I'm trying to get a value from a form, then display it on posting of the form. I can get the value to appear in the second text field, once I have chosen an option using the Ajax Auto-Select, but how do I get that value shown stored into a variable for display on posting? This is what I have been trying -
if ($_POST['action'] == 'getentity') {
$value= $entity;
$content .= '<div>'.$value.' hello</div>';
}
<form method="post" action="?">
<input type="text" name="TownID_display" size="50" onkeyup="javascript:ajax_showOptions(this,\'getEntitiesByLetters\',event)">
<input type="text" name="TownID" id="TownID_display_hidden" value="'.$entity.'" />
<input type="hidden" name="action" value="getentity" />
<input type="submit" name="submit" value="Find"/>
Many thanks for any help.
Try
<input type="text" name="TownID" id="TownID_display_hidden" value="<?php $value = $entity; echo $entity; ?>" />
and its better to use like this
if ($_POST['action'] == 'getentity') {
$value= $_POST['TownID'];
$content .= '<div>'.$value.' hello</div>';
}
it should work.
Related
I'm having some difficulty updating a quantity value in a multi-dimensional array I've created and really hoping you can help me correct where I've gone wrong;
.
The background
I've got two "items" which both have a simple form tag followed by a hidden input field with a unique value (1 for the first item, 2 for the second).
The button will just point back to this same page using the POST method.
The div on the right of the page will then load a "basket" which will use these post values and add them to an array.
When the "add" button is used again the value should update to +1 rather than create another sub_array.
.
What is currently happening
Currently when I click "add" the first time it adds the array as expected;
However when clicking "add" for the second time it adds a second array rather than +1'in the quantity.
On the third time clicking "add" it does actually now find the original value and update it as I expected, if I click again and again it will continue to update the quantity
It just seems to be that second time I click "add".
.
The Script
<?php session_start();
function in_array_r($needle, $haystack, $strict = false) {
foreach ($haystack as $item) {
if (($strict ? $item === $needle : $item == $needle) || (is_array($item) && in_array_r($needle, $item, $strict))) {
return true;
}
}
return false;
}
if (ISSET($_POST["prod"]))
{
if(in_array($_POST["prod"],$_SESSION["cart"])==TRUE)
{
$_SESSION["cart"][0] =
array($_POST["prod"],$_POST["name"],$_SESSION["cart"][0][2]+1);
}
else{
echo 'running else';
$_SESSION["cart"]=array($_POST["prod"],$_POST["name"],1);}}
if ($_POST['e']=='1')
{
$_SESSION['cart'] = '';
}
echo '<br /><br />';
print_r($_SESSION["cart"]);
}
Sample form
<form action="test.php" method="post" enctype="application/x-www-form-urlencoded">
MAST-O-MIR<br/>
img<br/>
£2.00<br/>
<input type="hidden" value="1" name="prod" />
<input type="hidden" value="MAST-O-MIR" name="name" />
<button class="plus-btn" type="Submit">Add</button>
</form>
Also, what you might well notice from my script is that when you "add" the second item it will actually overwrite the first one by creating the array from scratch so if you can help me with either or both of these I really would appreciate the expertise!
Many thanks to all in advance!
