We Keep Coding

Use array to post html form with every aray value ony by one

I am wondering if the folowing is possible.
I fetch a list of names as an array from a SQL database. I need all those names to be posted with a html form one by one. This action should be activated with one button. When the button is clicked the names should be posted one by one untill all names are posted, then stop. So probebly jquery or javascript is needed but that is new for me. I have been searching but I can not find anything that can help me accomplisch this.
I am sorry for asking this question and my language (english is not my main language) but I don't know if this is even possible and I cant find any corresponding topics while researching..
PS: I can not use Ajax for the post !!
Example to get the names:
$stmt = $mysqli->prepare("SELECT username FROM example WHERE examplefield = 1");
$stmt->execute();
$result = $stmt->get_result(); //only works when nd_mysli is set on the server!
while ($rowid = $result->fetch_assoc())
{
$arrayusername[] = $rowid['username'];
}
I need all the names from the $arrayusername[] to be posted with below form one by one by pressing the following button
<input type="button" value="Post all names one by one"
onClick="sendallvalues('???') "class="example_c" />
// The button should do the following post name 1, end. post name 2 end, post name 3 end. stop script when all names are posted.
<form method="post" target="_example" action="https://www.example.nl">
<input type="hidden" value="<?= $arrayusername[] ?>" name="postvalue" >
</form>
// should be hidden to the user and is only ment for the name atrribute to have a place ! All the stuff needs to happen by pressing that one button !
<script>
function sendallvalues(???) {
//I have no Idea where to begin to make this happen.. But it should post the form one by one with one value at the time untill all names are posted.
}
</script>

Maybe you can explain more detailed what you need.
The value attribute should have one name and you create different input fields.
<form method="post" target="_example" action="https://www.example.nl">
<input type="hidden" value="<?= $arrayusername[] ?>" name="postvalue" >
<input class="example_s" type="submit" value="Post name">
</form>
should be
<?php
$arrayusername = array('Name1','Name2');
echo '<form method="post" target="_example" action="https://www.example.nl">';
foreach ($arrayusername as $key => $value) {
echo '<input type="hidden" value="' . $value .'" name="name-' . $key . '" >';
}
echo'<input class="example_s" type="submit" value="Post name">';
echo '</form>'
?>

PHP echo javascript to change iframe source

I have a text input search form for support tickets. If the user inputs a ticket id (checked using a PHP if statement), the background color of the search box is changed to yellow (using a PHP echo) to highlight the search text entered. This part works.
I would like to also change the source value of an iframe as part of the PHP if condition like this:
<td align="left">
<input type="text" name="ticket" id="ticket" placeholder="Ticket" value="<?php echo $this->state->ticket;?>" class="input-mini search-query" onchange="document.adminForm.submit();"
<?php if ($this->state->ticket != "")
{
echo 'style="background-color: yellow;"';
echo '<script language="javascript">';
echo 'document.getElementById("ticketIframe").contentWindow.document.location.href="http://myurl.com"';
echo '</script>';
}?>/>
</td>
The javascript function (document.getElementById...) displays on the page rather than executing. I tried a simple javascript alert, but got the same result.
Suggestions?

Need to close the input tag before you insert script tag. You simply are generating invalid html that looks like:
<input <script></script> />
You get unexpected results when html is invalid

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.

Javascript Validation On Button Not Working

Hello I have got a newsletter form. Not too sure what's wrong with the validation but when a user clicks the subscribe button without entering any values into the fields it submits anyway and takes you to the external url. Is there a way of making it so that both fields have to be filled in order for the subscribe button to work?

Just include the clause required="true" into the inputs that must be filled in the html, like this:
print '<tr><td>' . $myNameLabel . '</td><td><input required = "true" class="modns inputbox ' . $mod_class_suffix . '" type="text" name="m_name'.$unique_id.'" size="' . $nameWidth . '"';

ajax display text on submit

I have a simple form with a submit button (below). I am trying to let the user type in the text box then when he/she clicks submit the page will refresh and echo out what they typed in a div. The data the user types is stored in a database before being echoed. The problem is that when I click submit, the input doesnt show immediatly. I have to click refresh for it to show and when I do my browser gives me a popup (safari) asking to resend the data. This will result in duplicate data inserted in the DB. I have a feeling I need to use javascript and I could also make it more elegant with a fadeIn, but I dont know how to do that. I guess I'm asking if there's a way to use javascript to take a user's text and insert it into a mysql DB and also display it after submit is clicked all on 1 or 0 (prefereably) refreshes. thanks
Here's my code:
<form method='POST' action='index.php'>
<input type='text' name='text' id='text'>
<input type ='submit' value='submit' name='submit'>
</form>
<?php
$input=$_POST['text'];
$put=mysql_query("INSERT INTO table VALUES ('$input')");
echo "<div id='area'>";
//i connect to the DB and echo out the data here
echo "</div>";
?>

I would put the php statements before you're actual html code and would modify you're code like this
<?php
if (isset($_POST['text']))
{
$input = mysql_real_escape_string($_POST['text']);
$put=mysql_query("INSERT INTO table VALUES ('$input')"); //At this point the info has been put inside the DB
echo "<div id='area'>";
//i connect to the DB and echo out the data here
echo mysql_query("SELECT * FROM table");
echo "</div>";
}
?>
<form method='POST' action='index.php'>
<input type='text' name='text' id='text'>
<input type ='submit' value='submit' name='submit'>
</form>
The reason why you don't see it is that the HTML is loaded before you php I think. So I would do a php page where all the sql treatement would be done and once that is done recal you index.php and in there query you're information from the database.

Setting aside SQL injection attacks your code is vulnerable with, the standard practice is to respond to the POST with a redirect back to where the form was. In your case, it will be the page which runs SELECT from table.

No need to use AJAX here. Just make sure that you do one of the following:
Make sure you SELECT from the DB after you have INSERTed the data.
Better, is that if ($_SERVER['REQUEST_METHOD'] == 'POST'), then rather than performing an extra SQL query, just display the POSTed value since you already know it.
A side note on your code sample. If you don't have magic_quotes enabled, it's susceptible to SQL injection, so make sure you properly escape user input before using it in queries.