update checkboxes after submit - php

I am currently running into an issue, where I have this form consisting of checkboxes. I get the values of user preferences for the checkboxes from a database. Everything works great, and does what is supposed to do, however after I change and check some boxes and then hit the submit button, it will still show the old values to the form again. If I click again in the page again it will show the new values.
The code is shown below with comments.
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk
FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name
FROM categories
INNER JOIN portals on categories.portal_id=portals.portal_id
ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>
<?php
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
if(isset($_POST['submit'])){
if(!empty($_POST['categories'])){
$cats= $_POST['categories'];
$result = mysqli_query($conn,$qry_del_usrcats); //delete all
for ($x = 0; $x < count($cats); $x++) {
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`)
VALUES ('".$_SESSION['user_id']."', '".$cats[$x]."');";
$result = mysqli_query($conn,$qry_add_usrcats);
}
echo "success";
}
elseif(empty($_POST['categories'])){ //if nothing is selected delete all
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
I am not sure what is causing to do that. Something is causing not to update the form after the submission. However, as i said everything works great meaning after i submit the values are stored and saved in the DB, but not shown/updated on the form. Let me know if you need any clarifications.
Thank you

Your procedural logic is backwards and you're doing a bunch of INSERT queries you don't need. As #sean said, change the order.
<?php
if(isset($_POST['submit'])){
if(isset($_POST['categories'])){
$cats= $_POST['categories'];
// don't do an INSERT for each category, build the values and do only one INSERT query with multiple values
$values = '';
for($x = 0; $x < count($cats); $x++) {
// add each value...
$values .= "('".$_SESSION['user_id']."', '".$cats[$x]."'),";
}
// trim the trailing apostrophe and add the values to the query
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`) VALUES ". rtrim($values,',');
$result = mysqli_query($conn,$qry_add_usrcats);
echo "success";
}
elseif(!isset($_POST['categories'])){ //if nothing is selected delete all
// you may want to put this query first, so if something is checked you delete all, so the db is clean and ready for the new data.
// and if nothing is checked, you're still deleting....
$qry_del_usrcats="DELETE FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name FROM categories INNER JOIN portals on categories.portal_id=portals.portal_id ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>

Typically this occurs due to the order of your queries within the script.
If you want to show your updated results after submission, you should make your update or insert queries to be conditional, and have the script call itself. The order of your scripts is fine, but you just need to do the following:
Take this query:
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
and put it inside the if statement so it looks like this:
if (isset($_POST['submit'] {
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
$result = mysqli_query($conn,$qry_del_usrcats);
[along with the other updates you have]
}
Also, you will need to move this entire conditional above the form itself; typically any updates, inserts, or deletes should appear year the top of the form, and then call the selects afterward (outside of the conditional)

Related

insert checkbox value into mysql database

I want to enroll the student and insert the student id into Mysql database after I check and submit the checkbox value, but I already tried so many ways but still cannot...
This is the php code
<?php
if (isset($_POST['submitxd'])) {
foreach ($_POST['enrol'] as $items) {
$insert = $link->query("INSERT INTO student_course(studentID) values ('$items')");}
}
?>
This is the html code
$result = $link->query("SELECT * FROM student WHERE programmeName = '$programme' AND intake = '$intake'");
while ($row = mysqli_fetch_array($result)) {
echo "<tr>
<td>".$row['studentID']."</td>
<td>".$row['studentName']."</td>
<td>".$row['studentGender']."</td>
<td>".$row['studentContact']."</td>
<td>
<input type='checkbox' name='enrol[]' value='".$row['studentID']."'>
</td>
</tr>";
}
check whether your array contains values or not:
echo "<pre>";
print_r($_POST['enrol']);
echo "</pre>";
if not, you should write html code properly i.e. check form tag and its action path carefully and before submitting the form, remember to check out the checkbox