I tried to debug your code and a possible solution could be the following:
<?php
session_start();
if(!isset($_SESSION["cart"]))
{
$_SESSION["cart"]=[];
}
if (isset($_POST["prod"]))
{
$prod_id=$_POST["prod"];
//let suppose $_POST['prod'] is your item id
$found=false;
for($i=0;$i<count($_SESSION['cart']);$i++)
{
if(isset($_SESSION['cart'][$prod_id]))
{
echo "found! so add +1";
$_SESSION['cart'][$prod_id][2]+=1;
$found=true;
break;
}
}
if($found==false)
{
echo 'not found! so create a new item';
$_SESSION["cart"][$prod_id]=array($_POST["prod"],$_POST["name"],1);
}
}
if (isset($_POST['e']) && $_POST['e']=='1')
{
$_SESSION['cart'] = '';
}
echo '<br /><br />';
print_r($_SESSION["cart"]);
?>
<form action="cart.php" method="post" enctype="application/x-www-form-urlencoded">
MAST-O-MIR<br/>
img<br/>
£2.00<br/>
<input type="hidden" value="1" name="prod" />
<input type="hidden" value="MAST-O-MIR" name="name" />
<button class="plus-btn" type="Submit">Add</button>
</form>
<form action="cart.php" method="post" enctype="application/x-www-form-urlencoded">
MAST-O-MIR<br/>
img<br/>
£2.00<br/>
<input type="hidden" value="2" name="prod" />
<input type="hidden" value="MAST-O-MIR" name="name" />
<button class="plus-btn" type="Submit">Add</button>
</form>
Another way to do it is using associative arrays.
The following code creates a cart array in $_SESSION using the item name as key(so you don't need to loop over the cart array to find the item) and
an array with properties as name=>value for each item.
session_start();
if(!isset($_SESSION["cart"]))
{
$_SESSION["cart"]=[];
}
//let's suppose you have unique names for items
if (isset($_POST["prod"]))
{
$name=$_POST["name"];
if(isset($_SESSION['cart'][$name]))
{
echo "found! so add +1";
$_SESSION['cart'][$name]['quantity']+=1;
}
else
{
echo 'not found! so create a new item';
$_SESSION["cart"][$name]=array("id"=>$_POST["prod"],"name"=>$_POST["name"],"quantity"=>1);
}
}
if (isset($_POST['e']) && $_POST['e']=='1')
{
$_SESSION['cart'] =[];
}
echo '<br /><br />';
print_r($_SESSION["cart"]);
?>
<form action="cart2.php" method="post" enctype="application/x-www-form-urlencoded">
MAST-O-MIR<br/>
img<br/>
£2.00<br/>
<input type="hidden" value="1" name="prod" />
<input type="hidden" value="MAST-O-MIR" name="name" />
<button class="plus-btn" type="Submit">Add</button>
</form>
<form action="cart2.php" method="post" enctype="application/x-www-form-urlencoded">
MAST-O-MIR<br/>
img<br/>
£2.00<br/>
<input type="hidden" value="2" name="prod" />
<input type="hidden" value="MAST-OMIR" name="name" />
<button class="plus-btn" type="Submit">Add</button>
</form>
It's hard to test your code without a sample form but I guess both of your problems may be solved by replacing:
$_SESSION["cart"][0] = array($_POST["prod"], $_POST["name"], $_SESSION["cart"][0][2]+1);
For:
$_SESSION["cart"][0][2]+= 1;
By the way, try to properly indent your code when you are going to post it. It is hard to read.
I am trying to get the value of a form (text field) with _POST, and store it to a text file but it doesn't work. Here's my code:
HTML
<form>
<input type="text" name="test" id="test" value="">
<input type="button" onclick="location.href='example.com/index.php?address=true';" value="ODOSLAŤ" />
</form>
PHP
if (isset($_GET['address'])) {
$email = $_POST['test'];
$myfile = fopen("log.txt","a") or die("Error.");
fwrite($myfile, "\n".$email);
// this prints nothing
echo $email;
}
I can't get the value of that text field. Nor GET nor POST doesn't work for me. What am I doing wrong?
You are missing a post method in your form.
<form method="post" action="example.com/index.php?address=true">
<input type="text" name="test" id="test" value="">
<input type="submit" value="ODOSLAŤ" />
</form>
You also have to change the following line.
if (isset($_GET['address'])) {
With
if (isset($_POST['address'])) {
You want to post after triggering an action as FreedomPride mentioned
Conclusion
You want to declare for example a post methodif you know that you would like to post that data in the future.
