What wrong with my code? uniqid() is for generating a unique code and it will be stored to the name attribute then a condition is stated if it is clicked it will generate it's working.. Could some help me please with this one? Thanks in advance..
<html>
<form method="POST">
<?php
$attname = uniqid();
echo "<input type='submit' name='$attname' class='btn btn-success' value='Post to Peónline'/>";
echo $attname;
if(isset($_POST[$attname])){
echo 'its working';
}
?>
</form>
<html>
This isn't going to work.
When you refresh the page the $attname value will change. This will happen when you submit the form. So the actual name you're checking on will change and not be the same as the new $attname.
Put the following after your echo $attname; line:
print_r($_POST);
Also for this to work properly you'll need to nest the <input> tag in a <form> tag e.g.:
<form method="POST">
<input>...</input>
</form>
The $attname will change after page refresh.
You need to store that name somewhere.
Add another hidden element.
...
echo "<input type='hidden' name='elemname' value='<?php echo $attname;?>'/>";
...
And after submit,
if (isset($_POST['elemname'])) {
$elemname = $_POST['elemname'];
$val = isset($_POST[$elemname]) ? $_POST[$elemname] : '';
echo $val;
}
Related
I have a problem when developping a page in php, when click on a submit button another submit button. appear.
my problem appeared at the isset of this submit button, echo $b doesn't work; here is my code:
<form method='POST'>
<input type="submit" name="s1">
</form>
</body>
</html>
<?php
if(isset($_POST["s1"])){
$b=2;
echo "<form method='POST'><input type='submit' name='s2'></form>";
if(isset($_POST["s2"])){
echo $b;
}
}
I've already tried to make $b a global variable, but no change :(
Thank's for your help.
The flow of your code is first send the value of input s2, if this was set with some value echo your second form and set b variable to 2, when your second form is printed out on the browser the b value will have lost, so your must store this value in some place, this could be done through a hidden field in your second form like this:
echo "<form><input type='text' name='s2'><input type='hidden' name='b' value='$b'></form>"
if(isset($_POST["s2"])){
echo $_POST['b'];
}
Hope this helps you!
I'm wondering what's the easiest way to make let's say a form with user/pass and submit button with php but after you submit it goes back to the same page instead of going to another php page.
I'm thinking of if/else statement but a bit confused how to set it after lots tries but still not getting the result wanted
weird I did all those you guys said before I posted..but...>.<"
let's say just something simple like this...
but I also want to set if nothing is entered then sumbit is clicked there'll be an error....should be something easy but I don't know why I can't seem to figure it out
<?php
function userPass()
{
echo "<form method='post' action=" . $_SERVER['PHP_SELF'] . ">";
echo "<input type='text' name='user' /><br/>";
echo "<input type='text' name='pass' /><br/>";
echo "<input type='submit' value='Login' />";
}
if(empty($_POST["user"]))
{
userPass();
}
if(!(empty($_POST["user"])))
{
if($_POST["user"] == "comp")
{
echo "Welcome comp";
}
else
{
echo "Wrong user";
}
}
?>
The other answers are right; you only need to send the user back around to your current page in the "action" property. You can test to see if they did so using "isset".
Something that I'm surprised hasn't been mentioned yet is that your input is not being sanitized, and you're setting yourself up for disaster. Huge injection vulnerability in your action attribute:
$_SERVER['PHP_SELF']
If you don't sanitize this, then someone can just modify the URL that they see and your poor PHP script won't know any better than to process that as your SELF constant.
In other words, you absolutely must use an htmlspecialchars() function to html-encode that parameter. With that, your action should look more like htmlspecialchars($_SERVER['PHP_SELF']).
if current page is index.php, use index.php in form tag as value of action.
like this:
<form action="index.php" method="post">
</form>
u can check for submitted form by putting:
if(isset($_POST)){
...
}
at top of page
Just use below syntax
<form method="post" action="">
You can check whether post is set using isset() method.
<?php function userPass() {
echo "<form method='post' action=" . $_SERVER['PHP_SELF'] . ">";
echo "<input type='text' name='user' /><br/>";
echo "<input type='text' name='pass' /><br/>";
echo "<input type='submit' value='Login' />"; }
if(empty($_POST["user"]))
{
userPass();
}
*if(!(empty($_POST["user"])))*
{
if($_POST["user"] == "comp")
{
echo "Welcome comp";
}
else
{
echo "Wrong user";
}
}
?>
This part of your code is wrong, you should type : if(!empty($_POST["user"]))
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
This is exactly how you work with your form to reload the page when click the submit button inside this form.
ADDITIONAL:
Add required to all input you wanted to be required or check if it's empty.
