I have retrieve data from lvl1itquepaper and display it with 4 radio button and save the user input to another table call lvl1itresult but when I press save, no matter which option I choose, the answer that will be saved into database sure is option 4 even if the student select other option.
<?php
include('../dbconnect.php');
session_start();
?>
<!DOCTYPE html>
<html>
<head>
<title>Online Examination System</title>
</head>
<body>
<div id="container">
<h1>Level 1 IT Question Paper</h1>
<h2>Please read the question carefully and answer it confidently. Good Luck All!</h2>
<?php
if(isset($_POST['Submit']))
{
$sql="SELECT * from lvl1itquepaper";
$run_que = mysqli_query($mysqli, $sql);
$check_que = mysqli_num_rows($run_que);
while ($row=$run_que->fetch_assoc())
{
$questionno = $row['questionno'];
$question = $row['question'];
$option1 = $row['option1'];
$option2 = $row['option2'];
$option3 = $row['option3'];
$option4 = $row['option4'];
$ans_array = array($option1, $option2, $option3, $option4);
$student_ans = $row['option1'];
$student_ans = $row['option2'];
$student_ans = $row['option3'];
$student_ans = $row['option4'];
$sql="Insert into lvl1itresult (questionno, question, studentans, username) values ('.$questionno.', '$question', '$student_ans', '".$_SESSION['login_user']."')";
$submit = $mysqli->query($sql);
}
}
?>
<form method= "post">
<?php
echo "Welcome, ";
$sql="SELECT * from lvl1itstudent WHERE username= '".$_SESSION['login_user']."'";
$find_student = mysqli_query($mysqli, $sql);
$check_student = mysqli_num_rows($find_student);
if ($check_student>0){
while($row = $find_student->fetch_assoc())
{
echo $row['username'];
}
}
echo "<br><br><br><br>";
$sql="SELECT * from lvl1itquepaper";
$run_que = mysqli_query($mysqli, $sql);
$check_que = mysqli_num_rows($run_que);
if($check_que>0){
while ($row=$run_que->fetch_assoc())
{
$questionno = $row['questionno'];
$question = $row['question'];
$option1 = $row['option1'];
$option2 = $row['option2'];
$option3 = $row['option3'];
$option4 = $row['option4'];
$ans_array = array($option1, $option2, $option3, $option4);
shuffle($ans_array);
echo "".$questionno. "." .$question."<br>";
echo "<input type='radio' name='.$questionno.' value='".$ans_array[0]."'>".$ans_array[0]."<br>";
echo "<input type='radio' name='.$questionno.' value='".$ans_array[1]."'>".$ans_array[1]."<br>";
echo "<input type='radio' name='.$questionno.' value='".$ans_array[2]."'>".$ans_array[2]."<br>";
echo "<input type='radio' name='.$questionno.' value='".$ans_array[3]."'>".$ans_array[3]."<br><br>";
}
}
else {
echo "there is no data in database";
}
?>
<input type="submit" value = "Submit" name= "Submit" style= "width:60px; height:30px";>
</form>
</div>
</body>
$student_ans = $row['option1'];
$student_ans = $row['option2'];
$student_ans = $row['option3'];
$student_ans = $row['option4'];
Because of this, you keep pointlessly overwriting the same variable. You should only store the option they selected in $student_ans.
You may want to review your logic a bit, you're moving things in and out of arrays and different variables for no reason.