As FreedomPride also mentioned :
You are using a GET, but if a user does not input anything your script won't work, there by it is recommended to use a POST
You are not submitting your form.
Change your code as follows....
<form method="post" action="example.com/index.php?address=true">
<input type="text" name="test" id="test" value="">
<input type="submit" value="ODOSLAŤ" />
</form>
You'll get what you want....
This is what you're missing as Tomm mentioned.
<form method="post" action="yourphp.php">
<input type="text" name="address" id="test" value="">
<input type="submit" name="submit"/>
</form>
In your PHP, it should be post if it was triggered
if (isset($_POST['address'])) {
$email = $_POST['test'];
$myfile = fopen("log.txt","a") or die("Error.");
fwrite($myfile, "\n".$email);
// this prints nothing
echo $email;
}
Explanation :-
In a form, an action is required to pass the action to the next caller with action. The action could be empty or pass it's value to another script.
In your PHP you're actually using a GET. What if the user didn't input anything. That's why it's recommended to use POST .
I am trying want to ouput a string ,stored in a variabel, in a label on a page when i click on a button.
But i can't find out how. Still a beginner.
<form action="Test.php" method="post">
Output text: <input type="label" name="word" />
<input type="submit" method="submit" value="Print!" />
</form>
<?php
$word = "test";
if (isset($_POST['submit']))
{
//something that gives the label value $word//
}
?>
There are a few things wrong with your code.
Let me outline them.
Your submit input should have a name attribute, since your conditional statement is based on it if (isset($_POST['submit'])){...}, something I've modified to check if the input is not left empty, using PHP's empty() function.
The input type you have for your "Output text" is invalid, it should be type="text" and not type="label", there is no type="label".
method="submit" for your submit button is invalid for a few reasons. Method belongs in <form> and there is no method="submit".
You then need to assign a POST variable from the input:
such as:
$word = $_POST['word'];
Plus, from what looks to me that you're executing the entire code from within the same page, you can just do action="", unless your code is set in 2 seperate files.
In regards to what you want to achieve: You can then echo the input (if one was entered) using a ternary operator and giving it (the input) a value.
I.e.:
value="<?php echo isset($_POST['word']) ? $_POST['word']: '' ?>"
Here:
<form action="" method="post">
Output text: <input type="text" name="word" value="<?php echo isset($_POST['word']) ? $_POST['word']: '' ?>" />
<input type="submit" name="submit" value="Print!" />
</form>
<?php
if ( isset($_POST['submit']) && !empty($_POST['word']) )
{
$word = $_POST['word'];
echo $word;
}
?>
If you want to use a "label" for your input, then use:
<label for="word">Output text:
<input type="text" name="word" />
</label>
You should also guard against XSS attacks (Cross-side scripting) using:
http://php.net/strip_tags
http://php.net/htmlentities
http://php.net/manual/en/function.htmlspecialchars.php
I.e.:
$word = strip_tags($_POST['word']);
$word = htmlentities($_POST['word']);
$word = htmlspecialchars($_POST['word']);
A few articles you can read on XSS:
http://en.wikipedia.org/wiki/Cross-site_scripting
https://www.owasp.org/index.php/XSS_%28Cross_Site_Scripting%29_Prevention_Cheat_Sheet
Your code should look like this
<form action="Test.php" method="post">
Output text: <input type="label" name="word" value ="<?php echo isset($label['data'])?$label['data']: '' ?>" />
<input type="submit" method="submit" value="Print!" />
</form>
<?php
$word = "test";
$label = array();
if (isset($_POST['submit']))
{
//something that gives the label value $word//
$label['data'] = $word;
}
?>
This should work
Regards
Ahmad rabbani
First of all your input element has type = label, it doesn't mean anything. Change it to type=text
And you are submitting value but not printing it. So in input field you have to print it also.
Look below code.
