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prevent result, show through AJAX, after reload browser

I have a insert query through ajax. It is working correctly. But when I reload browser then result disappears from div section and if I insert form through ajax again then result is showing.
I have a file first.php (in which, form is present), a AJAX code and a firstcall.php where query will be execute.
My first.php (html form) is:
<form class="reservation-form mb-0" action="" method="post" autocomplete="off">
<input name="name1" id="name1" class="form-control" type="text" placeholder="Enter Name" required aria-required="true">
<input name="age" id="age" class="form-control" required type="number" placeholder="Enter Age" aria-required="true">
<input type="checkbox" id="checkbox" class="checkbox1" name="namec[]" value="<?php echo $value['id']; ?>" >
<input type="button" class="pull-right btn btn-warning" value="Submit" id="submit">
</form>
Here data should be display:
<div class="col-md-5">
<div class="panel panel-primary" id="showdata">
<!-- Here is the results, but when reload browser then result disapper-->
</div>
</div>
AJAX is:
<script type="text/javascript">
$(document).ready(function(){
$("#submit").click(function(){
var name1 = $("#name1").val();
var age = $("#age").val();
var chkArray=[];
$('.checkbox1:checked').each( function() {
chkArray.push($(this).val());
} );
var selected;
selected = chkArray.join(',') ;
if(selected.length > 1){
$.ajax( {
url:'firstcall.php',
type:'POST',
data:{name1: name1,age: age,namec: chkArray},
}).done(function(data){
$("#showdata").html(data);
});
}
else{
alert("Please at least one of the checkbox");
}
});
});
</script>
firstcall.php is:
<div class="panel panel-primary" id="showdata">
<?php
foreach($_POST['namec'] as $selected){
echo $selected;
$_SESSION['name1']=$_POST["name1"];
$_SESSION['age']=$_POST["age"];
echo $name1=$_SESSION['name1'];
echo $age=$_SESSION['age'];
$query=mysql_query("insert into patient_details (p_name,p_age,g_number) values ('$name1','$age','$selected')") or die(mysql_error());
}
?>

First of all fix your query to use MySQLi, instead of MySQL, check this or the PHP manual
Also don't ever add direct $_POST or $_GET variables into your mysql query, filter them first using mysqli_real_escape.
$name1 = mysqli_real_escape($link, $_POST["name1"]);
$age = mysqli_real_escape($link, $_POST["age"]);
After that, to show the data when the page reloads, you need to load the data with the page, easiest way to do that is just add in your HTML PHP tags with an echo command inside, adding your variables.

If I understand your question correctly, you want the Ajax result to also show on page load?
Right now you only execute the JS after a click (so on page load/relaod you will just see the html), you might want to execute it after page load aswell (so execute the script without the .click)
You could create a function once, and call it when the page is ready and on click.

Only text values being posted to php file but not files

I have the following code for an HTML form:
<form id="idForm" method="post" enctype="multipart/form-data">
<input type="text" name="first" value="Bob" />
<input type="text" name="middle" value="James" />
<input type="text" name="last" value="Smith" />
<input name="image" type="file" />
<input type="submit" name="submit" value="Submit" />
</form>
and the following Javascript code on the same page where the form is located. Also, this code is written below the form code:
$("#idForm").submit(function() {
alert("hi");
var url = "submit.php";
$.ajax({
type: "POST",
url: url,
data: $("#idForm").serialize(),
success: function(data) {
alert(data);
}
});
return false;
});
And here is the PHP code that I'm using to see if the form data is being posted:
if (isset($_FILES['image'])) {
$image = $_FILES['image'];
echo $image;
}
if (isset($_POST['first'])) {
$name = $_POST['first'];
echo $name;
}
With the above code, it seems as though the attached file is not being sent to the PHP file. Only the values are being sent, since they are echoed when I inspected the element using Google Chrome dev tool.
I want to be able to post both text value and files to the PHP file for processing.
Why am I getting this problem? How do I fix this?
Also, as part of the solution, I want to prevent the page from being refreshed when the submit button is clicked.
Thanks much!

Unfortunately input[type='file'] cannot be submitted by .serialize(), you have to manipulate by formData.
You will find a very good formData example in here

Get value from input text in php without pressing Enter

Hay I'm new to php and I have made php code like this :
<?php
session_start();
echo 'Hellow Hisoka';
?>
<form name="form" method="post">
<div class="col-sm-3">
<label class:>Nama :</label>
</div>
<div class="col-sm-9">
<input type="text" name="nama_tamu" id='nama_tamu' class="form-control" placeholder="Nama Lengkap">
<?php
$myValue = $_POST['nama_tamu'];
?>
</div>
<br/><br/>
<?php
echo $myValue;
?>
</form>
When I want to show the echo message, I need to hit enter on my keyboard first in order to get the value of $_POST['nama_tamu'];. My question is can I get the value of nama_tamu input without pressing enter, or maybe without using POST or GET and then assign it to $myvalue?

