Am puzzled about how to make a system that can automatically mark.
Am using a variable $count in PHP and is the one storing data from the checked radio button.
I have not tried to check on how to do it and am just afraid of how to do it.
I have the following codes which out puts the data in radio buttons
<?php
$get=mysql_query("select * from courses where courseid='".$cid."'");
while ($picked=mysql_fetch_array($get)) {
echo "<b>The questions for ".$picked['coursename']." are:</b><br><br>";
}
$query=mysql_query("select * from papers where cid='".$cid."'");
if (mysql_num_rows($query)>0) {
$count=1;
while ($row=mysql_fetch_array($query)) {
echo "Qn ".$count.". ".$row['question']."<br><input type='radio' name='".$count."' value='A'>".$row['A']."<br><input type='radio' name='".$count."' value='B'>".$row['B']."<br><input type='radio' name='".$count."' value='C'>".$row['C']."<br><input type='radio' name='".$count."' value='D'>".$row['D']."<br><br>";
$count++;
}
}else{
echo "No papar has been set yet for this paper.";
}
?>
<input type="submit" name="but" value="Submit Assessment">
and it outputs the following:
And the database is like:
Related
I have been searching for help from various forums and similar posts, but without any progress.
I have three pages, one that lets me insert information about projects into my database, a second that show the images and names of every project in the database, and a third page which I want to have a function that shows the image and name of the selected project in the second page.
Code on the second page(dashboardadmin.php):
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
$conn = mysqli_connect("localhost","root","","wildfire");
if(mysqli_connect_errno())
{
echo mysqli_connect_error();
}
$sql= "SELECT pid, project_name, image, image_type FROM project";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
echo "<form action='omprojekt.php' method='post'>
<div id='comp' name='comp'>
<img src=pic.php?pid=".$row['pid']." width=100xp height=100xp/>"." ".$row['project_name']."
</div>
<input type='submit' name='submit' value='Choose' />
</form>";
}
}
else {
echo "0 results";
}
mysqli_close($conn);
?>
Code on the third page (omprojekt.php):
<?php
/* Tried both of the $val variables but of course only one at a time. This is only to show you what I have tried. */
$val = isset($_POST['comp']) ? $_POST['comp'] : '';
$val = $_POST['comp'];
if(isset($_POST['submit'])){
echo "$val";
}
?>
In the last code you can see that I have two $val variables, but I have only used one of them at a time in my codes. The purpose of showing both of them here is to show you that I have tried both of them.
What I want to do is to make the third page show the image and name of the selected project in the second page. As you see, I have tried to retrieve the content from the DIV using the same "name". The problem is that the third page(omprojekt.php) doesn't show any content at all, and not even any errors.
You're expecting the div to submit like an input, but it won't, because it's not an input. So put it in an input.
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
// Don't use and id attribute because you're in a loop and you might have multiple id's with the same value.
echo "<form action='omprojekt.php' method='post'>
<div>
<img src=pic.php?pid=".$row['pid']." width=100xp height=100xp/>"." ".$row['project_name']."
</div>
<input type='hidden' name='pid' value='".$row['pid']."'>
<input type='hidden' name='project_name' value='".$row['project_name']."'>
<input type='submit' name='submit' value='Choose' />
</form>";
}
}
Then on the next page,
$val = (isset($_POST['pid']) && isset($_POST['project_name'])) ?
"<img src=pic.php?pid={$_POST['pid']} width=100xp height=100xp/> {$_POST['project_name']}" : '';
For the sake of completeness, there are a few other things wrong with your code.
1) The width and height attributes on the iamge should have quotes, and do not accept "px", they are just numbers. If you want to use "px" you should use style instead. <img src='' style='width:20px; height:20px;' />
2) You should be escaping user input before running it through your query.
Data will only be sent from a <form> to the action script if it exists in an <input...> HTML tag. You cannot pick data out of randon <DIV> tags etc.
