i have a multiple choice quiz which sends a possible answer via URL like this
<form method="post" action="test2score.php?q=<?php echo $quiz_id; ?>">
on the test2score.php page it reads and tries to compare the value like this
$ans = mysql_result(mysql_query('select c_answer from quiz where quiz_id = "'.$_GET['q'].'"'),0);
if ($_POST['answer'] == $ans){
echo "You got it right!";
}else if ($_POST['answer'] <> $ans){
echo "You got it wrong!";
}
where c_answer is the correct answer stored in db, but even when i select the right answer and post it it still echoes "you got it wrong"
any help would be appreciated.
Your form method="post" is post and your receiving values in $_GET['q'] Kindly correct either of the one
Or use
$_REQUEST['q']`
Use a hidden field q and post that value to your action page and receive that value using
$_POST['q'] and use that in your query.
Thanks.
Why are you using a $_GET request in the action of a form that is $_POST? Since you aren't worried about your users seeing the value of q then why not just put a hidden input in the form with the results of q:
<input type="hidden" value=<?php echo $quiz_id ?> name="q" />
Then in your query, check for $_POST['q']
Seems the saner, more logical way to achieve this
instead of passing quiz id from form action pass it from hidden form field
like below
set form action to
action="test2score.php"
and take hidden form field as
<input type="hidden" name="q" value=<?php echo $quiz_id; ?> />
and get that value like below
$_POST['q'];
Related
using codeigniter mvc pattern I create form in view that take only two values form user
<form action="<?php base_url(); ?> blogs/new_post" method="POST">
<label>Title</label>
<input type="text" name="post_title" />
<label>discription</label>
<input type="text" name="post_detail" />
<input type="submit" value="post" />
</form>
now when i submit the form, data goes to the controller now here confusion created in my code that i can't able to understand i use three cases in controller fist is if i use !empty($_POST) in controller and in view weather i fill the form or not fill the form message displayed in controller is post
my question is why always displaying post why not displaying not a post when i fill nothing in the form
if(!empty($_POST)) {
echo "post";
} else {
echo "not a post";
}
my second question is same related to the first condition now i use isset instead of !empty
public function new_Post() {
if(isset($_POST)) {
echo "post";
} else {
echo "not a post";
}
}
in this case either i fill form or not fill form when i submit the form the result always same that is "post"
and in third case if i use !isset the the result is always not a post eiter i fill or not fill the form
hope so you will understand my problem when i comes to if(!empty($_POST)) this condition then my mind is confuse what is the purpose of $_post
This is because you are using the whole global variable $_POST to check empty and isset(). $_POST is not empty when you click on the button. You just Print_r the $_POST it will have the value of submit button. You need to print the value of $_POST on click and see the values in the array
Well, in CodeIgniter (and most of the framework) doesn't allow you to access $_POST directly due to security reason.
You must access $_POST values through $this->input->post()
For more information read Input Class from CodeIgniter's docs
https://www.codeigniter.com/user_guide/libraries/input.html
$_POST is exist when you click on the button:
so empty and isset cant work Correctly
the value at the first is
Array
(
)
SO U should check
if($_POST)
instead
if(!empty($_POST)) OR if(isset($_POST))
AND best way is u check
if ($this->input->post())
I am creating a script and I came across a problem. I am having to use $_POST and $_GET on the same page, which I don't think makes sense.
I get the value for user id using the GET method from another page through a link and I have to input data from a form into the database in the current page.
CODE SNIPPET 1 : (GET METHOD on page page.php)
echo "<td align=center width=90px height=10px><strong><a href='sessions.php?id=".$userid."' style=text-decoration:none><font color='red' size='5pt'>$i</font></a></strong></td>";
CODE SNIPPET 2 : (POST METHOD on sessions.php going to same page sessions.php)
if($_SERVER["REQUEST_METHOD"]=="GET"){
$id=$_GET["id"];
if($_SERVER["REQUEST_METHOD"]=="POST")
$suggestions=$_POST["suggest"];
Can I get the userid from the url and the POST the values from the form in the same page?
If not, is there a way I can do this better?
You can use one hidden input field in your form in session.php and store $_GET['id'] in value attribute of that input field and it will be pass with your form submit to session.php <input type="hidden" name="id" value="<?php echo $_GET['id']">
you can get the value from GET by adding the following code in sesssion.php
$id="";
if (isset($_GET['id'])) {
$id=$_GET['id'];
}else{
$id="";
}
then in the form located at session.php place a input field like following
<input type="hidden" value="<?php echo $_GET['id']" name="suggest">
then add the following code in the same page
if($_SERVER["REQUEST_METHOD"]=="POST"){
if($id){
$suggestions=$_POST["suggest"];
// run your sql query for insert.