You will need to use Javascript. You can use the Jquery events :
<script>
$( "#nama_tamu" ).keyup(function() {
alert( $this.val() );// alerting the value of the input field
});
</script>

Web development is all about communication. In this case, communication between two (2) parties, over the HTTP protocol:
The Server - This party is responsible for serving pages.
The Client - This party requests pages from the Server, and displays them to the user. In most cases, the client is a web browser.
Each side's programming, refers to code which runs at the specific machine, the server's or the client's.
You cannot get values without submitting for the user has not entered any yet. PHP is a server side language. To get values before submit and do certain actions with them you will need javascript (a client side programming language).
The simplest method to get a value is using the getElementById().
var something = document.getElementById('someid');
<input type="text" name="something" id="someid">
You can also use jQuery:
var something = $('#someid').val();
Conclusion
The simple answer to your question is: This is not possible.
Why not? I hear you asking. Because PHP doesn't know the values of your form before you send the form to your webserver.

Use keyup().
function check(id)
{
document.getElementById("result").innerHTML = id;
}
<input type="text" name="test" id="test" onkeyup="check(this.value);">
Your value: <span id="result"> </span>
$(document).ready(function() {
$("#check").keyup(function(){
alert($(this).val());
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="check" >

For this purpose you should use .keyup function/event. Following are the snippet :
$(document).ready(function () {
$("#nama_tamu").keyup(function(){
$("#enterdata").html($("#nama_tamu").val());
$.ajax({
type: 'POST',
url: "getdata.php",
data: "nama_tamu="+$("#nama_tamu").val(),
success: function(res)
{
$("#outputdata").html(res);
}
});
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<?php
session_start();
echo 'Hellow Hisoka';
?>
<form name="form" method="post">
<div class="col-sm-3">
<label class:>Nama :</label>
</div>
<div class="col-sm-9">
<input type="text" name="nama_tamu" id='nama_tamu' class="form-control" placeholder="Nama Lengkap">
<?php
$myValue = $_POST['nama_tamu'];
?>
<br> You press following character:<div id="enterdata"> </div>
</div>
<br/><br/>
<div id="outputdata"></div>
<?php
echo $myValue;
?>
</form>
Also create one file for the required output.
Now in getdata.php file
echo $nama_tamu=$_POST['nama_tamu'];

Send multiple headers or multiple submitting for the form?

When the user inserts two numbers (from-to), I want to send (((full path))) for each number.
how would I accomplish this? Is there a way in PHP, javascript or any other language. Would I send many headers or multiple submissions for the form?
P.S:
The (go.php) page that receive and process each number individually.. I can't put hands on it to make changes, I only must send individual numbers to it because that's how the other page is coded.
This what i've tried:
<form action="test.php" method="post">
<input type="text" name="first">
<input type="text" name="second">
<input type="submit" name="submit">
</form>
<?php
$f=$_POST['first'];
$s=$_POST['second'];
for($i=$f; $i<=$s; $i++){
header('location:go.php?f='.$i);
}
?>

you can not send header twice you can do this like below
<?php
$f=$_POST['first'];
$s=$_POST['second'];
header('location:go.php?first='.$i.'&second='.$s);
?>
go.php
you can catch these two variables using get method
<?php
$first=$_GET['first'];
$second=$_GET['second'];
//rest of the code
?>

Why not use javascript to submit so many times:
// supposing you have jQuery
for (var i = Number($(':input[name=first]').val()),
end = Number($(':input[name=second]').val());
i <= end; i++) {
$.get('go.php', { f: i }, function (response){
// do something with response
});
}

You can't do it via simple single script. You can do it via multithreading or curl. I am giving you example of curl:
one.php
<form action="test.php" method="post">
<input type="text" name="first">
<input type="text" name="second">
<input type="submit" name="submit">
</form>
<?php
$f=$_POST['first'];
$s=$_POST['second'];
for($i=$f; $i<=$s; $i++){
//write curl code to execute two.php with url : http:// yoursite.com/two.php?f=$i
}
?>
two.php
<?php
header('location:go.php?f='.$_REQEST['f']);
?>
Hope this explanation help you. Best of Luck.

jQuery detecting if php $_POST isset

I'm new to jQuery, and just wondering if there's a way to detect if a form has been submitted on page load, and if so parsing some code.
I have:
<form action="/login/register" method="post">
<input type="text" value="Username" name="regName"/>
<input type="password" value="Password" name="regPass"/>
<input type="submit" value="Register"/>
</form>
and:
<?php
if (isset($_POST['regName']))
{
//check inputs and add account
}
?>
But i was wondering if i could check if the form had been submitted via jQuery?
Thanks.

Just have PHP set a javascript variable to true if a POST submission was made. Then jQuery just needs to check its value.
<script type="text/Javascript">
var was_posted = <?php echo ('POST' === $_SERVER['REQUEST_METHOD']) ? 'true' : 'false'; ?>
</script>