How to GET URL parameters with PHP inside if(isset) statement for update form

I know you can get the URL parameters by using if(isset($_GET['id'])){
$id = $_GET['id']; and this works great, if I then close that 'if' statement and create a new one for the update/append-entry form on this same page (its the display page for each ind. database entry), I can't get the id that's been passed to the URL in this next if statement.
It is the if-statement that submits a new row ("condition") to my child table ("conditions") that has a foreign key that connects it to the parent table of health providers (the display page gets the provider's id from the search engine selection and displays their information, then has this "append entry" form at the bottom if someone wants to add a new health condition treated by this doctor.)
Everything works if I give it the right id number directly in my PHP ($id = 56), but not if I try to grab it from the URL INSIDE this second if statement ($id = $_GET['id'];).
if (isset($_GET['providerid']))
{
$providerid = $_GET['providerid']; /*THIS WORKS GREAT*/
$sql = "SELECT *, GROUP_CONCAT(DISTINCT conditions.condition_name
SEPARATOR ', ') AS all_conditions FROM `providers` INNER JOIN
`conditions` ON `providers`.`id` = `conditions`.`prov_id` WHERE
`prov_id`= $providerid";
$data = mysqli_query($connection, $sql) or die('error');
if(mysqli_num_rows($data) > 0){
$numresults = mysqli_num_rows($data);
while($row = mysqli_fetch_assoc($data)){
$providerid = $row['id'];
$providerfirstname = $row['provider_first_name'];
/*etc....*/
$conditions = $row['all_conditions'];
/*table displays provider info:*/
echo "<br><h1>".$providerfirstname." ".$providerlastname."
</h1>";
echo '<TABLE id="myTable" width="350px" border="1">';
//etc....(cutting out details)
echo '<tr><td><div id="myTable"><a href="#" id="addNew">Add+ .
</a></div></td><td><b>Conditions Treated:</b></td> .
<td>'.$conditions.'</td></tr>';
/*If user wants to add a new condition treated by provider that's
not listed:*/
echo '<tr><td></td><td>Add a new condition:</td><td><form
action="profilebackup.php" method="POST"><input type="text"
size="40" name="newcond" value="" placeholder="Add a Condition" /> .
<input type="submit" name="add1" value="Add"><br></td></tr>';
}
echo '</TABLE>';
}
else {
echo "0 results";
}
}
/*sends added conditions to child table, EVERYTHING WORKS EXCEPT
GETTING ID (WHICH WORKED ABOVE):*/
if(isset($_POST['add1'])){
$providerid = $_GET['providerid'];
$condition = mysqli_real_escape_string($connection,
$_POST['condition']);
$insql = "INSERT INTO `conditions` (condition_name, prov_id) VALUES
('$condition','$providerid')";
if(mysqli_query($connection, $insql)){
echo "Thank you! Your provider has successfully been submitted
to the database!";
}
else {
echo "Sorry, there was an problem submitting your provider to
the database." . $insql . mysqli_error($connection);
}
}
Everything works EXCEPT the GET function in the second if-statement. It returns error code that it does not have the correct foreign key constraint: "Cannot add or update a child row: a foreign key constraint fails" because it's not grabbing the id.
You are submitting a form via post to profilebackup.php but the handler for this form is trying to get the providerid. You need to send the providerid as a query string in the form's action if you want to be able to access it via $_GET.
<form method="post" action="profilebackup.php?providerid=<?php echo $providerid; ?>" ...
The proper way to do this would be to include a hidden input in the form and then access it via $_POST consistently.
<form method="post" action="profilebackup.php">
<input type="hidden" name="providerid" value="<?php echo $providerid; ?>">
...
</form>
And the handler code.
<?php
...
// Notice that now you can use $_POST consistently.
if (isset($_POST['add1'])) {
$providerid = $_POST['providerid'];
...
The problem lines in $sql statement where it take variable $providerid as string, not a php variable. $_GET works fine even if sending a different protocals such as POST, PUT, DELETE,... Something like this should work as expected:
$sql = "SELECT *, GROUP_CONCAT(DISTINCT conditions.condition_name
SEPARATOR ', ') AS all_conditions FROM `providers` INNER JOIN
`conditions` ON `providers`.`id` = `conditions`.`prov_id` WHERE
`prov_id`= " . $providerid ;