So you could do this by using a hidden input field like this ( this is only one way )
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
echo "<form action='omprojekt.php' method='post'>
<div>
<img src=pic.php?pid=".$row['pid'] .
" style="width:100px;height:100px" /> " .
$row['project_name']."
</div>
<input type='hidden' name='comp' value='" . $row['pid'] . "' />
<input type='submit' name='submit' value='Choose' />
</form>";
}
}
Now when you get to omprojekt.php the $_POST['comp'] variable will exist.
You can have as many hidden input fields as you like so if you want to pass the project_name as well just add another hidden field.
For the last 4 hours I've been struggling to get something to work. I checked SO and other sources but couldn't find anything related to the subject. Here is the code:
<?php
$email=$_SESSION['email'];
$query1="SELECT * FROM oferte WHERE email='$email'";
$rez2=mysql_query($query1) or die (mysql_error());
if (mysql_num_rows($rez2)>0)
{
while ($oferta = mysql_fetch_assoc($rez2))
{
$id=$oferta['id_oferta'];
echo "<input type='radio' name='selectie' value='$id' id='$id'> <a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a>";
echo "</br>";
}
echo "</br>";
//echo "<input type=\"button\" id=\"cauta\" value=\"Vizualizeaza\" onclick=\"window.location.href='oferta.php?id={$oferta['id_oferta']}'\" />";
//echo " <input type=\"button\" id=\"cauta\"value=\"Modifica\" onclick=\"window.location.href='modifica.php?id={$oferta['id_oferta']}'\" />";
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
echo "</form>";
echo "</div>";
}
else
{
}
?>
The while drags all of the user's entries from the database and creates a radio button for each one of them with the value and id (because I don't really know which one is needed) equal to the entry's id from the db. I echoed that out and the id is displayed as it should so no problems there.
The delete script works ok as well so I won't attach it unless you tell me to. All good, no errors, until I try to delete an entry. Whatever I choose from the list of entries, it will always delete the last one. Note that I have two other inputs echoed out, those will be the "view" and "modify" buttons for the entry.
I really hope this is not JavaScript related because I have no clue of JS. I think this will be of major help to others having this problem. Please let me know if I need to edit my question before downrating. Thanks!
After edit:
This is the delete script, which as I said earlier works fine.
<?php
if (isset($_GET['id']))
{
$id = $_GET['id'];
echo $id;
require_once('mysql_connect.php');
$query = "DELETE FROM oferte Where id_oferta = '$id'";
mysql_query($query) or die(mysql_error());
//header('Location: oferte.php');
}
else
{
//header('Location: oferte.php');
}
?>
The session is started as well, like this:
<?php
session_start();
?>
The reason the last $id is deleted is because this line is outside/after the while loop:
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
You want to move this line inside the loop so that you have a button that executes delete for each radio button.
Update:
To have links to delete and
echo "<input type='radio' name='selectie' value='$id' id='$id'> ";
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a> ";
echo "<a href='delete.php?id=$id'>delete</a>";
Also I do not think the radio button is needed here at all since you are not really doing anything with it. You could simply echo out the value of your choice and have these links as follows:
echo $oferta['denumire_locatie'] . ' '; // replace $oferta['denumire_locatie'] with something of your choice
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a> ";
echo "<a href='delete.php?id=$id'>delete</a>";
echo "<br />";
The problem, in this case, is JavaScript related, yes. What I recommend you to do is to simply add a Remove link for each item.
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a>";
echo " - <a href='delete.php?id={$oferta['id_oferta']}'>Remove</a>";
echo "</br>";
Your $id is outside your while() loop.
The last one is getting deleted because the $id has the last one's value when the loops is exited.
Include all your code :
echo "</br>";
//echo "<input type=\"button\" id=\"cauta\" value=\"Vizualizeaza\" onclick=\"window.location.href='oferta.php?id={$oferta['id_oferta']}'\" />";
//echo " <input type=\"button\" id=\"cauta\"value=\"Modifica\" onclick=\"window.location.href='modifica.php?id={$oferta['id_oferta']}'\" />";
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
Inside your while loop.
