I have made a yes/no quiz where the game tries to guess what creature you are thinking of based on these questions, there are only 16 possible answers.
What I am trying to is make it so that if the user closes the site part-way through the quiz then when they return to the quiz they will taken to the same question they left off on. Below is what I have done so far but it doesn't work properly (see for yourself at http://s504518.brunelweb.net/Creatures.php) You must answer yes to the first question and then close the browser as I have only done a session for that question so far. Here is part of the what I have done so far
//These are used for saving the session
$q2 = "<div class='questions'><p>{$questions[0][0]}</p></div>";
$q3 = "<div class='questions'><p>{$questions[0][1]}</p></div>";
$f2 = "<div class='questions'><form method ='GET' action='Creatures.php'>
<input type='submit' name='answer1' value='Yes' class='buttons' />
<input type='submit' name='answer1' value='No' class='buttons' />
</form></div>";
$f3 = "<div class='questions'><form method ='GET' action='Creatures.php'>
<input type='submit' name='answer2' value='Yes' class='buttons' />
<input type='submit' name='answer2' value='No' class='buttons' />
</form></div>";
//If start button has not been pressed, display nothing
if (!isset($_POST['start'])){
} //If start button has been pressed, display questions
else
{
//Display the first question
echo $firstquestion;
echo "<div class='questions'><form method ='GET' action='Creatures.php'>
<input type='submit' name='yes1' value='Yes' class='buttons' />
<input type='submit' name='no1' value='No' class='buttons' />
</form></div>";
}
//Question 2
if (isset($_SESSION['form']))
{
echo $_SESSION['question'];
echo $_SESSION['form'];
}
else{
if ($_GET['yes1']) //If answer to Q1 is yes then display this
{
echo "<div class='questions'><p>{$questions[0][0]}</p></div>";
showquestion(1);
$_SESSION['question'] = $q2;
$_SESSION['form'] = $f2;
}
}
What I am trying to achieve here is, if a session has already been set then display the question and form from that session and if it hasn't display nothing until they click the start button. I'm not sure why it doesn't work properly and any help would be much appreciated.
You need to create a session with session_start(), put this into the first line of php:
session_start();
As php.net tells me, it will also use already generated sessions. Just be shure you have no html-output before starting the session
Related
I have got a while loop that runs through all the records of the database printing them on a table. Now i also have some checkboxes within that very loop that I want to use to submit a form when clicked. Now when I click the checkbox it will indeed submit the form thanks to a Jquery script I found, BUT whenever i submit it it submits with the ID of the first record of the table.This Image
shows the table, as you see the first record has ID 34. Now every checkbox I click will send the $id 34.
This does not happen with normal submit buttons.
Is there a way I can submit with the individual userID's
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
I'm sorry if i'm not very clear with the explanation it is quite hard to explain this situation. Thank you guys!
Probably your problem is that you are submiting the same form always and its because you create a form for each row but it has the same id
For you the easy way is to put each form with the id cointaining the unique value of the row and doing submit with that.
Something like this
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete_<?php echo $id; ?>' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete_<?php echo $id; ?>").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
It looks like the <form> your creating has a static id, so ALL forms will have id='resComplete'. The jQuery submit function will grab the first element with id='resComplete' and submit it. You need to make it unique for every form and make the onchange='$("#resComplete").submit();' code match it.
Eg.
<?php
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete-<?php echo $id; ?>' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete-<?php echo $id; ?>").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
Better yet, use jQuery to find out what form it's in by chaning the onchange to something like:
<input id='complete' type='checkbox' name='complete' value='1' onchange='$(this).closest('form').submit();' <?php if($complete == 1){echo "checked";}?>>
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 having difficulty deleting rows from my database.
I have a delete button in a form that when it is clicked then performs a DELETE FROM query but it doesnt work and Im wondering whether my theory is completly wrong (the theory being that having a form and submit button to INSERT data into a database works so why not use that to delete stuff? This is the code
$league_id = $_GET['id'];
$delete_entry = "<form action=\"".$_SERVER["REQUEST_URI"]."\ method=\"post\">
<input type=\"submit\" name=\"ooops\" value=\"Delete Entries\"></p>
</form>";
if ($_POST['ooops']) { //if the data is rubbish then delete and start again...