Pass value from one PHP file to another

I'm trying to pass a value which is input into a box on one Php page (itinerary.php) into another Php page ('submit.php') so it can, from there, be saved into a database. But I can't quite figure out how to get it across. I've tried using GET as you can see from code below, but I am already using a GET statement to receive and acknowledge another value from that very same page 'submit'. I guess I am overcomplicating it, but my knowledge of Php is still pretty limited at this stage so any ideas would be appreciated!
This is an extract from the itinerary.php file (it sits within a Bootstrap/Html framework. Note the entry which contains the input box for the sequence number).
<h3><br>YOUR ITINERARY</h3>
<?php
//Display contents of itinerary
if(!empty($_SESSION['itinerary'])){
//Retrieve details of each location in array from database
$query = "SELECT * FROM locations WHERE loc_id IN (";
foreach ($_SESSION['itinerary'] as $loc_id=>$value)
{$query.=$loc_id.',';}
$query = substr($query, 0, -1).')ORDER BY loc_id ASC';
$result = mysqli_query($db, $query);
echo'<table><tr><th colspan="5">LOCATIONS IN YOUR ITINERARY</th></tr>';
//Display locations in array
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$loc_id = $row['loc_id'];
echo"<br><td>{$row['loc_name']}</td></tr><br>
<td>{$row['loc_desc']}</td>
<td><p>Seq. No</p><input type=\"text\" size=\"3\" name=\"sequence\" value=????????>NOT SURE WHAT TO ADD INTO THIS LINE TO RETAIN THE INFO THAT IS INPUT</td>
<td><a href=remove_loc.php?value=$loc_id>Remove location</a><br></br></td>
</tr><br></table>";
}//while
print_r($_SESSION);
mysqli_close($db);
}//if
else {echo '<p><br>Your itinerary is empty.<br></p>';}
echo '<br><p>
<a href=submit.php?submit>Save itinerary</a>
<a href=clear_itin.php?clear>Clear itinerary</a>
Your details
Logout
</p>';
?>
And this is where I am trying to receive it and then use it in a SQL command to add to the database. (Again, this is within a Bootstrap framework)You can ignore the first SQL Insert statement as it is passing other info in successfully anyway..
<div class="row">
<div class="col-md-9">
<?php
if (isset($_GET['sequence'])){
$sequence = $_GET['sequence'];
}//if
if (isset($_GET['submit'])){
$query = "INSERT INTO itineraries (user_id, date_created) VALUES (".$_SESSION['user_id'].", NOW())";
$result = mysqli_query($db, $query);
$itinerary_id = mysqli_insert_id($db);
$retrieve_locs = "SELECT * FROM locations WHERE loc_id IN (";
foreach ($_SESSION['itinerary'] as $id=>$value)
{$retrieve_locs.=$id.',';}
$retrieve_locs = substr($retrieve_locs, 0, -1).')ORDER BY loc_id ASC';
$result = mysqli_query($db, $retrieve_locs);
//Store items in itin_locs db
while ($row = mysqli_fetch_array ($result, MYSQLI_ASSOC)){
//This is the command I have been trying to use, commented out. The second one below this works fine, but obv doesn't input a sequence number.
//$insert_locs = "INSERT INTO itin_loc (itinerary_id, loc_id, sequenceNo) VALUES ($itinerary_id, ".$row['loc_id'].", $sequence)";
$insert_locs = "INSERT INTO itin_loc (itinerary_id, loc_id) VALUES ($itinerary_id, ".$row['loc_id'].")";
$insert_result = mysqli_query($db, $insert_locs);
echo mysqli_error($db);
if ($insert_result === FALSE) {
die("Query failed!" . mysql_error() . $insert_locs);}
}//while
mysqli_close($db);
echo"<p>Your itinerary is saved!. Itinerary number is #".$itinerary_id."</p><br>";
$_SESSION['itinerary']= NULL;
echo '<p>
Your details
Logout
</p>';
}//if
else {echo '<p>Your itinerary is empty.<br></br></p>';}
?>
Use a form on your HTML page to hold all the input fields in the table.
Also instead of anchor "a" tag use a submit button to submit the form to other page.
Use some thing like:
Send value of submit button when form gets posted