When the rendered html reaches the browser, it will be something like this:
<input type='radio' name='selectie' value='1' id='1'> <a href='oferta.php?id=1'>TEXT</a>
<input type='radio' name='selectie' value='2' id='2'> <a href='oferta.php?id=2'>TEXT</a>
<input type='radio' name='selectie' value='3' id='3'> <a href='oferta.php?id=3'>TEXT</a>
<input type='radio' name='selectie' value='4' id='4'> <a href='oferta.php?id=4'>TEXT</a>
<br/>
<input type="button" id="sterge" value="Sterge" onclick="window.location.href='delete.php?id=5'" />
With this markup you won't be able to accomplish what you want without using javascript to update the onclick attribute whenever you select a radio button.
On the other hand, instead of using the client-side onclick event you can use the button's default behaviour, which is to submit the form.
You'll just have to set the action attribute:
<form method="post" action="http://myurl.php">
and write the myurl.php page which will just read the posted variable $_POST['selectie'] and call the delete method with the posted id.
I am working on a solution that will help me submit specific images from a list of images to a MySQL Database.
My database consists of the following:
Database
id(INT)
photo(BLOB)
caption(VAR)
Code
I am first retreiving images from a list and giving them each a submit button.
foreach ($media->data as $data) {
echo $pictureImage = "<img src=\"{$data->images->thumbnail->url}\">";
echo "<form action='tag.php' method='post'>";
echo "<input type='submit' name='submit' value='Click Me'>";
echo "</form>";
}
$pictureImage parses the data URL and then puts it into an actual image.
The submit button is below each of those images.
I am then making it so that when the submit button is pressed, it is added to the database.
if(isset($_POST['submit'])) {
//Database code would be above the following
$sql="INSERT INTO $usertable (image) VALUES ('$pictureImage')";
}
Problem
I am running into an issue where the last image in my list is the one being submitted to the database, rather than the image with the corresponding submit button. How do I make it so that it is grabbing the photo with the corresponding submit button?
Any help would be appreciated greatly.
You need to include any identifier to the image inside the form in order to store it.
For example you can try building the forms like this:
foreach ($media->data as $data) {
echo $pictureImage = "<img src=\"{$data->images->thumbnail->url}\">";
echo "<form action='tag.php' method='post'>";
echo "<input type='hidden' name='imageid' value='{$data->images->thumbnail->url}'>";
echo "<input type='submit' name='submit' value='Click Me'>";
echo "</form>";
}
And when you want to store the image on the form submit you can actually pick up the identifier of the image (in this case I used the URL you posted):
if(isset($_POST['submit'])) {
$sql="INSERT INTO $usertable (image) VALUES ('$_POST[imageid]')";
}
You are not posting any data when you click on your submit buttons.. i would suggest that for each form you would have a hidden field with the url of the image.
something like this:
<form action='tag.php' method='post'>
<input type="hidden" value="{$data->images->thumbnail->url}" name="pic"/>
<input type='submit' name='submit' value='Click Me'>
</form>
Problem
foreach ($media->data as $data) {
echo $pictureImage = "<img src=\"{$data->images->thumbnail->url}\">";
echo "<form action='tag.php' method='post'>";
echo "<input type='submit' name='submit' value='Click Me'>";
echo "</form>";
}
By the time this loop is done, you will have the last image in the loop as the value for $pictureImage.
So when it gets to this point, you're $pictureImage is still... the last image.
if(isset($_POST['submit'])) {
//Database code would be above the following
$sql="INSERT INTO $usertable (image) VALUES ('$pictureImage')";
}
Solution
I'm not exactly sure what data value you want to save because right now it looks like the whole tag. But whatever it is, you need to put it into a form field first ...
foreach ($media->data as $data) {
..
echo "<form action='tag.php' method='post'>";
echo "<input type='hidden' name='imageurl' value='<img src=\"{$data->images->thumbnail->url}\">' />";
..
