $key => $writer) { ?>
<option value="<?php echo $key; ?>"><?php echo $writer; ?></option>
<?php } ?>
</select>
<p class="guidelines" id="guidelines_<?php echo $i; ?>">Writer not found in database? Add them here.</p>
</li>
<?php } ?>
<li id="buttons">
<input type="hidden" name="form_id" value="2">
<input type="hidden" name="submit" value="1">
<input id="nextButton" class="button_text" type="submit" name="submit" value="Next">
</li>
</ul>
</form>
Neither of the alert windows pop-up when I click Next and if I stick random debugg-ing alert windows into the validation script, the only ones I was able to get to show up were the ones I stuck at the beginning of the code. The furthest I got them to work was the first assignment of a after that, not even the debuggers would show. I'm assuming I'm doing something incorrectly in the assignment or there is something wrong with the values in the dropdown? Either way, I'm stumped.

I'd say the problem is in your HTML:
<select "creditdropdown_<?php echo $i; ?>">
You are specifying, presumably, the name attribute but you forgot the name= part. Try this:
<select name="creditdropdown_<?php echo $i; ?>">
In your validation function when you get to this line:
var a = document.forms["addcredits"]["creditdropdown_"+i].value;
You'll find you have an error because document.forms["addcredits"]["creditdropdown_"+i] is undefined because of the missing name= in the html and undefined can't have a .value property.
(By the way, debugging with alert() is useful very occasionally, but you're much better off using the built in developer tools in Chrome - just hit (I think) ctrl-shift-J to get the JS console - or download FireBug for Firefox. Or use IE's developer toolbar. Use console.log() instead of alert(), and/or step through the code until you find the problem.)