PHP: How to post hidden field with style='display:none' - php

This doesn't get posted if it's hidden like this, how do i get around it?
<input name="Pid" style="display:none"/>
and PHP.
$Pid = preg_replace ('#[^0-9 ]#i', '', $_POST['Pid']);

Kindly use it as:
<input name="Pid" type="hidden" />

You can use hidden field simply setting type as hidden
<input type="hidden" value="" name="pid"/>

If you really need your input type to remain, like text or checkbox, use a <div style='display:none;'></div> tag to hide your input object.
Sometimes, this is required if your submit code does not include checks like
if (isset($_POST['Your_Object_name'])) {...}
Sometimes you don't want to show them, but neither do you want your code to issue error messages.
Your code will then be as follows
<div style='display:none;'>
<input name="Pid" type="text" />
</div>
Et voila

Related

Is it possible to use placeholders on <input type="submit">?

I'm currently working on at the displaying of information from a database. I was making a summary site where you can only see the important things of a table. After that i made the first element as an <input type="submit"> in a <form>, so u can click it and come to the detail site. My problem is now: The value of this input type has to be my ID, so i can query correctly on me detail site. I was wondering if it is possible to use something like a placeholder, so that the ID is the value, but on the input type is written other text.
My input:
<form method="post" action="Details.php">
<input type="submit" placeholder = "test" name="Overview" onclick="parent.location='Details.php'" value="<?php echo $data[$i1]; ?>">
</form>
How it currently looks
I want it that the input type still has the same value, but is displaying on the website something else like "test".
Greetings!
No, but buttons can have different values and labels.
<button name="foo" value="bar">baz</button>
Since you are using a form-tag per row, you can add a hidden input-field in the form and set the value of the submit-button to whatever you like.
<form method="post" action="Details.php">
<input type="hidden" name="id" value="<?php echo $data[$i1]; ?>" />
<input type="submit" name="Overview" value="test" />
</form>

How to send a get form wthout losing get parameters

I have a form located at a url containing get parameters,my form is also using this method.When the form is submitted it rewrites the previos get parameters.
Is there a simple way to rewrite only my form parameters?
I have in mind a Javascript solution ,however I want to know if there is a simpler way?Using HTML/PHP perhaps?
As far as I know, u u are not interested in using JS, then using form's hidden element is only way u have like this-
<form action="demo_form.asp">
First name: <input type="text" name="fname"><br>
<input type="hidden" name="country" value="Norway">
<input type="submit" value="Submit">
</form>
<p>Notice that the hidden field above is not shown to a user.</p>
The question is how u can use it with PHP, right?
The solution is here-
//In PHP
if( isset($_GET['fromPerson']) )
{
echo $fromPerson;
}
So combined HTML and PHP code will be like this (assuming a get element from prevous page is named fromPerson)-
<form action="demo_form.asp">
First name: <input type="text" name="fname"><br>
<?php
if( isset($_GET['fromPerson']) )
{
echo '<input type="hidden" name="country" value=".$_POST['fromPerson'].">';
}
?>
<input type="submit" value="Submit">
</form>
Lets say you get a parameter p1 from a get request, it should look like this:
http://server.com/?p1=123
In your form, you can add hidden fields that would have the same effect when you submit, like this:
<form method="GET">
<input type="hidden" value="<?php echo $_GET["p1"]; ?>" name="p1">
</form>
That way you can resend the variables as many times as you need.
I'm not sure I understand your question... Can you post your code?
I assume you mean something like this?
in index.php
<input type="hidden" name="id" value="<?php echo $id; ?>" />
in return.php
Edit