Do slight modification to form
echo "<input type='hidden' name='qno' value='".$questionno."'";
echo "<input type='radio' name='ans' value='".$ans_array[0]."'>".$ans_array[0]."<br>";
echo "<input type='radio' name='ans' value='".$ans_array[1]."'>".$ans_array[1]."<br>";
echo "<input type='radio' name='ans' value='".$ans_array[2]."'>".$ans_array[2]."<br>";
echo "<input type='radio' name='ans' value='".$ans_array[3]."'>".$ans_array[3]."<br><br>";
and then on posting form fetch them using:
$_POST['ans'] for answer field and $_POST['qno'] for question
This way you will only have the option selected by student
Related
I have searched all over this website, but not yet found the answer for this. Pr maybe I am not able to apply it correctly. I have a form that grabs all photos with a certain GALLERY_id attached to it. The backend user can then change the title of the photo and change the tags. After submitting the form the query should update all rows. Here is what I have so far which does not doe anyting:
THE FORM
if(isset($_GET['id']))
{
$id=$_GET['id'];
$result = $db->prepare("SELECT * FROM photos WHERE gallery_id = :gallery_id ");
$result->bindParam(':gallery_id', $id);
$result->execute();
echo '<form action="" method="POST">';
echo "<ul id='photos'>";
for ($i = 0; $row = $result->fetch(); $i++)
{
$id = $row['id'];
$title = $row['title'];
$tags = $row['tags'];
$src = $row['src'];
echo "<li><a class='lightbox' href='images/$src'><img src='images/$src' id='$id' alt='$title' /></a><br />";
echo "<input type='text' name='photo_title' value='$title' /><br />";
echo "<input type='text' name='photo_tags' value='$tags' />";
echo "<input type='hidden' name='photo_id' value='$id' />";
echo "</li>";
}
echo "</ul>";
}
?>
<div style="clear:both"></div>
<input type="submit" name="changeTitle" value="Save"/>
</form>
UPDATE QUERY
if (isset($_POST['changeTitle']))
{
foreach ($_POST as $p)
{
$id=$p['photo_id'];
$title=$p['photo_title'];
$tags=$p['photo_tags'];
$sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
$query = $db->prepare($sql);
$query->execute(array($title, $tags, $id));
}
}
Since you have multiple html fields with same names, you have to submit them as an arrays:
echo "<input type='text' name='photo_title[]' value='$title' /><br />";
echo "<input type='text' name='photo_tags[]' value='$tags' />";
echo "<input type='hidden' name='photo_id[]' value='$id' />";
After submitted, loop through any array variable like
foreach ($_POST['photo_id'] as $key => $photo_id) {
$id = $photo_id;
$title = $_POST['photo_title'][$key];
$tags = $_POST['photo_tags'][$key];
$sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
$query = $db->prepare($sql);
$query->execute(array($title, $tags, $id));
}
I have searched all over this website, but not yet found the answer for this. Pr maybe I am not able to apply it correctly. I have a form that grabs all photos with a certain GALLERY_id attached to it. The backend user can then change the title of the photo and change the tags. After submitting the form the query should update all rows. Here is what I have so far which does not doe anyting:
THE FORM
if(isset($_GET['id']))
{
$id=$_GET['id'];
$result = $db->prepare("SELECT * FROM photos WHERE gallery_id = :gallery_id ");
$result->bindParam(':gallery_id', $id);
$result->execute();
echo '<form action="" method="POST">';
echo "<ul id='photos'>";
for ($i = 0; $row = $result->fetch(); $i++)
{
$id = $row['id'];
$title = $row['title'];
$tags = $row['tags'];
$src = $row['src'];
echo "<li><a class='lightbox' href='images/$src'><img src='images/$src' id='$id' alt='$title' /></a><br />";
echo "<input type='text' name='photo_title' value='$title' /><br />";
echo "<input type='text' name='photo_tags' value='$tags' />";
echo "<input type='hidden' name='photo_id' value='$id' />";
echo "</li>";
}
echo "</ul>";
}
?>
<div style="clear:both"></div>
<input type="submit" name="changeTitle" value="Save"/>
</form>
UPDATE QUERY
if (isset($_POST['changeTitle']))
{
foreach ($_POST as $p)
{
$id=$p['photo_id'];
$title=$p['photo_title'];
$tags=$p['photo_tags'];
$sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
$query = $db->prepare($sql);
$query->execute(array($title, $tags, $id));
}
}
Since you have multiple html fields with same names, you have to submit them as an arrays:
echo "<input type='text' name='photo_title[]' value='$title' /><br />";
echo "<input type='text' name='photo_tags[]' value='$tags' />";
echo "<input type='hidden' name='photo_id[]' value='$id' />";
After submitted, loop through any array variable like
foreach ($_POST['photo_id'] as $key => $photo_id) {
$id = $photo_id;
$title = $_POST['photo_title'][$key];
$tags = $_POST['photo_tags'][$key];
$sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
$query = $db->prepare($sql);
$query->execute(array($title, $tags, $id));
}
I'm trying to make a quiz application that shows answers with radio buttons on the sides. When you press the next button a set of new answers will appear and replace the .