This is a code that I created to control learning the required input fields satisfy the requirements. when this is so, the data will be sent to the database. if it does not meet the requirements, there will be a message may be shown at the top of the page
<?php
if (!isset($_POST['submitform'])) {
}
else
{
$_SESSION['firstname'] = $_POST['firstname'];
$_SESSION['lastname'] = $_POST['lastname'];
$_SESSION['email'] = $_POST['email'];
$_SESSION['mobile'] = $_POST['mobile'];
$_SESSION['telephone'] = $_POST['telephone'];
$_SESSION['place'] = $_POST['place'];
$_SESSION['street'] = $_POST['street'];
$_SESSION['housenumber'] = $_POST['housenumber'];
$_SESSION['gender'] = $_POST['gender'];
if (empty($_POST['firstname']) or empty($_POST['lastname']) or empty($_POST['email']) or empty($_POST['mobile'])or empty($_POST['telephone']) or empty($_POST['place'])
or empty($_POST['street']) or empty($_POST['housenumber']) or !isset($_POST['gender']))
{
echo "Sending denied";
}
else
{
require 'database.php';
header('Location: succes.php');
} ?>
I hope this is helpful information for you
this code will help you
<form action="<?= $_SERVER['REQUEST_URI'] ?>" method="post">
or you could just do this:
<form action="" method="post">
Thank you ....
You can define in the form submission:
<form method=get action=index.php >
You can also leave action out altogether, and it will automatically post/get to that same file.
I think that all have missed the actual question, i.e. if there is a way to stay in the same page after submitting a form. The answer is NO. Once you submit a form -- whether you use the POST or GET method, and of course assuming that you are using $_SERVER['PHP_SELF'] or empty action, etc. -- a new page is opened automatically. This holds at least for PHP. (I have tried a lot of different ways using XAMPP.) I don't now about other scripting languages.
I have a PHP code which generates a dynamic list inside a form like the following, note that the list is built dynamically from database:
echo '<form name="List" action="checkList.php" method="post">';
while($rows=mysqli_fetch_array($sql))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='hidden' name='SessionID' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
What I need is to POST the data corresponding to the user choice when he clicks on the button for that row to another page.
If we use hyperlinks with query strings there will be no problem as I'll receive the data from the other page using a GET request and the hyperlinks would be static when showed to the user.
Also I need to obtain the user input from a textbox which is only possible with POST request.
Simply from the other page (checkList.php) I need these data for further processing:
$SessionID=$_POST['SessionID'];
$Code=$_POST['code'];
As I have a while loop that generates the fields, I always receive the last entry form the database and not the one corresponding to the line (row) that the user chosed from the LIST.
I'm going to recommend that you clean up the names of variables so that your code can
at least tell us what it's supposed to do. It should be rare that someone looks at your code
and has a lot of trouble trying to see what you're trying to accomplish :P, ESPECIALLY when you need help with something ;]. I'm going to try some things and hope that it makes doing what you want easier to comprehend and perhaps get you your answer.
It's good to try your best to not echo large amounts of HTML unnecessarily within a script , so firstly I'm going to remove the
echos from where they are not necessary.
Secondly, I'm going to use a mysql function that returns an easier to process result.
$user = mysqli_fetch_assoc($sql)
Third, I don't know if form having a name actually does anything for the backend or frontend of php, so I'm
just going to remove some of the extra crust that you have floating around that is either invalid HTML
or just doesn't add any value to what you're trying to do as you've presented it to us.
And yes, we "note" that you're building something from the database because the code looks like it does =P.
I'm also sooo sad seeing no recommendations from the other answers in regard to coding style or anything in regard to echoing html like this :(.
<?php while($user = mysqli_fetch_assoc($sql)): ?>
<form action="checkList.php" method="post">
<input type='password' name='code' value='<?php echo $user['code'] ?>' />
<input type='hidden' name='SessionID' value='<?php echo $user['id'] //Whatever you named the field that goes here ?>' />
<input type='submit' value='Take Survey' />
</form>
<?php endwhile; ?>
i not sure this is correct
echo '<form name="List" method="post">';
while($rows=mysqli_fetch_array($result))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='button' value='Take Survey' onclick=show($rows[0])>";
echo "<br>";
}
and javascript
<script>
function show(id)
{
alert(id);
window.location="checkList.php?id="+id;
}
</script>
On checkList.php
$id=$_GET['id'];
echo $id;
You can just check in checkList.php whether $_POST['code'] exists and if exists retrieve $_POST['SessionID'] which will be generated from database. But one thing, if You have all hidden fields in one form, they all will be sent, so You need to think how to correct that - maybe seperate forms for each hidden field, submit button and maybe other POST fields.
And afterwards, You will be able to get data in the way You need - $SessionID=$_POST['SessionID'];
I suppose it is the easiest way to solve that.