Is POST method used to send user defined inputs only

Well, I have 2 php files. The first one user form.php has an html form to collect user inputs and send those data to other php file mysql_insert.php via POST method.
The problem is that along with 6 user inputs, I also have to transfer one more variable count(as discussed in code below) which is not input from user.
I tired following approach:
user form.php--->
<?php
$count=file_get_contents("cnt.txt","r");
$count=$count+1;
echo"Welcome!<br/>You are student number $count.<br/>";
?>
<html>
<p>Fill in the following information to save your marksheet to the database:</p><br/>
<form action="mysql_insert.php" method="POST">
Name:<input type="text" name="name" value=""/><br/><br/>
Marks(out of 100)<br/>
Subject 1:<input type="text" name="sub1" value=""/><br/>
Subject 2:<input type="text" name="sub2" value=""/><br/>
Subject 3:<input type="text" name="sub3" value=""/><br/>
Subject 4:<input type="text" name="sub4" value=""/><br/>
Subject 5:<input type="text" name="sub5" value=""/><br/><br/>
<p name="count" value="$count"></p>
<input type="submit" value="Submit"/>
</form>
</html>
I simply named an empty P element as count and set its value as $count to send it along with input variables(correct me if there is something wrong here, I am very novice to php).
And on the receiving end I used following code:
mysql_insert.php--->
<?php
require("connect.php");
$name=$_POST['name'];
$s1=$_POST['sub1'];
$s2=$_POST['sub2'];
$s3=$_POST['sub3'];
$s4=$_POST['sub4'];
$s5=$_POST['sub5'];
$count=$_POST['count'];
.
.
.
?>
Now on the line $count=$_POST['count']; the browser is throwing the error:
Notice: Undefined index: count in C:\xampp\htdocs\Vikas-117-PHP\level 3\mysql_insert.php on line 10
It seems the count is not being posted to this file.
Please guide me where I am wrong.
P.S.: I can of course use the file_get_contents() in the mysql_insert.php and get the count value directly in this file, but I am considering that way as my last option.
So please help if the non user-input variable can be posted via forms???
Thanks a million!!!
you have to put your $count variable in a hidden field in your form.. The data which are put in form inputs only get posted in form submit..
<input type="hidden" name="count" value="<?php echo $count ?>">
<p name="count" value="$count"></p>
you can not use directly any html tag for post data only use html form fields for posting data.
for use html tag values post you need to use js/ajax/php
<p name="count"><?php echo $count;?></p>
or better you use
<textarea name="count"><?php echo $count;?></textarea>
or for data not showing use hidden field
and get by name this p value or use class or id for get and post data using ajax
<p name="count" value="$count"></p>
This is not an input element so you cant post directly. You can however use jquery to get the value and POST using js.
You need to have the value inside input type text or hidden if you want to do the way you are doing now.
Change this line : <p name="count" value="$count"></p>
to the : <input type="hidden" name="count" value="<?=$count?>" />
You can use hidden elements like the following
use
<input type="hidden" value="<?php echo $count; ?>" name="count"/>
instead of
<p name="count" value="$count"></p>
Instead of
<p name="count" value="$count"></p>
Use
<input type="hidden" name="count" value="<?php echo $count;?>">
this line will not work : <p name="count" value="$count"></p>
you can use <input type="hidden" name="count" value="<?php echo $count; ?>" />

handle html input elements with numeric id javascript

i have html form
<form name="updatefrm" id="updatefrm" action="" method="post">
<input type="hidden" name="98" id="98" value="" />
<input type="hidden" name="99" id="99" value="" />
<input type="hidden" name="100" id="100" value="" />
<input type="hidden" name="101" id="101" value="" />
<input type="hidden" name="102" id="102" value="" />
<input type="hidden" name="updateqty" id="updateqty" value="1" />
</form>
now i want to assign value to this elements
i m using following javascript code for element with id 98
document.updatefrm."98".value = elements[0].value;
but its giving me error in console.
can any one help me with this ?
You should use document.formname to access forms as not all browsers support this(actually I'm not sure if any does). use document.forms.formname instead. Also to access a numeric property of an object use bracket notation instead of dot notation.
document.forms['updatefrm']['98'].value = elements[0].value;
Why can't I have a numeric value as the ID of an element?
// Change your numeric id.
var myValue=document.getElementById('your_changed_id').value;
alert(myValue)
Go for simplicity..Instead of writing this complex code.why not try a simple code which does the same thing
document.getElementById('your_changed_id').value
You can also use getElementById to set the value. Beside this also check Javascript Naming Convention
document.getElementById('98').value = "YOUR VALUE";
You should not give numbers as ID include any kind of character..
Try like this:
document.forms["updatefrm"]["s98"].value = elements[0].value;
Fiddle

