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how to keep correct checkbox value check array after submit/document.form[0] submit or refresh page SOLVED

<script>
function refreshPage()
{
document.forms[0].submit();
}
</script>
<?php
foreach($values as $value)
{
?>
<input type = "checkbox" name = "food[]" value = "<?php echo($value['dinner']);?>">
<?php
}
?>
<button name = "pass" onClick = "refreshPage()">refresh</button>
so I have n amount of check box (depending on how many values user has)I want to keep them the check box check after the page refresh (by clicking on refresh button). I tried searching and saw similar form but the answer did not work for me.
For example the most popular answer was:
<input type="checkbox" name="txtCheck" value="your value"
<?php if(isset($_POST['txtCheck']))
echo "checked='checked'"; ?>
/><br />
this had two problem after i hit the refresh button value would not be save until i hit it twice to save(i want it to save after one click. Also sometime it save after one click but if i hit the refresh 3 time values are lost I want it to be save no matter how many time user refresh)
the second problem was it check all of the box would be check. I only want to keep the one user has check to be save.
I looked at various other possible solution but nothing worked
so if you could please help much would be appreciated. Also I need the value to be kept. I am using the checkbox value somewhere else
Edit something like this but for my array food[]. This only works for invidiual values
<input type="checkbox" name="small" class="checkbox" <?php if
($_POST['small']) echo 'checked'; ?> /> Small

Targetting button outside of a form in action.php?

I currently have 2 forms, which adds data to my database. They both work, but they each have their own submit button - and I only want one submit button to collect the data from all the forms.
I found some javascript that should set the deal;
$("#submitbutton").click(function(){
$("#form1").submit();
$("#form2").submit();
});
The button I'm targetting is outside of both forms and looks like this:
<input id="submitbutton" type="button" value="add">
I'm pretty sure the reason why it doesn't work, is because of the way my php is written. I'm targetting the submit button in each form to excecute the php.
You can see the forms and php below.
One of the forms allows you to upload a picture;
<form id="form1" action="addimage.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" />
<br/><br/>
<input type="submit" name="sumit" value="Upload" />
</form>
The action file contains this php;
<?php
if(isset($_POST['sumit']))
{
if(getimagesize($_FILES['image']['tmp_name'])== FALSE)
{
echo "Please select an image.";
}
else
{
$image= addslashes($_FILES['image']['tmp_name']);
$name= addslashes($_FILES['image']['name']);
$image= file_get_contents($image);
$image= base64_encode($image);
saveimage($name,$image);
}
}
displayimage();
function saveimage($name,$image)
{
$con=mysql_connect("localhost","root","");
mysql_select_db("ssdb",$con);
$qry="insert into pictures (name,image) values ('$name','$image')";
$result=mysql_query($qry,$con);
if($result)
{
//echo "<br/>Image uploaded.";
}
else
{
//echo "<br/>Image not uploaded.";
}
}
function displayimage()
{
$con=mysql_connect("localhost","root","");
mysql_select_db("ssdb",$con);
$qry="select * from pictures";
$result=mysql_query($qry,$con);
while($row = mysql_fetch_array($result))
{
echo '<img height="300" width="300" src="data:image;base64,'.$row[2].' "> ';
}
mysql_close($con);
}
?>
The other form lets you choose between multiple categories collected from my database;
<form id="form2" action="checkbox.php" method="post">
<label for="Category">Category</label>
<br />
<!-- iterate through the WHILE LOOP -->
<?php while($row = mysqli_fetch_array($result_category)): ?>
<!-- Echo out values {id} and {name} -->
<input type="checkbox" name="category[]" value=" <?php echo $row['id']; ?> "><?php echo $row['name'] . '<br />'; ?>
<?php endwhile; ?>
<input type="submit" name="Submit" value="Submit" class="btn btn-default"/>
</form>
And has the following php;
<?php
include("config.php");
$checkbox = $_POST['category'];
if($_POST["Submit"]=="Submit")
{
for ($i=0; $i<sizeof($checkbox);$i++) {
$query = "INSERT INTO placecategory (category_id) VALUES ('".$checkbox[$i]."')";
mysql_query($query) or die(mysql_error());
}
echo "Category is inserted";
}
?>
I've tried targetting the new button I made that should excecute the javascript, but it doesn't seem to work because that button is out of the form.
Is there a way to target the button outside of the form so the php excecutes when that is clicked? Or how can I rewrite this?
Any help is appreciated!

