I have a text box which picks up a value from my database and is a read only text box. it contains some information which i want to send to my datbase but nothing is being recorded. Once the textbox isnt a read only, the data is successfully stored. Is there any way i can keep my textbox disabled and still send the data to my database?
<form action="refnoadded.php?public=<?php echo $id; ?>" method="post">
<span id="sprytextfield1">
<input type="text" name="ref" value="<?php echo $newpass ?>" disabled />
<input type="button" onClick="history.go(0)" value = "Generate Reference">
<br />
<span class="textfieldRequiredMsg">Reference Number Required</span>
</span>
<br />
<input type="submit" value="Add Reference" />
</form>
any ideas?
disabled doesn't send data to server
use
<input readonly value="my value" name="myname">
HTH!
have the data in a second hidden input:
<input type="hidden" name="ref_hidden" value="<?php echo $newpass ?>" />
the user won't see the difference and you will get your value send when submitting the form.
Disabled textfields don't submit their information to $_POST or $_GET. You can simply use the form element
<input type="hidden" name="rev_hidden" value="<?php print $password; ?>" />
This is the standard (correct) way to pass hidden information in the form.
Another use for this element is if you want to pass a "formsubmitted" variable.
However, if you want to create a value and have it uneditable by the user, do it on the server side. Create the value when you create the database, since users can relatively-simply send other data in the place of what you've generated.
Related
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>
I generate a form, which mostly consists of input fields that are already populated with values from the db.
So I do this currently like so:
<input id="misc" name="misc" value="<?php echo $workout['misc']; ?>" />
But when I try and do this:
<input id="misc" name="misc" value="<?php echo set_value($workout['misc']); ?>" />
along with a validation rule, the form does reload itself, the error message does display BUT the form is reset
What am I doing wrong?
As per the manual:
Permits you to set the value of an input form or textarea. You must supply the field name via the first parameter of the function. The second (optional) parameter allows you to set a default value for the form. Example:
<input type="text" name="quantity" value="<?php echo set_value('quantity', '0'); ?>" size="50" />
The above form will show "0" when loaded for the first time.
Hence in your case:
<input id="misc" name="misc" value="<?php echo set_value('misc', $workout['misc']); ?>" />
OR
<input id="misc" name="misc" value="<?php echo set_value('misc'); ?>" />
Documentation:
https://ellislab.com/codeigniter/user-guide/helpers/form_helper.html
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 call a PHP script from my HTML form submit.
Then I process the values retrieved from this HTML form in the PHP script and now I want to display these values in another HTML form that I am calling from this PHP script.
How do I retrieve this variable?
Note: I tried echo but I think I actually want to display this value in the form (HTML) and not break it again in a PHP tag.
I'm not sure what you mean by "not breaking it again with a PHP tag". HTML on its own cannot access PHP variables. This is because PHP is a server-side language and HTML is a client side language. But here is the simplest way to print php variables retrieved from a form.
<form>
<p>Your name: <?php echo $_POST['name'] ?></p>
</form>
This is assuming that there was a form that submitted a field called 'name' using POST.
You can process the name in a php script at the top of the file and then simply echo it when you're printing the html. This way, you won't have too much php code mixed in with the HTML (which makes it look cleaner).
Once you got the values in the PHP script, are you calling a new script? If so, you might wanna save the values in $_SESSION["varible_name"]. If not, you just have to echo it.
It depends on how you are accessing your form data, either through $_POST or through $_GET. For simplicity, I'll assume your using $_GET and modify this example for more clarity.
So lets say you have a form hosted on welcome.php:
<form action="welcome.php" method="get">
Name: <input type="text" name="fname" />
Age: <input type="text" name="age" />
<input type="submit" />
</form>
Now the results will be returned back to the same page, so you want to modify the page to say:
<form action="welcome.php" method="get">
Name: <input type="text" name="fname" value="<?php echo $_GET["fname"]; ?>"/>
Age: <input type="text" name="age" value="<?php echo $_GET["age"]; ?>" />
<input type="submit" />
</form>
Though you'll notice that we're using the same page, and we can only have one version of the page, so we want to render if upon the condition that our variable has been set.
if (isset($_GET["fname"])){
//code to print second form
}
else{
//code to print first form
}
Or, in another way (using the ternary operator):
<form action="welcome.php" method="get">
Name: <input type="text" name="fname" value="<?php echo ((isset($_GET["fname"]))?$_GET["fname"]:""); ?>"/>
Age: <input type="text" name="age" value="<?php echo ((isset($_GET["age"]))?$_GET["age"]:""); ?>" />
<input type="submit" />
</form>