You can try something like this:
while($rows=mysqli_fetch_array($sql))
{
$index = 1;
echo "<input type='password' name='code' id='code'>";
//Attach $index with SessionID
echo "<input type='hidden' name='SessionID_$index' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
On checkList.php
<?php
$num = explode('_', $_POST['SessionID']);
$num = $num[1];
//in $num you will get the number of row where you can perform action
?>
$form = 1;
while($rows=mysqli_fetch_array($sql))
{
echo '<form name="List_$form" action="checkList.php" method="post">';
echo "<input type='password' name='code' id='code_$form'>";
echo "<input type='hidden' name='SessionID' id='SessionID_$form' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
echo '</form>';
$form++;
}
Not sure but this may be a bug in php.
Here is my dummy code.
<script language="javascript">
function fb_logout() {
document.getElementById("hidden_logout_id").value
alert(document.getElementById("hidden_logout_id").value);
document.getElementById("form_1").submit();
}
</script>
echo "<form id=\"form_1\" action=\"" . $_SERVER['PHP_SELF'] . "\" method=\"post\">";
echo "<input type=\"hidden\" id=\"hidden_logout_id\" name=\"hidden_logout_name\" value=\"1\"/>";
echo "<input type=\"text\" id=\"h_logout2\" name=\"h_logout2\" value=\"1\"/>";
echo "</form>";
var_dump($_POST);
echo "<span onclick=\"fb_logout();\">Logout</span>";
When doing a post trough the click on the span the values of the hidden and the text inside the form do not get posted: var_dump($_POST) shows an empty variable. But it's strange that if I remove the hidden from the form (or I just place it outside the form) it works and it passes in the $_POST the value of the input text remaining in the form.
Does this has to do with that the hidden is modified from an event outside the form?
Any help is appreciated.
This code may help..
<?php
if(isset($_POST['hidden_logout_name'])) var_dump($_POST);
?>
<script language="javascript">
function fb_logout() {
alert(document.getElementById("hidden_logout_id").value);
document.getElementById("hidden_logout_id").value = "hidden_logout_value";
document.getElementById("form_1").submit();
}
</script>
<form id="form_1" action="" method="post">
<input type="hidden" id="hidden_logout_id" name="hidden_logout_name" value=""/>
<input type="text" id="h_logout2" name="h_logout2" value="1"/>
</form>
<button onclick="fb_logout();">Logout</button>
document.getElementById("hidden_logout_id").value
you have not done any action for this line either assign value to this or assign its value to variable.
var logout_value=document.getElementById("hidden_logout_id");
alert(logout_value.value);
document.getElementById("form_1").submit();
May this help you.
look for the
hidden_logout_name
instead of
hidden_logout_id
so your value is in the
$_POST['hidden_logout_name']
I want to send data from one page to the other via a form in PHP. In the initial page I have included the following PHP script that generates an input form with hidden fields. The names of these hidden fields are generated by the PHP:
<?php
echo "<form class='available-form' name='available_os' method='get' action='process-results.php'>";
echo "<input type='hidden' name='$software'></input>";
echo "<input type='hidden' name='$version'></input>";
echo "<input type='submit' name='available-button' value='Find Available Libraries for this Software'></input>";
echo "</form>";
?>
In the second page, named process-results.php, I would like to get the names of these hidden fields via the $_GET method but of course using $_GET[$software] and $_GET[$version] wouldn't work...Can someone tell me if there is a solution to this issue or if there is a better alternative? Thanks in advance
Instead of
"<input type='hidden' name='$software'></input>";
you should use
"<input type='hidden' name='software' value='".$software."'></input>";
for each. This way, you can use $_GET['software'] to retrieve the value. Do this for each of your hidden inputs.
I think you may want something like:
<form ... >
<input type="hidden" name="software" value="<?php echo $software ?>" />
<input type="hidden" name="version" value="<?php echo $version ?>" />
</form>
and then
$_GET['software'];
$_GET['version'];
I'm not sure what you're trying to accomplish, but this looks odd to me. Isn't the below code more of what you're looking for?
<?php
echo "<form class='available-form' name='available_os' method='get' action='process-results.php'>";
echo "<input type='hidden' name='software' value='$software'></input>";
echo "<input type='hidden' name='version' value='$version'></input>";
echo "<input type='submit' name='available-button' value='Find Available Libraries for this Software'></input>";
echo "</form>";
?>
That way you will get a query string in form of ?software=yoursoftwarename&version=yourversion and it will be available via $_GET["software"] and $_GET["version"] on the next page.
You could iterate over each of the items in the $_GET array on process-results.php. The problem is that the keys for the value will be whatever $software and $version are set to on the first page. Try something like this:
foreach($_GET as $key=>$string) {
// Do stuff with them
}
Add
enctype="multipart/form-data"
To the tag... so it looks like
<form enctype="multipart/form-data" method......
If you really need to have the dollar-sign inside the name, escape it:
echo "<input type='hidden' name='\$software'>";
or put the string in single-quotes:
echo '<input type="hidden" name="$software">';
Otherwise PHP is looking for a variable named "$software", if you look inside the browser-source you will see that the name-attributes are empty(except you're having those variables defined somewhere).