Not sure I understand completely what you're trying to do but try this with HTML5 add form="form1" and form="form2" for your second one and let me know if this works.
<form id="form1" action="addimage.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" />
<br/><br/>
<input type="submit" name="sumit" value="Upload" />
</form>
<input type="submit" name="sumit" value="Upload" form="form1" />

Taking in account your complementary comment, then your proposed javascript sequence would work, assumed you have:
suppressed buttons such as <input type="submit"... /> from both forms
added the new button <button name="submitbutton">...</button> outside of the forms
modified the PHP parts accordingly, referencing $_POST['submitbutton'] instead of sumit and Submit respectively
You didn't report how it currently don't work, so the above steps target only the changes precisely related to the fact you replace two in-form buttons by a unique out-form one.
But something may turn wrong elsewhere. I notably noticed the non-usual way (at least for me) you get and process the image. Also you must ensure that your javascript part have really been executed.
Let me know if it doesn't work yet, than adding precise description of what turns wrong.
Edit, based on your provided comments and live preview.
Fully testing is not really possible because the server PHP part is out of reach, bue we can use Firebug to debug JS part.
So adding a breakpoint we can observe that, as soon as $("#form1").submit(); has been executed, the server returns a new page with the "Place has been added!" message.
In the other hand, without any breakpoint set, the server returns returns a new page with the "Categorie inserted!" message.
Though the OP didn't show the addingplace.php part of the current live preview, and comparing with the checkbox.php part, we can guess that:
In the reduced execution part, the first step addingplace.php did work as expected.
What we observe while whole execution merely means that all three parts have worked as expected, but each one having its returned message overwritten by the next one, but for the last one.
In other terms, when you comment "it only seems to submit the last form", this is a false impression based on what you only can see.
To ensure this is true or not you should control what is really updated or not in your database.
Let me know.
That said, it must be noted that this illustrates how the couple server-browser works in those circumstances: as already pointed by #David in a comment under the OP, to submit a form causes the browser to immediately receive a new page which overwrites the current one.
In the current example, it works because there are only few forms, and all three are submitted in a very reduced time, almost instantly: so the 3rd submit() can get executed before the 1st one's returned page comes.
So the recommended way to achieve the whole work in your example is merely to use only one form, with its unique submit button. BTW I wonder why you wanted to have this more complicated structure with three forms: what is the expected benefit?
Last point, outside of the precise issue you reported: you should pay attention to how you're coding. There is a lot of inconstencies in the HTML part, e.g.: a <html><body> part inside the already existing <body>; an exotic <br></br>; also you kept a supplemental button in the 1st form.

php form wont submit with javascript

I have a dashboard page, and for the easiest solution I've made the entire page a form (as there are several drop downs scattered across the whole page). I want to implement a feature that can submit the form every 30 minutes, be it with JavaScript, jQuery or anything else, but when I've tried it just refuses to execute the code, so I tried going back to something basi such as submitting the form when the drop-down is changed via "OnChange".
Here is an example snippet of a seperate page with some code from my dashboard page. This in itself should be working but I just can't see why it won't execute the code, maybe I'm missing something obvious? Can you help me fix this:
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript">
function submit_my_form(myfield)
{
myfield.form.submit();
}
function submitForm() {
document.getElementById("branchForm").submit();
}
</script>
</head>
<body>
<?php
$branch_array = array(
array(1,"BRANCH 1",1, "http://BRANCH.BRANCH1:1"),
array(2,"BRANCH 2",1, ""),
array(3,"BRANCH 3",3, "http://branch3:3"),
);
?>
<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="POST" name="branchForm" id="branchForm">
<select name="query" id="query" class="select" onChange="submit_my_form(this);">
<?php
foreach ($branch_array as $x) { // populate select box with branches available from array
echo '<option value ="'.$x[0].'"';
if (isset($_POST['query']) && $_POST['query'] == $x[0])
{
echo ' selected="selected" >'.$x[0].". ".$x[1].'</option>';
} else {
echo ' option="'.$x[0].". ".$x[1].'">'.$x[0].". ".$x[1].'</option>';
}
}
?>
</select>
<input type="submit" name="submit" value="Refresh" class="btnRefresh" />
<input type="button" value="go" name="click" onClick="submitForm();" />
</form>
<?php echo "<br/>Output: {$_POST['query']}"; ?>
</body>
</html>
Nothing happens when the drop-down is changed, and nothing happens when the "go" button is pressed.