I've managed to make four questions pop up as intended and four new ones popup when I press the next button.
Right now theres one problem, my first set of four answers (with qid = 1) does not dissapear, which is weird since the other set of answers with qid = 2 and 3 does and replace eachother whenever I press the next button.
How do I make it so that the new answers appear and replace the old answers?
Here is my code so far PHP:
$qid1 = 1;
$sql1 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid1'");
while($row=mysqli_fetch_assoc($sql1))
{
echo "<input type='radio' name='answer1' value='".$row['Point']."'>"
.$row['answer'] ."<br>";
}
echo "<input type='submit' name='forward1' value='next'>";
$qid2 = 2;
$sql2 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid2'");
while($row2=mysqli_fetch_assoc($sql2)){
if (isset($_POST['forward1'])) {
echo "<input type='radio' name='answer2' value='".$row2['Point']."'>"
.$row2['answer'] ."<br>";
}
}
echo "<input type='submit' name='forward2' value='next'>";
$qid3 = 3;
$sql3 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid3'");
while($row3=mysqli_fetch_assoc($sql3)){
if (isset($_POST['forward2'])) {
echo "<input type='radio' name='answer3' value='".$row3['Point']."'>"
.$row3['answer'] ."<br>";
}
}
echo "<input type='submit' name='forward3' value='next'>";
You need to separate your inputs with form tags. For each of your loops do something like this..
echo "<form>";
$sql3 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid3'");
while($row3=mysqli_fetch_assoc($sql3)){
if (isset($_POST['forward2'])) {
echo "<input type='radio' name='answer3' value='".$row3['Point']."'>"
.$row3['answer'] ."<br>";
}
}
echo "<input type='submit' name='forward3' value='next'>";
echo "</form>";
Try this..
<?php
$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) || die("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');
// let's put the qid in a session var
session_start();
$qid = isset($_SESSION['qid']) ? $_SESSION['qid']+1 : 1;
$_SESSION['qid'] = $qid;
ob_start();
echo "<form>";
$sql1 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid'");
while($row1=mysqli_fetch_assoc($sql1))
echo "<input type='radio' name='answer1' value='{$row1['Point']}'>{$row1['answer']}<br>";
echo "<input type='submit' name='forward1' value='next'>";
echo "</form>";
$output = ob_get_clean();
?>
?><!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
</head>
<body>
<?php echo $output; ?>
</body>
</html>
Hi i have a simple form that uploads files and i want to get the id value using $POST and put in the database but my code is incorrect. Here is my code. I just want to ask if im doing the passing and fetching of $POST correctly? thanks
echo "<form action='process.php' method='post' enctype='multipart/form-data' id='uploadfile'>";
echo "<input type='hidden' name='MAX_FILE_SIZE' value='100000' />";
echo "<select name='selectedValue'>";
echo '<option selected="selected">' .'Choose a User'. '</option>';
foreach ($registeredUsers as $key => $value) {
$registered = JFactory::getUser($value);
echo '<option name="id" value="'.$registered->id.'">'.$registered->name.'</option>';
}
echo "</select>";
echo "Choose a file to Upload:";
echo "<input name='uploadedfile' type='file' /> <br/>";
echo '<input type="submit" name="submit" value="Upload">';
echo "</form>";
Here is the process.php file.
