Form keeps selecting multiple radio buttons - php

I have a feeling there is a simple solution to this problem but I can't seem to figure it out.
I'm using a mix of jQuery, PHP and HTML to generate this dynamic form of mine.
Each result SQL finds will be passed into the form, it will display a person along with a radio button. So if there are 3 returned results in the DB, the form will generate 3 people and one of them will have the radio button already checked. My problem is, I only want to be able to select 1 of the 3 radio buttons generated. It keeps multi selecting and I don't know what's causing it or how to stop/fix it.
Example:
Note - showing snipet to illustrate my point
<?php
$group_query = mysqli_query($conn,"SELECT * FROM Groups WHERE gp_ID = '26'")
or die("Error: ".mysqli_error($conn));
?>
<form action="#" class="form-horizontal" method="post">
<?php
$counter = 1;
$passenger_query = mysqli_query($conn,"SELECT * FROM Customers AS C
WHERE C.gp_ID = '".$group_Results['gp_ID']."'")
or die("Error: ".mysqli_error($conn));
while($passenger_Results = mysqli_fetch_array($passenger_query)){
?>
<input type="radio"
class="the_Leader"
name="the_Leader_<?php echo $counter; ?>"
value="<?php echo $passenger_Results['customer_ID'] ?>"
<?php echo $passenger_Results['customer_ID'] == $group_Results['leader_ID'] ? 'checked' : ''; ?>
/>
<?php
echo $passenger_Results['name'];
$counter++;
}//end while loop
?>
</form>
tl;dr dynamic form generated using PHP/MySQL, one of the results is already checked. Want to be able to change it but form multi selects radio buttons.

You are assigning different name= attributes to the buttons when you write name="the_Leader_<?php echo $counter; ?>". To be in a button group, all buttons must have the same name.
You want the names to be the same and the values to be different; the latter lets you know which button was clicked when the form is submitted.

Related

display a table when a button is pressed [duplicate]

I'm having trouble adding the value of type="button" form elements to a mySql database, and I'm wondering if I'm missing something.
Edit - It doesn't look like the information for that element is being passed from the html to the php because it's not echoing a value. My only problem is with this one element and the rest of the form is being submitted properly.
I'm using this for an online quiz which builds a user profile based upon images they've selected, and am setting the images as background images for the button elements, and I'm trying to do this in straight html (as opposed to using javascript together with radio buttons or check boxes).
<input type="button" name="quiz_start" value="jeans" style="background: url(files/start1.jpg) no-repeat; width:54px;height:140px; cursor:pointer; border:none; color: transparent; font-size : 0">
I've simplified the php code for purposes of asking the question (including specifying the user id and limiting it to only one field). I've also included the full code below.
<?php
//Start session & connect to database
$user_id = 3;
$qry = "INSERT INTO style(user_id, quiz_start) VALUES('$user_id','$_POST[quiz_start]')";
$result = #mysql_query($qry);
header("location: page2.html");
exit();
?>
The full query is:
$fieldlist=$vallist='';
foreach ($_POST as $key => $value) {
$fieldlist.=$key.',';
$vallist.='\''.($value).'\',';
}
$fieldlist=substr($fieldlist, 0, -1);
$vallist=substr($vallist, 0, -1);
$fieldlist.=', user_id';
$vallist.=','.$user_id;
$setlist='';
foreach ($_POST as $key=>$value){
$setlist.=$key .'=\''.$value.'\',';
}
$setlist=substr($setlist, 0, -1);
$result = mysql_query('UPDATE style SET '.$setlist.' WHERE user_id='.$user_id);
if (mysql_affected_rows()==0) {
$result = mysql_query('INSERT INTO style ('.$fieldlist.') VALUES ('.$vallist.')');
}
header("location: page2.html");
exit();
?>
Seeing that you are unable to echo $_POST['quiz_start'] that means your value is not actually set. This is because when you use a class button as in <input type='button'> your form is not actually submitted like <input type='submit'>
One solution would be to change your button to an actual submit and format that... or you need to call a javascript function with an onClick from your button as in:
<input type="button" onClick="myfunction()">
For reference to what I am talking about look at this post.
If as you say the rest of the form values are submitting fine but just the button value is not working you have a few different possible solutions depending on your preference.
Use a select field or checkbox for people to select a type in which you can pass your data.
Submit your form in javascript with <input type="button" onClick="myfunction()"> then running your update query in javascript.
Finally if you still want to run your query in PHP you can run a javascript function to make an AJAX call to return JSON information in which you can define a php variable after the page has loaded in which you can then plug into your update query.
Try this:
<?php
//Start session & connect to database
$user_id = 3;
$qry = "INSERT INTO style(user_id, quiz_start) VALUES('".$user_id."','".$_POST['quiz_start']."')";
$result = #mysql_query($qry);
header("location: page2.html");
exit();
?>
Since it seems that input type="button" can't capture the user's selection, I wanted to share a really simple way I figured out for doing this with radio buttons or check boxes using only html.
All you need to do is set the input element to style="display:none", and surround both the image and input element with a label tag so that users can click anywhere on the image to select the element :-)
<label for="quiz_start">
<img src="files/start1.jpg" />
<input style="display:none" type="radio" id="quiz_start" name="quiz_start" value="jeans">
</label>