}
and then retrieve it in the POST part of your script:
if(isset($_POST['submit'])) {
$imageurl = $_POST['imageurl'];
//Database code would be above the following
$sql="INSERT INTO $usertable (image) VALUES ('$imageUrl')";
}
This way, it won't always be the last value in your list, rather, it will be the one you selected.
I'm new to PHP-to-Mysql so I'm using the mysql_* function which others have stated to be unsupported anymore. But I find this place the best to ask my question since I can't find anyone with a similar question.
So basically, my problem is how to make a dropdown list stay which is inside an IF/ELSE block. When it is passes a value (submit) through another IF/ELSE block, it basically disappears so I'm trying a code where I don't have to re-write the entire dropdown code inside the next IF/ELSE block. Since what I'm trying is to show a textbox depending on what option is selected inside the dropdown (it has an auto-submit function). Is this possible? Or I have to resort to other libraries (jQuery,javascript etc.)?
Here's a part of the code if it helps:
elseif($_POST['question'] == edit ){
$res = mysql_query("SELECT questionID FROM questions");
echo "<form action='' method='post'>
Edit Question No.
<select name='question_select' onchange=this.form.submit()>
<option value=null selected>--</option>";
while($row = mysql_fetch_array($res))
{
echo "<option value=\"".$row['questionID']."\">".$row['questionID']."</option>";
}
echo "</select></form>";
}
//TEXTBOX TO BE DISPLAYED WHEN A NUMBER IS SELECTED IN EDIT EXISTING
if (isset($_POST['question_select'])){
$res = mysql_query("SELECT question FROM questions WHERE questionID='{$_POST['question_select']}'");
$row = mysql_fetch_assoc($res);
echo "Editing Question ",$_POST['question_select'],"<br>
<form action='' method='post'>
<textarea name='edited_question' rows='4' cols='50'>",$row['question'],"</textarea><br>
<input type='submit' name='save' value='Save'>
<input type='hidden' name='question_num' value='{$_POST['question_select']}'>
<input type='submit' name='cancel' value='Cancel'>
</form>";
}
//PASSING OF VALUES WHEN SAVE IS PRESSED
if (isset($_POST['save'])){
$edited_question = trim($_POST['edited_question']);
mysql_query("UPDATE questions SET question='$edited_question' WHERE questionID='{$_POST['question_num']}'") or die(mysql_error);
header("Location:admin_questions.php");
}
use jquery for that.it is easy and simple to use.
for depanding dropdown,jquery code is:
$(document).ready(function(e) {
$("#abc").change(function()
{
var firstfeild= $("#firstfeild").val();
$.post("getdata.php",{"firstfeild":firstfeild},function(data)
{
$("#abc").html(data);
});
});
});
and in getdata.php, just write mysql query.
function renderDropdown($res)
{
echo "
<select name='question_select' onchange=this.form.submit()>
<option value=null selected>--</option>";
while($row = mysql_fetch_array($res))
{
echo "<option value=\"".$row['questionID']."\">".$row['questionID']."</option>";
}
echo "</select>";
}
if(1==1)
{
echo "<form action='' method='post'>Edit If Question No.";
renderDropdown($res);
echo "</form>";
}
else
{
// $someOtherRes
echo "<form action='' method='post'>Edit Else Question No.";
renderDropdown($someOtherRes);
echo "</form>";
}
i have few answers to quiz stored in db. i am displaying them as radio buttons on form like this
$row = mysql_fetch_array(mysql_query('select * FROM quiz order by rand() limit 1'),MYSQL_ASSOC);
<?php
foreach ($row as $key => $value) {
echo "<input type='radio' name='answer'>".$value."</input><br />";
}
?>
but when i post this form the echo on other end is not getting a radio button answer i selected. please help
echo "<input type='radio' name='answer' value='".$value."'>".$value."</input><br />";
The only things that I can see is that
a) There is no form tag outside of the radio
b) You aren't giving the radio a value such as:
<input type='radio' name='answer' value="1"> Male
<input type='radio' name='answer' value="2"> Female