I have a page which is sending the variable idClient to another page.
Lets say on the step1.php i fill the idClient input and submit it via GET.
It goes to step2.php?idClient=1, then i'am hidding it:
<input type="hidden" id="idClient" value="<?php echo $_GET['idClient']; ?>">
But after that, when i submit that second page to another, it doesn't send the idClient variable.
Can anyone give me a hint?
You have to change id="idClient" to name="idClient".
id attribute is only used in the client side, and will never passed to the server side.
Try this:
<input type="hidden" id="idClient" name="idClient" value="<?php echo $_GET['idClient']; ?>">
use POST to access idClient
<input type="text" id="idClient" name="idClient" value="<?php echo $_GET['idClient']; ?>">
Related
I have a form located at a url containing get parameters,my form is also using this method.When the form is submitted it rewrites the previos get parameters.
Is there a simple way to rewrite only my form parameters?
I have in mind a Javascript solution ,however I want to know if there is a simpler way?Using HTML/PHP perhaps?
As far as I know, u u are not interested in using JS, then using form's hidden element is only way u have like this-
<form action="demo_form.asp">
First name: <input type="text" name="fname"><br>
<input type="hidden" name="country" value="Norway">
<input type="submit" value="Submit">
</form>
<p>Notice that the hidden field above is not shown to a user.</p>
The question is how u can use it with PHP, right?
The solution is here-
//In PHP
if( isset($_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
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; ?>" />
I am taking input from user in a form. Can I get that value(user input) to calculate other fields in the same form.
<input size="12" id="inputField" name="inputField" autofocus="" type="date" onblur="return dateValidate(this)"/>
Can I collect this form input later in the form and use it to calculate other fields. I was trying to using to use $_POST to retrieve the value but I am not sure if thuis is the right thing.
You cannot do it in PHP before submitting the form. You can easily do so in JavaScript. However you could simply add some AJAX code to send the value to your PHP script when user enters something in the box, and parse the response accordingly. PHP works on server side, and does not interact with user without any server side request
You can certainly do it in 2 ways...
1st way
<?php
$first_digit = '';
$second_digit = '';
$third_digit = '';
if(isset($_POST['calculate'])) {
$first_digit = $_POST['first_digit'];
$second_digit = $_POST['second_digit'];
$third_digit = $first_digit + $second_digit;
}
?>
<form method="POST">
<input type="text" name="first_digit" value="<?php if(isset($first_digit)) echo $first_digit; ?>" />
<input type="text" name="second_digit" value="<?php if(isset($second_digit)) echo $second_digit; ?>" />
<input type="text" name="third_digit" readonly value="<?php if(isset($third_digit)) echo $third_digit; ?>" />
<input type="submit" name="calculate" value="Calculate" />
</form>
2nd Way
Total the variables in the code and instead of showing the result output in a text box you can instead calculate and echo out the result, or you can store that in the database.
Note: Be sure you keep the third input[type=text] as readonly as it is showing you the calculated value so probably you don't want your users to change
Yes you can get input from user to calculate other fields in the same form, you should use scripts like OnChange() or onClick() to have this.
I am trying to create posts with a userid = to the currently logged in users id.
On page 1 I use $_SESSION to set a variable
$_SESSION['Userid'] = $row_getUserStuff['userid'];
I can call $_SESSION['Userid'] on both page 1 and page 2 with
echo($_SESSION['Userid']);
which outputs the users id, my problem is when I try to push that data to the database.
In the form I put
<input type="hidden" name="userid" value="<?php $_SESSION['Userid']?>" />
But when I try to post it I get the error
Column 'userid' cannot be null
Don't put the session data in the form. It is pointless for the browser to have to send the server something the server already knows, and anyone can change it in an injection attack.
Instead, just access $_SESSION['Userid'] when inserting the value into the database.
try this:
<input type="hidden" name="userid" value="<?php echo $_SESSION['Userid']?>" />
You're not actually telling PHP to output the session variable, you're just referring to it. But since it's a session variable, it doesn't need to be in the form at all, I assume there's a $_POST['userid'] somewhere in your code to handle the form, just replace that with $_SESSION['Userid'].
try
<input type="hidden" name="userid" value="<?php echo $_SESSION['Userid']; ?>" />
I added the echo statement and the ; so that no php error will be generated and the $_SESSION['Userid'] is properly sent to the browser.
Replace this line:
<input type="hidden" name="userid" value="<?php $_SESSION['Userid']?>" />
to
<input type="hidden" name="userid" value="<?php echo $_SESSION['Userid']?>" />
I'm submitting data from pagex.php to pagey.php via jQuery post.
pagex.php contains
$('#btn').click(function(e) {
e.preventDefault();
var x = 'variable1';
var y = 'variable2';
$.post("/pagey.php", { var1: x, var2: y}, function(data) {});
});
pagey.php contains
<form action=....>
<input type="text" name="x" value="<?php echo $_POST['var1'] ?>" />
<input type="text" name="y" value="<?php echo $_POST['var2'] ?>" />
<input id="submit" type="submit" value="submit" />
<script type="text/javascript">$('#submit').trigger('click')</script>
So basically when i post the values from pagex.php to pagey.php, i want to automatically submit the form on pagey.php . The jQuery line at the end of pagey.php will trigger an automatic click to the submit button. However jQuery is not triggering the submit click. it works if i access paygey.php directly (i tried it with pre defined variables) but not by doing $.post from pagex. I was assuming that by using $.post from pagex, pagey should automatically get the values and run the jQuery submit. What is the problem here.
JavaScript (which powers jQuery) is not run on the server, it's run from your users browser. So from my understanding, in order to run that little bit of script you will have to actually send your users to pagey.php
<input type="text" name="x" value="<?php echo $_POST[var1] ?>" />
<input type="text" name="y" value="<?php echo $_POST[var2] ?>" />
should be
<input type="text" name="x" value="<?php echo $_POST['var1'] ?>" />
<input type="text" name="y" value="<?php echo $_POST['var2'] ?>" />
You need to fix your array indices.
you should post directly to the action url on pagey... what is the value of pagey if its a simple form that auto posts.
simple answer is to do a form post on document.ready in pagey...
I think the bigger question is why are you posting data to pagey if you just re-post it to another page using your form action?
Try posting the data directly to the action page and let us know if that works.
I bet this will be useful to some of you. Regards.
<?php
echo
"<script type='text/javascript'>
$(document).ready(function() {
$('#submit').trigger('click');
});
</script>";
?>