}
}
I've created a test system that has multiple steps (using jquery) allowing users to check checkboxes to select their answers, with a summary page and a final submission button... all within a form. I now want to create the scoring system.
1) Firstly this is the code (within a loop) that grabs the answers from Wordpress for each question:
<input type="checkbox" name="answer<?php echo $counter; ?>[]" value="<?php echo $row['answer']; ?>" />
2) In Wordpress next to each answer is a dropdown with a yes or no option to mark whether the answer is right or wrong. This is output in the following way:
<?php $row['correct']; ?>
3) Each correct answer the user checks should be worth 1 point. The passmark is determined by the field:
<?php the_field('pass_mark'); ?>
4) I want it to update a hidden field with the score as the user checks the correct answer:
<input type="hidden" value="<?php echo $score; ?>" name="test-score" />
How can I update the hidden field with the user score as the user is checking the correct answer? I'm not sure what to try with this to even give it a go first!
Ok, everyones spotted a big hole in this. I'm completely open to doing it a hidden way so people can't check out the source. The type of user this is targeted at wouldn't have a clue how to look at the source but might as well do it the right way to start with!
The whole test is within a form so could it only update the hidden field on submit?
I still need some examples of how to do it.
In my opinion you should use sessions for that purpose, because any browser output may be saved and viewed in ANY text editor. This is not right purpose oh hidden input elements. You use hidden inputs when you need to submit something automatically, but never use it when processing some important data.
Mapping your questions and answers via id will allow you not to reveal real answers and scores in HTML.
Just a very simple example how to do that:
<?php
$questions = array(
125 => array("text"=>"2x2?", "answer"=>"4", 'points'=>3),
145 => array("text"=>"5x6?", "answer"=>"30", 'points'=>2),
);
?>
<form method="post">
<?php foreach ($questions as $id => $question): ?>
<div><?php echo $question['text'] ; ?></div>
<input type="text" name="question<?php echo $id ; ?>"/>
<?php endforeach ; ?>
<input type="submit" value="Submit"/>
</form>
/* In submission script */
<?php
if (isset($_POST['submit'])){
foreach($questions as $id => $question){
if (isset($_POST["question{$id}"])){
$answer = $_POST["question{$id}"] ;
if ($answer === $question['answer']){
$_SESSION['score'] += $question['points'] ;
}
}
}
}
Spokey is right - the user would be able to cheat if your score it on the client side like using the method you suggested.
Instead, either user a JQuery $.post call to post each answer and then store the score in a PHP Session. Or just wait until the entire form is submitted and evaluate the score of the form as a whole on the server side.
* Update *
You have to submit the form to a script that can evaluate the form. So say it gets submitted to myForm.php
In myForm.php, get the post vars:
$correct_answers = $however_you_get_your_correct_answers();
//Assuming $correct_answers is a associative array with the same keys being used in post -
$results = array();
if($_POST){
foreach ($_POST as $key=>$value) {
if ($_POST[$key] == $correct_answers[$key]){
$results[$key] = 'correct';
}
else $results[$key] = 'incorrect';
}
}
This is untested, but it should work.
This is more of a technique question rather than maybe code. I am having a php form with many fields (items to select). Naturally some of the items might be selected and some not. How do I know which ones are selected when i post the data from page 1 to page 2? I thought of testing each one if empty or not, but there are just too many fields and it doesn't feel at all efficient to use or code.
Thanks,
UPDATE EDIT:
I've tried the following and maybe it will get me somewhere before I carry on testing the repliers solutions...
<html>
<body>
<form name="test" id="name" action="testprocess.php" method="POST">
<input type="text" name="choices[shirt]">
<input type="text" name="choices[pants]">
<input type="text" name="choices[tie]">
<input type="text" name="choices[socks]">
<input type="submit" value="submit data" />
</form>
</body>
</html>
and then second page:
<?php
$names = $_POST['choices'];
echo "Names are: <br>";
print_r($names);
?>
This gives out the following:
Names are: Array ( [shirt] => sdjalskdjlk [pants] => lkjlkjlk [tie]
=> jlk [socks] => lkjlkjl )
Now what I am going to try to do is iterate over the array, and since the values in my case are numbers, I will just check which of the fields are > 0 given the default is 0. I hope this works...if not then I will let you know :)
I think what you're looking for is this:
<form action="submit.php" method="POST">
<input type="checkbox" name="checkboxes[]" value="this" /> This
<input type="checkbox" name="checkboxes[]" value="might" /> might
<input type="checkbox" name="checkboxes[]" value="work" /> work
<input type="submit" />
</form>
And then in submit.php, you simply write:
<?php
foreach($_POST['checkboxes'] as $value) {
echo "{$value} was checked!";
}
?>
The square brackets in the name of the checkbox elements tell PHP to put all elements with this name into the same array, in this case $_POST['checkboxes'], though you could call the checkboxes anything you like, of course.