$delete_lge_sql = "DELETE FROM st_position WHERE league_id = '$league_id'";
$delete_lge_res = mysqli_query($statto, $delete_lge_sql)
or die(mysqli_error($statto));
}
When I click Delete Entries the page reloads and the URL looks like this
page.php?ooops=Delete+Entries
Many thanks for any help
You missed a double-quote in this statement
$delete_entry = "<form action=\"".$_SERVER["REQUEST_URI"]."\ method=\"post\">
<input type=\"submit\" name=\"ooops\" value=\"Delete Entries\"></p>
</form>";
Change it to
$delete_entry = "<form action=\"".$_SERVER["REQUEST_URI"]."\" method=\"post\">
<input type=\"submit\" name=\"ooops\" value=\"Delete Entries\"></p>
</form>";
Try :
$delete_entry = "<form action=\"".$_SERVER["REQUEST_URI"]."\ method=\"post\">
<input type=\"hidden\" name=\"ooops\" value=\"1\" />
<input type=\"submit\" value=\"Delete Entries\">
</form>";
This way, your $_POST['ooops'] variable will be equal to "1". Just make your test on this value and it will be ok.
I have a PHP code which generates a dynamic list inside a form like the following, note that the list is built dynamically from database:
echo '<form name="List" action="checkList.php" method="post">';
while($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>';
$form++;
}
the below code works perfectly in FF and CHROME but not in IE. Please help. I have commented out my santize functions as i thought they might be affecting it, but it still does the same.... nothing in IE.
Thank you in advance for any assistance.
<?php
//IF UPDATE BUCKET CHANGE STATUS...
if(isset($_POST['updatebucket'])){
$complete = $_POST["complete"];
$bucketid = $_POST["bucketid"];
//$complete = sanitizeone($_POST["complete"], "plain");
//$complete = strip_word_html($complete);
//$bucketid = sanitizeone($_POST["bucketid"], "plain");
//$bucketid = strip_word_html($bucketid);
if ($complete=="1")
$complete = "0";
else
$complete = "1";
$updatebucket = "UPDATE membersbuckets SET complete = '$complete' WHERE userid = '$userid' AND bucketid = '$bucketid'";
mysql_query($updatebucket);
}
?>
and the front end....
<? if ($complete=="1") {
echo "<form action='' method='post' name='updatebucket'><input name='complete' type='hidden' value=" .$complete. " /><input name='userid' type='hidden' value=" .$userid. " /><input name='bucketid' type='hidden' value=" .$bucketid. " /><input type='image' name='updatebucket' value='updatebucket' src='images/tick.png' /></form>";
}else{
echo "<form action='' method='post' name='updatebucket'><input name='complete' type='hidden' value=" .$complete. " /><input name='userid' type='hidden' value=" .$userid. " /><input name='bucketid' type='hidden' value=" .$bucketid. " /><input type='image' name='updatebucket' value='updatebucket' src='images/cross.png' /></form>";
}
?>
Dan
You should post your front-end, not back-end (since it's pretty much not browser-dependant).
Your HTML probably isn't valid.
Edit:
Yep, IE doesn't take value for image type of input. It only sends the x & y (field_name_x, field_name_y) and totally discards the original "value" attribute.
Try with a hidden input instead.
It seems that input type='image' doesn't send the value when used from IE. You'll need another hidden field:
<input type='hidden' name='updatebucket' value='updatebucket' />
<input type='image' src='images/tick.png' />
That way, the updatebucket parameter will be posted to the server, regardless of the browser used.
The assumption here was that all browsers handle HTML forms the same way (and they don't); that's why I keep Eric Lawrence's excellent Fiddler around - it can diff two HTTP requests, so you'll see the difference between the browsers immediately.
An alternative would be to check for $_POST[{image-element-name}_x}] (in this case $_POST['updatebucket_x']. All browsers will send the x/y coordinates of the image element as updatebucket.x & updatebucket.y, and PHP silently (and frustratingly) alters the updatebucket.x to updatebucket_x. Then again, you only need this is clicking different input type=submit / type=image elements would alter the action taken, otherwise the previous solution of a hidden element as previously suggested would do.