POST method and arrays

This is my first php project. I have created a website where users can upload their picture and then view the pictures of other users, one person at a time (similar to the old hotornot.com). The code below works as follows:
I create an array (called $allusers) containing all members except for the user who is currently logged in ($user).
I create an array (called $usersiviewed) of all members who $user has previously either liked (stored in the likeprofile table) or disliked (stored in the dislikeprofile table). The first column of likeprofile and dislikeprofile has the name of users who did the liking/disliking, second column contains the name of the member they liked/disliked.
I use the array_diff to strip out $usersiviewed from $allusers. This is the list of users who $user can view (ie, people they have not already liked or disliked in the past).
Now the problem is when I click the like button, it updates the likeprofile table with the name of the NEXT person in the array (i.e., not the person who's picture I am currently looking at but person who's picture appears next). Additionally, if I refresh the current page, the person who's profile appears on the current page automatically gets 'liked' by me. I would really appreciate any advice on this.
<?php
// viewprofiles.php
include_once("header.php");
echo $user.' is currently logged in<br><br>';
echo <<<_END
<form method="post" action="viewprofiles.php"><pre>
<input type="submit" name ="choice" value="LIKE" />
<input type="submit" name ="choice" value="NEXT PROFILE" />
</pre></form>
_END;
$allusers = array();
//Create the $allusers array, comprised of all users except me
$result = queryMysql("SELECT * FROM members");
$num = mysql_num_rows($result);
for ($j = 0 ; $j < $num ; ++$j)
{
$row = mysql_fetch_row($result);
if ($row[0] == $user) continue;
$allusers[$j] = $row[0];
}
//Create the $i_like_these_users array, comprised of all users i liked
$result = queryMysql("SELECT * FROM likeprofile WHERE user='$user'");
$num = mysql_num_rows($result);
for ($j = 0 ; $j < $num ; ++$j)
{
$row = mysql_fetch_row($result);
$i_like_these_users[$j] = $row[1];
}
//Create the $i_dislike_these_users array, comprised of all users i disliked
$result = queryMysql("SELECT * FROM dislikeprofile WHERE user='$user'");
$num = mysql_num_rows($result);
for ($j = 0 ; $j < $num ; ++$j)
{
$row = mysql_fetch_row($result);
$i_dislike_these_users[$j] = $row[1];
}
//Create the $usersiviewed array, comprised of all users i have either liked or disliked
if (is_array($i_like_these_users) && is_array($i_dislike_these_users))
{
$usersiviewed = array_merge($i_like_these_users,$i_dislike_these_users);
}
elseif(is_array($i_like_these_users))
{
$usersiviewed = $i_like_these_users;
}
else
{
$usersiviewed = $i_dislike_these_users;
}
// this removes from the array $allusers (i.e., profiles i can view) all $usersviewed (i.e., all the profiles i have already either liked/disliked)
if (is_array($usersiviewed))
{
$peopleicanview = array_diff($allusers, $usersiviewed);
$peopleicanview = array_values($peopleicanview); // this re-indexes the array
}
else {
$peopleicanview = $allusers;
$peopleicanview = array_values($peopleicanview); // this re-indexes the array
}
$current_user_profile = $peopleicanview[0];
echo 'check out '.$current_user_profile.'s picture <br />';
if (file_exists("$current_user_profile.jpg"))
{echo "<img src='$current_user_profile.jpg' align='left' />";}
// if i like or dislike this person, the likeprofile or dislikeprofile table is updated with my name and the name of the person who liked or disliked
if (isset($_POST['choice']) && $_POST['choice'] == 'LIKE')
{
$ilike = $current_user_profile;
$query = "INSERT INTO likeprofile VALUES" . "('$user', '$ilike')";
if (!queryMysql($query)) echo "INSERT failed: $query<br />" . mysql_error() . "<br /><br />";
}
if (isset($_POST['choice']) && $_POST['choice'] == 'NEXT PROFILE')
{
$idontlike = $current_user_profile;
$query = "INSERT INTO dislikeprofile VALUES" . "('$user', '$idontlike')";
if (!queryMysql($query)) echo "INSERT failed: $query<br />" . mysql_error() . "<br /><br />";
}
?>
Because when you refresh page it sends previus value of
Form again...and problem when u like a user it being liked next user.. There there is something in yor for loop while fetching row ...insted of for loop try once while loop ...i hope it will solve ur problem
You are calculating the $iLike variable with the currently loaded user and then updating the database with that user.
You should probably change your application logic a bit:
pass the user ID of the user you liked or did not like as a POST parameter in addition to the like/didn't like variable
move the form processing logic to the top of your page (or better yet separate out your form processing from HTML display)
Also, it's best not to use the mysql_* extensions in PHP. Use mysqli or PDO.
Try to make two different forms. One with "LIKE", another with "NEXT" to avoid liking from the same form
When you submit your form - your page refreshes, so in string $current_user_profile = $peopleicanview[0]; array $peopleicanview doesn't have user from previuos page (before submitting) you have to attach it, e.g. in hidden field
<form method="post" action="viewprofiles.php">
<input type="hidden" name="current_user" value="$current_user_profile" />
<input type="submit" name ="choice" value="like" />
</form>
<form method="post" action="viewprofiles.php">
<input type="submit" name ="go" value="next" />
</form>
and INSERT it later
"INSERT INTO likeprofile VALUES" . "('$user', '".$_POST['current_user']."')"
ps remove <pre> from your form
Lets start by simplifying and organizing the code.
<?php
// viewprofiles.php
include_once("header.php");
//if form is sent, process the vote.
//Do this first so that the user voted on wont be in results later(view same user again)
//use the user from hidden form field, see below
$userToVoteOn = isset($_POST['user-to-vote-on']) ? $_POST['user-to-vote-on'] : '';
// if i like or dislike this person, the likeprofile or dislikeprofile table is updated with my name and the name of the person who liked or disliked
if (isset($_POST['like']))
{
$query = "INSERT INTO likeprofile VALUES" . "('$user', '$userToVoteOn ')";
if (!queryMysql($query))
echo "INSERT failed: $query<br />" . mysql_error() . "<br /><br />";
}
if (isset($_POST['dislike']))
{
$query = "INSERT INTO dislikeprofile VALUES" . "('$user', '$userToVoteOn ')";
if (!queryMysql($query))
echo "INSERT failed: $query<br />" . mysql_error() . "<br /><br />";
}
//now we can create array of available users.
$currentProfileUser = array();
//Create the $currentProfileUser array,contains data for next user.
//join the 2 other tables here to save php processing later.
$result = queryMysql("SELECT `user` FROM `members`
WHERE `user` NOT IN(SELECT * FROM `likeprofile` WHERE user='$user')
AND `user` NOT IN(SELECT * FROM `dislikeprofile` WHERE user='$user')
and `user` <> '$user'
LIMIT 1");
//no need for a counter or loop, you only need the first result.
if(mysql_num_rows > 0)
{
$row = mysql_fetch_assoc($result);
$current_user_profile = $row['user'];
}
else
$current_user_profile = false;
echo $user.' is currently logged in<br><br>';
//make sure you have a user
if($current_user_profile !== false): ?>
<form method="post" action="viewprofiles.php">
<input type="hidden" name="user-to-vote-on" value="<?=$current_user_profile?>" />
<input type="submit" name ="like" value="LIKE" />
</form>
<form method="post" action="viewprofiles.php">
<input type="hidden" name="user-to-vote-on" value="<?=$current_user_profile?>" />
<input type="submit" name ="dislike" value="NEXT PROFILE" />
</form>
check out <?=$current_user_profile?>'s picture <br />
<?php if (file_exists("$current_user_profile.jpg")): ?>
<img src='<?=$current_user_profile.jpg?>' align='left' />
<?php endif; //end check if image exists ?>
<?php else: //no users found ?>
Sorry, there are no new users to view
<?php endif; //end check if users exists. ?>
You'll notice I changed the code a lot. The order you were checking the vote was the main reason for the issue. But over complicating the code makes it very difficult to see what's happening and why. Make an effort to organize your code in the order you expect them to run rather a vote is cast or not, I also made an effort to separate the markup from the logic. This makes for less of a mess of code to dig through when looking for the bug.
I also used sub queries in the original query to avoid a bunch of unnecessary php code. You could easily have used JOIN with the same outcome, but I think this is a clearer representation of what's happening. Also please use mysqli instead of the deprecaded mysql in the future, and be aware of SQL injection attacks and makes use of real_escape_string at the very least.
Hope it works out for you. Also I didn't test this code. Might be a few errors.