Updating hidden input depending on what user has checked

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.

check if checkbox is checked in php

I have a check box list which I fill it with data from my table.Here is the code:
<?php
mysql_connect("localhost","root","");
mysql_select_db("erp");
$a="Select * from magazine";
$b=mysql_query($a);
$c=mysql_fetch_array($b);
while($c=mysql_fetch_array($b))
{
print '<input type="checkbox"/>'.$c['den_mag'];
echo "</br>";
}
if(isset($_POST['den_mag']))
{
echo "aaaa";
}
?>
It's a simple query and for each data just show it with a checkbox.Now what I want is when I press a checkbox the value of that checkbox to be shown in a table.So if I have check1 with value a , check2 with value b and I check check1 the value a to be outputted to a table row.How can I achieve that? how cand I get which checkbox is checked?
A few notes:
Try to avoid using SELECT * queries. Select the fields you are going to use:
$sql= '
SELECT
id,
den_mag
FROM
magazine
';
Use better variable names. $a and $c make your code harder to follow for others, and for yourself when you come back at a later time. Use more descriptive variable names like $query_object and $row. Your code should read almost like an essay describing what you're doing.
In your form, use an array of elements. By giving the input a name like selected_magazines[], you will end up with an array in your post data, which is what you want -- multiple selections
Use the row ID as the value of the checkbox element. Your array in POST will then be a list of all the IDs that the user selected
Separate your logic from your HTML generation. The top portion of your script should take care of all logic and decisions. At the bottom, output your HTML and avoid making logical decisions. It makes for a script that is easier to follow and maintain, as well as debug.
Here is a sample script incorporating these ideas with the details you've given:
<?php
// FILE: myfile.php
mysql_connect("localhost","root","");
mysql_select_db("erp");
if(isset($_POST['selected_magazine'])) {
// $_POST['selected_magazine'] will contain selected IDs
print 'You selected: ';
print '<ul><li>'.implode($_POST['selected_magazine'], '</li><li>').'</li></ul>';
die();
}
$sql= '
SELECT
`id`,
`den_mag`
FROM
`magazine`
';
$query_object=mysql_query($sql);
$checkboxes = array();
while($row = mysql_fetch_array($query_object)) {
$checkboxes[] = '<input name="selected_magazine[]" value="'.$row['id'].'" type="checkbox" /> '.$row['den_mag'];
}
?>
<form action="myfile.php" method="post">
<?php print implode('<br>', $checkboxes); ?>
<input type="submit" value="Submit" />
</form>
<input name="test" type="checkbox" />
<?php
if(isset($_REQUEST['test'])){
// selected
}
?>
When you give input-type elements (input, textarea, select, button) a name attribute (like I did), the browser will submit the state/value of the element to the server (if the containing form has been submitted).
In case of checkboxes, you don't really need to check the value, but just that it exists. If the checkbox is not selected, it won't be set.
Also, you need to understand the client-server flow. PHP can't check for something if the client does not send it.
And finally, someone mentioned jQuery. jQuery is plain javascript with perhaps some added sugar. But the point is, you could in theory change stuff with jQuery so that it gets (or doesn't get) submitted with the request. For example, you could get jQuery to destroy the checkbox before the form is submitted (the checkbox won't be sent in this case).
Here you go :
<html>
<input name="test" value="true" type="checkbox" />
</html>
<?php
$Checkbox1 = "{$_POST['test']}";
if($Checkbox1 == 'true'){
// yes, it is checked
}
?>

Extract value from select/option tag in php?

I am not exactly how to do it and how to word the question, so i shall try my best (PS: i'm new to web dev, so please be clear in your answers if you could).
So, I have got a drop down menu with the list names, which are taken from my database. In that database i have a table with names column (the ones that are rendered to the dropdown box) and relevant information to those names. Now, I want that relevant information to appear below in a tag when a user choose one of those names. I also cannot use the form.submit() method because my submit button is already taken for something else.
Here is the code to that bit:
<form name="name_choice" method="post" action="index.php">
<select name="names" onchange="form.some_method()">
<option value="NULL" selected="selected">--Select name--</option>
<?php
for ( $i = 0; $i < $numrows; $i++ ) { //for all the columns, iterate and print out
$id_names = mysql_result($result, $i);
echo "<option value='".$id_names."'>".$id_names."</option>";
}
?>
</select>
</form>
So the bit above works fine, but the "some_method()" is my problem, i don't know what to trigger to display the text in the div below the drop down box (code is below for it):
<div class="information"> <!--if the name is chosen ONLY!-->
<?php
if($_POST['names'] == "NULL") {
echo '<p>Please select an option from the select box.</p>'; //this bit is for testing
}
else {
echo '<p>You have selected: <strong>', $_POST['names'], '</strong>.</p>';
//and then how to echo the relevant information?:(
}
?>
</div><!--end of possible info-->
onchange is a JavaScript event. PHP can't do realtime processing of form data, as it sits on the server and the form is on the client. You can sort of do it by using AJAX and passing the form data as the user types, but that would be a lot more work than is needed. Take a look at JavaScript form validation posts to get yourself headed on the correct path.

implement a button to send information to another php file?

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.

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