if(isset($_POST['selectedValue']))
{
$selectedValue = $_POST['id'];
$db = JFactory::getDbo();
$query = $db->getQuery(true);
$query = "INSERT INTO sample_table ('user_id') VALUES ('$_POST[id]')"
$db->setQuery($query);
$result = $db->execute();
}
echo "<form action='process.php' method='post' enctype='multipart/form-data' id='uploadfile'>";
echo "<input type='hidden' name='MAX_FILE_SIZE' value='100000' />";
echo "<select name='id'>";
echo '<option selected="selected">' .'Choose a User'. '</option>';
foreach ($registeredUsers as $key => $value) {
$registered = JFactory::getUser($value);
echo '<option value="'.$registered->id.'">'.$registered->name.'</option>';
}
echo "</select>";
echo "Choose a file to Upload:";
echo "<input name='uploadedfile' type='file' /> <br/>";
echo '<input type="submit" name="submit" value="Upload">';
echo "</form>";
process.php
if(isset($_POST['id']))
{
$selectedValue = $_POST['id'];
$db = JFactory::getDbo();
$query = $db->getQuery(true);
$query = "INSERT INTO sample_table ('user_id') VALUES ('$selectedValue')";
$db->setQuery($query);
$result = $db->execute();
}
Okay. So you had some problems, you've set the "name" on the options, which is wrong, you have to set that on the .
Second, you missed a semicolon at the end on $query.
And last, i've chaged to check if the a value has been chosed from the dropdown and to set selectedValue to that, and use that in the query.
I am fetching the data from the MySQL table in a while loop, and inserting the form data into another MySQL table in action page in foreach, but I do not get the correct value of the radio button, I am attaching my code, please help.
<?php
$i = 1;
$j = 1;
while ($row = mysqli_fetch_array($questions)) {
?>
<div class="control-group">
<label class="control-label" for="focusedInput">(<?php echo $i; ?>)
<?php
$questionid = $row['question_id'];
$question = $row['question'];
?>
<input type="hidden" name="questionid[]" value="<?php echo $questionid; ?>" />
<input type="hidden" name="question[]" value="<?php echo $question; ?>" />
<?php echo $row['question']; ?></label>
<div class="controls">
<?php
if ($row['answer_type'] == "Ratings") {
echo "
<p>
Low<input type='radio' name='rating$i' value='1' id='rating_0'>
<input type='radio' name='rating$i' value='2' id='rating_1'>
<input type='radio' name='rating$i' value='3' id='rating_2'>
<input type='radio' name='rating$i' value='4' id='rating_3'>
<input type='radio' name='rating$i' value='5' id='rating_4'>High
</p>
";
$i++;
} else if ($row['answer_type'] == "Comments") {
echo "<textarea name='answer[]' cols='' rows=''></textarea>";
$j++;
}
echo "<br />";
?>
</div>
</div>
<?php } ?>
Action File Code
foreach($_POST['questionid'] as $key=>$questionid){
$questionid = $_POST['questionid'][$key];
$answer = $_POST['answer'][$key];
$result3 = mysqli_query($con, "select question,answer_type from questions where question_id=$questionid;");
while($row = mysqli_fetch_array($result3)) {
$question = $row['question'];
$answer_type = $row['answer_type'];
if($answer_type == "Comments") {
$query2 = "insert into review_details (review_id,survey_id,question_id,question,answer_freeresponse) values(1,$_SESSION[surveyid],$questionid,'$question','$answer')";
$result2 = mysqli_query($con,$query2);
if(!$result2) {
echo mysqli_error($result2);
}
}
else if($answer_type == "Ratings") {
$query2 = "insert into review_details (review_id,survey_id,question_id,question,answer_rating) values(1,$_SESSION[surveyid],$questionid,'$question',$key)";
$result2 = mysqli_query($con,$query2);
if(!$result2) {
echo mysqli_error($result2);
}
}
}
$i++;
}
Output
I want to store the ratings displayed as a radio button, I guess its taking the counter incremented as variable $key, I don't know how to store the values of the radio button.
I guess you will get your radio button value as,
$ratingKey = "rating".$key;
$rating = $_POST[$ratingKey];
Than use $rating in your insert query instead of $key.
INSERT into review_details (review_id,survey_id,question_id,question,answer_rating)
values(1,$_SESSION[surveyid],$questionid,'$question',$rating)