PHP INSERT a variable number of records to mysql from a html form

I have a HTML form that retrieves a varying number product types that the user inputs stock figures. This data then needs to be INSERTED to a new table.
Here is the PHP query that populates the form.
require_once 'config.php';
$i = 1;
$sql = "SELECT * FROM dealer_product WHERE customer_code='$custcode' ORDER BY prod_code";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
$prodcode = $row['prod_code'];
echo "<tr><td><input type='text' name='prod".($i++)."' value='" . $prodcode . "'/></td><td><input type='number' name='openstock".($i++)."'/></td><td><input type='number' name='sold".($i++)."'/></td></tr>";
}
mysql_close($con);
?>
I know how to INSERT a set number of multiple records, but how do I INSERT a varying number of records?
Thanks in advance. Sorry for my basic knowledge, I'm a network admin not PHP MYSQL.
Name your input fields as if they were arrays, e.g.:
<input name="prods[0]" />
You can then output a variable number of inputs in your HTML, even add more with JavaScript. PHP will convert the input to an array over which you can iterate:
<?php
foreach ($_POST['prods'] as $prod) {
/* Process $prod */
}
?>
I recommend that you go in the same way but using
while ($row = mysql_fetch_assoc($result))
And now you have not numbers as index that is more complicated way to see things, better see associative way that's the column name from your table in mysql.
And you're doing a lot of increments there doing $i++ a lot of times. I don't know if you're doing that intentionally but if not just increment $i once.
Also the number of row can be reached using this:
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
Look inside the manual
http://us1.php.net/mysql_fetch_assoc

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