Can i Mix $_get and $_post?

in order to edit my entries i want to:
<form id="pregunta" name="pregunta" class="form_pregunta" method="post" action="pregunta.php?id=26">
<h2>Titulo de la pregunta</h2><input name="q" id="q" class="q" value="este es mi títiulo " type="text">
<h2>Describe tu pregunta</h2>
<textarea name="texto" id="texto" style="width: 98%;"><p>esta es mi descripcion</p></textarea>
<h2>Etiquetas</h2>
<input name="tags" id="tags" onmouseover="mostrar_tooltip('nube_e','','0','70','')" onmouseout="ocultar_tooltip('nube_e')" value="dos,tres,una,">
<input name="responde_a" style="display: none;" id="responde_a" value="0">
<button name="pregunta" id="pregunta" type="submit">form_edit_question_button</button>
</form>
And then in file.php
i'd like to $_get['id'] and $_post['inputs']
but when i go:
if(isset($_POST['edit_pregunta'])){
echo 'lalalalalalalalalalalalalalala';
post_edit_pregunta();
}
it won't ever enter :S. is that normal or i'm missing something... i wanted not to have a hidden input with the id of the post i want to edit..
I'm not 100% sure, but forms don't send their name when submitted, much less their id.
You could do the following instead:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
<input type="hidden" name="edit_pregunta" value="anything">
... //inputs here
</form>
and your if should now enter.
It looks like you're checking for your form's "id" attribute. This is not sent when the form is submitted, only values in <input>, <select>, <textarea> and <button> are sent.
You should check for one of those.
Edit: Your button name is "pregunta", so that is the POST variable you should be checking for, eg
if(isset($_POST['pregunta'])){
Just to comment in general on mixing params in the form's "action" and inputs, you can mix them as long as the form method is "post". You cannot set GET params in the form's action and use the "get" method
<!-- Good -->
<form action="proc.php?id=123" method="post">
<input name="foo" value="foo">
<input type="submit">
</form>
<!-- Bad -->
<form action="proc.php?id=123" method="get">
<input name="foo" value="foo">
<input type="submit">
</form>
There should be no problem at all with having get and post variables in the same request, but are you sure your syntax is correct? If this is normal php, shouldn't you write
<form id="edit_pregunta" method="post" action="file.php?id=<?php echo $this->id; ?>">
... //inputs here
</form>
[Edit]
The problem is (if I'm correct and this is standard php) that you generate a form that looks something like this:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
... //inputs here
</form>
This will make id look like this: '$this->id' (including the '-signs). When what you want is something like this:
<form id="edit_pregunta" method="post" action="file.php?id=51">
... //inputs here
</form>
Then $_GET['id'] would be 51.
[Edit2]
Also, I think you need to change
if(isset($_POST['edit_pregunta'])){
with
if(isset($_POST['pregunta'])){
If I'm not mistaken the name of a form doesn't get sent to the server, however, the name of the submit-button does, but I might be wrong about that part.
Yes you can, I've done it several times.
Probably something else is wrong with your code.
Is there any control with name="edit_pregunta" or is it just the id of the form? IDs are not sent to the server.
Simply adding the id to the form will not create the $_POST['edit_pregunta'] you verify.
Instead, inside the form tag, add an <input name="foo" />; in the php script verify $_POST['foo']
While the HTTP spec doesn't disallow query parameters in POST methods, it is somewhat unusual. You'd be better off using a hidden input field in the form to pass any non-user values up to the script.
That said, the syntax for your form is wrong. You need to use "echo" to insert the value of $this->id into the action.
Use input type="submit" in place of button tag.
You need name for form submission and activate php script!
HTML Code:
<form action="change.php" method="POST">
<input type="password" name="p1" class="change_text" placeholder="New Password"/></br>
<input type="password" name="p2" class="change_text" placeholder="Re-Password"/></br>
<input type="submit" name="change" value="Change Password" id="change" />
</form>
PHP Code:
<?php
if (isset($_POST['change']) {
$p1=$_POST['p1'];
}
?>

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