Since you don't mind using jQuery, here you go:
var form = $("#formId");
$("select[name=selectField]").on("change", function(e){
form.trigger("submit");
});
http://jsfiddle.net/zt8zz1j5/
All it essentially does is listen for change on the select element, and simply triggers the submit event on the form. Of course you'll need to change the names and IDs to fit your code, but I'm sure you can handle that :), and remember to include jQuery in the document.
And for the button it's the same thing. You listen for the 'click' event and then trigger 'submit' on the form.

Cant submit form

Basically I've got a form with 5 radio buttons. No submit button. I want the form to run when the user clicks on any of the radio buttons, so this is what I did.
<input id="5" type="radio" name="star" onchange="this.form.submit();" <?php if ($row["star"] =="5") echo "checked";?> value="5"/>
a querystring is required for the form so I'm using a form action like this
<form name="submit" action="http://example.com/page.php?id=<?php echo $urlid;?>&title=<?php echo $title;?>" method="POST">
and my php is
if ($_POST["submit"]) {
$rating = $_POST['star'];
$title = $_GET['title'];
$verification = ($_GET['verification']);
} else {
//run the page like usual
}
After testing, I found that onclick, it runs the form action, but on the php side, it goes to "else" where is runs the page like usual instead. Any ideas?

Your PHP is checking if $_POST['submit'] contains a value. Your form does not contain a form element with the attribute name="submit", so therefore it fails and moves straight to the else statement.
If you want to check if the form was posted then you should instead check for:
if (!empty($_POST)) {}
or
if ($_SERVER['REQUEST_METHOD'] == 'POST') {}

The form element seems to have invalid attributes, missing a quote and space.
It's generally easier to write a little more code, and keep it clearer
<?php
$url = "http://example.com/page.php?id=". $urlid ."&title=". $title;
?>
<form name="submit" action="<?php echo $url; ?>" method="POST">

Since you are checking with -
if ($_POST["submit"]) { // this checks whether there is any item named 'submit' inside the POST or not
} else {
//run the page like usual
}
The easiest would be to put a hidden item with name submit so that the check validates to true-
<form .. >
....
<input type='hidden' name='submit' value='submit' />
</form>

Pass php session variables to javascript and back to php

I have the following code which basically the user enters a number and presses the search button which makes PHP lookup the value and return any records from the MySQL database (approval1.php?go). This works as expected so I left that code out. What I am trying to do is after the user reviews the records they can be approved by clicking the link. That link runs some javascript which takes the number entered from the PHP session. Once the javascript takes the user to the next page which also works I want it to pass that value back to php so that it may be displayed in the next form. How can this be done and also is there a simpler way to do this? Everything I find is either from php to javascript or the other way around. I can't seem to find anything that shows how to get the session variable back.
<!DOCTYPE HTML>
<?php
session_start();
?>
<html>
<head><title>Title</title></head>
</head>
<p><body>
<h3>Search:</h3>
<p>Please enter the reference number of the record you would like to review:</p>
<form method="post" action="approval1.php?go" id="searchform">
<input type="number" name="REF_NO">
<input type="submit" name="submit" value="Search">
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
function approve() {
passSession("<? echo $_SESSION['REF_NO']; ?>");
$.get("approval2.php");
return false;
}
</script>
Approve Request
</form>
</body>
</html>
</p>

I think what you're looking for is just this: Pass the approval as a GET variable.
Approve Request
Then, in the approval2.php page, receive those GET variables:
$ref_no = $_GET['ref_no'];
$approval = $_GET['approval'];

<?php
session_start();
<?php
<script type="text/javascript">
var vars=passSession("<? echo $_SESSION['session_id']; ?>");
</script>

You can set a hidden input field:
<form method="post" action="approval1.php?go" id="searchform">
<input type="number" name="REF_NO">
<input type="hidden" name="php_sess" value="<?php echo $_SESSION['REF_NO']; ?>">
<input type="submit" name="submit" value="Search">
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
function approve() {
passSession("<? echo $_SESSION['REF_NO']; ?>");
$.get("approval2.php");
return false;
}
</script>
Approve Request
</form>
So when you submit a form, $_REQUEST['php_sess'] will be populated in php.

I don't think you should use a $_SESSION variable, here. When you lookup something at approval1.php?go and you return some DB records, also return the id REF_NO of that record (e.g. use JSON to send two things at a time). Then, on approval, send this id again, simply by GET. To me, this much more consistent and prevents you from failure e.g. when users browse with two tabs, etc.
Example for approval1.php
<?php
$result = json_encode(array(
"REF_NO" => $id,
"content" => $content
);
echo $result;
?>
Then, access the JSON in Javascript. Use the REF_NO node to prepare your approval link.