<?php
$word = "test";
if (isset($_POST['submit']))
{
// whatever you do with $word
}
?>
<form action="Test.php" method="post">
Output text: <input type="text" name="word" value="<?php echo $word; ?>"/>
<input type="submit" name="submit" value="Print!" />
</form>
UPDATE
One more thing I forgot to mention that you are submitting form to Test.php and printing this to file, if this file's name is Test.php then not an issue, other wise leave action property blank, so it submit data to itself.
method = submit there is nothing like this. you can set button name to submit, like name= "submit".
Is this better then?
<form id="form1" name="form1" method="post" action="cart.php">
<input name="size" type="radio" value="Small">Small<br>
<input name="size" type="radio" value="Large">Large<br>
<input name="size" type="radio" value="XXL">XXL<br>
<input type="hidden" name="sizes" id="sizes" value=""/>
<input type="hidden" name="pid" id="pid" value="85" />
<input type="submit" name="button" value="Add To Cart"/></form>
Just a quick note that for retreiving the result I am using this code here:
if(isset($_POST['sizes'])){
$myvar = $_POST['sizes'];
echo "Your Size:", $myvar ;
}
You seem to be calling the radio buttons 'form1' but you have the same name for the first hidden field.
I would suggest that the form is submitting this field and ignoring the radio buttons for this reason.
Looking at your revised code, surely you need:
$myvar = $_POST['size'];
if ($myvar) {
echo "Size is $myvar";
}
'sizes' is never getting set to a 'true' value.
Note: the if ($myvar) works, in this instance, as effectively as the 'isset' statement but is simpler to code and read.
With JavaScript & jQuery:
$("#form1 input[name='size']").click(function(){
var size = $('input:radio[name=size]:checked').val();
$("#form1 input[name='sizes']").val(size);
});
That should do the trick.
I am learning PHP now, so pardon my silly question which I am not able to resolve.
I have created a simple web form where in I display the values entered by a user.
function submitform()
{
document.forms["myForm"].submit();
} // in head of html
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" onclick="submitForm()" />
</form>
and res.php contains:
foreach($_POST as $field => $value)
{
echo "$field = $value";
}
When I click on the submit button, I just get a blank page without any values from the form. Can anyone please let me know what am I missing?
There's no need for the javascript. This should do:
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" />
</form>
Let's start with fixing errors:
JavaScript is case-sensitive. I see that your function name is submitform and the form's onclick calls submitForm.
The javascript is not really necessary from what you've shown us, I would try this on a single php page and see if it works:
Create a test.php file for test purpose:
<?php
if($_POST){
foreach($_POST as $key=>$value){
echo "$key: $value<br />";
}
}else{
<form action="test.php" method="post">
<input type="text" value="1" name="name1" />
<input type="text" value="2" name="name2" />
<input type="submit" value="submit" name="Submit" />
</form>
}
?>
If it does work, slowly work your way into your current form setup to see what is breaking it. If it doesn't work, there's something larger at play.
There are 2 things you should do now.
Remove the JavaScript function to submit the form. It's not required (or necessary). The default behavior of a submit button is to... well... submit. You don't need to help it with JavaScript.
Enable error display by using error_reporting(E_ALL).
After you do both things, you should be able to debug and assess the problem much more easily.
Put your php code inside php tags!
<?php
foreach($_POST as $field => $value)
{
echo $field ." = ." $value.'<br />';
}
?>
If you do
<?php
print_r($_POST);
?>
what do you get?
If this still doesn't work, does your server parse php?
Create the file test.php and access it directly http://localhost/test.php or whatever your URL is
<?php
echo 'Hello';
?>
if this doesn't work..it's a whole diferent problem
You can submit an HTML form using PHP with fsubmit library.
Example:
require_once 'fsubmit.php';
$html = "<form action ='res.php' method='post'><input type='text' name='name'/></form>";
$form = new Fsubmit();
$form->url = 'http://submit_url_here.com';
$form->html = $html;
$form->params = ['name'=>'kokainom'];
$response = $form->submit();
echo $response['content'];