You should post your code so we would better understand what you want to do.
But from what I understood you are making a form with check boxes. If you want to see if the check boxes are selected, you can go like this:
if(!$_POST['checkbox1'] && !$_POST['checkbox2'] && !$_POST['checkbox3'])
This looks if all the three check boxes are empty.
Just an idea:
Create a hidden input field within your form with no value. Whenever any of the forms fields is filled/selected, you add the name attribute of that field in this hidden field (Field names are saved with a comma separator).
On doing a POST, you can read this variable and only those fields present in this have been selected/filled in the form.
Hope this helps.
Try this.....
<?php
function checkvalue($val) {
if($val != "") return true;
else return false;
}
if(isset($_POST['submit'])) {
$values = array_filter(($_POST), "checkvalue");
$set_values = array_keys($values);
}
?>
In this manner you can get all the values that has been set in an array..
I'm not exactly sure to understand your intention. I assume that you have multiple form fields you'd like to part into different Web pages (e.g. a typical survey form).
If this is the case use sessions to store the different data of your forms until the "final submit button" (e.g. on the last page) has been pressed.
How do I know which ones are selected when i post the data from page 1 to page 2?
is a different question from how to avoid a large POST to PHP.
Assuming this is a table of data...
Just update everything regardless (if you've got the primary / unique keys set correctly)
Use Ajax to update individual rows as they are changed at the front end
Use Javascript to set a flag within each row when the data in that row is modified
Or store a representation of the existing data for each row as a hidden field for the row, on submission e.g.
print "<form....><table>\n";
foreach ($row as $id=>$r) {
print "<tr><td><input type='hidden' name='prev[$id]' value='"
. md5(serialize($r)) . "'>...
}
...at the receiving end...
foreach ($_POST['prev'] as $id=>$prev) {
$sent_back=array( /* the field values in the row */ );
if (md5(serialize($sent_back)) != $prev) {
// data has changed
update_record($id, $sent_back);
}
}
I've got the following php code printing out the contents of a SQL table.
$query="select * from TABLE";
$rt=mysql_query($query);
echo mysql_error();
mysql_close();
?>
<i>Name, Message, Type, Lat, Lng, File </i><br/><br/>
<?php
while($nt=mysql_fetch_array($rt)){
if($nt[name] != null){
echo "$nt[id] $nt[name] $nt[message] $nt[type] $nt[lat] $nt[lng] $nt[file]";
}
}
?>
How would I implement a button so for each "row" if the button is clicked on that row it'll submit the information of that row to another php file?
I want it looking something like...
details details2 details3 BUTTON
details4 details5 details6 BUTTON
details7 details8 details9 BUTTON
details10 details11 details12 BUTTON
Where if BUTTON was hit on row 1 details1,2,3 would be sent to a php file, on row 2 detals 4,5,6 would be sent etc. How would I do this?
Thanks.
it's going to be something like that, depending on the data you need to send:
while($nt = mysql_fetch_array($rt)) {
if($nt[name] != null){
echo "$nt[id] $nt[name] $nt[message] $nt[type] $nt[lat] $nt[lng] $nt[file] ".'send request<br/>';
}
}
You can either use GET method and send a query string to the second php page and receive the variables there, like
next.php?variable1=value1&variable2=value2&...
or use POST method by making a hidden form for each row and assign a hidden field for each variable you want to send.
<form method="post" action"next.php">
<input type="hidden" name="variable1" value="value1" />
<input type="hidden" name="variable2" value="value2" />
.
.
.
</form>
or instead of sending all the values, just send the row ID (if any) using any of these two methods and run another query in next.php to get the information you need from database.
Instead of submitting the entire data, just send the ID and fetch the results from the database in the other script. If you want to have an input button, you can do
<form action="/other-script.php" method="GET">
<?php printf('<input type="submit" name="id" value="%s" />', $nt["id"]); ?>
</form>
but you could also just add a link, e.g.
printf('Submit ID', $nt["id"]);
If you really want to send the entire row values over again, you have to make them into form inputs. In that case, I'd send them via POST though.