Add to Database based on Checkbox

Beginners - Learning PHP
#1

Hi guys, i’m a total noob, trying desperately to create a little script just for a fun project in the hope to learn along the way. It’s probably very basic for most out there, but for me it is far from it. So what i have at the moment is simply information from a database displayed with check boxes beside it. What i would like is to add “DiscNo” to MYTABLE if the check box has been selected.

My Code so far :
[php]

<?php $con=mysqli_connect("localhost","kslc_admin","password","kslc_cdg"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT DISTINCT DiscNo, Description FROM CDG ORDER BY DiscNo"); echo " Toggle All
"; echo ""; while($row = mysqli_fetch_array($result)) { echo ""; echo ""; echo ""; echo ""; echo ""; echo ""; echo ""; } echo "
Tag Description DiscNo View Songs
" . $row['Description'] . "" . $row['DiscNo'] . "" . $row['TrackNo'] . "View Tracks
"; mysqli_close($con); ?>[/php]

I have absolutely no idea how to go about this, so any help at all will be greatly appreciated.
Thanks

#2

When you say “ADD DiscNo to MyTable” what do you mean?

You want to allow people to INSERT new discs? Or UPDATE Existing Discs?

Either way, I don’t see where you give people the opportunity to insert or update information in the HTML you provided.

You’re going to need some html Input fields.

#3

Hey thanks for your reply, sorry for poor information, i guess i can’t really explain what i don’t understand. However since posting i have been digging and experimenting.
Basically my code produces a list of Disc Numbers (DiscNo) with a checkbox beside each. What i want is to be able to let people choose 1 or more discs by check box and somehow add those to a column on a db table. I may have got a little further but not sure

[php]

<?php $con=mysqli_connect("localhost","kslc_admin","password","kslc_cdg"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT DISTINCT DiscNo, Description FROM CDG ORDER BY DiscNo"); echo " Toggle All
"; echo ""; echo ""; while($row = mysqli_fetch_array($result)) { echo ""; echo ""; echo ""; echo ""; echo ""; echo ""; echo ""; } echo "
Tag Description DiscNo View Songs
" . $row['Description'] . "" . $row['DiscNo'] . "" . $row['TrackNo'] . "View Tracks
"; echo ""; echo ""; mysqli_close($con); ?>[/php]

Now what happens is that a url is passed to post.php containing the checkbox name (which is also the DiscNo) it looks like this

http://mydomain.com/post.php?SF155=on&SF163=on&SF179=on&submit=Submit+

Where SF155, SF163 & SF179 are the checkbox names which i wish to save to a column on my db table.

I’m not sure if i am going the right direction and really stumped about how to advance… Do i make any sense ?

And no i don’t wish to allow people to insert new discs, basically i am allowing them to create their own list from my existing database of discs.

#4

Ok you’re heading in the right direction…

What is the table you’re saving it to and what are the column names?

#5

Awesome … thank the lord !!

Ehm the table is “members” and the column is “discno”

Thanks again.

#6

I need a little more info

Table Name: Members
Column Name: DiscNo

I’m sure you have more then just 1 column in this table.

#7

Yeh there are more columns but i didn’t think they were relevant “id” “usr” “discno” “pass” “email” …

But really all i need to know is how to pass the checkbox name into “discno”
Obvioulsy i plan to adapt with a WHERE condition to detect the session id of the signed in user but i think i can handle that bit myself … i hope lol

#8

Well it makes a big difference…

Because, from what I’m seeing now from what you shared it looks like you have your database set up incorrectly…

Table Name: Members
Column Name: DiscNo
Column Name: ID
Column Name: Usr
Column Name: Pass
Column Name: email

This tells me that you are going to store the information Comma separated? Which means if you ever wanted to do a query to see how many people choose DiscNo 5 you’re going to have speed optimization issues. It also effects your insert query and update queries, because you’re going to allow people to change their selection?

So I propose you break this out into two tables linked by the User ID.

Table Name: Members
Column Name: ID
Column Name: Usr
Column Name: Pass
Column Name: email

Table Name: Members_DiscNo_Selection
Column Name: User_ID
Column Name: DiscNo
Primary Key (User ID and DiscNo)

Now the database is in 3rd Normal Form and you have a proper relationship with the right constraints on the table.

So populated it will look like this

User ID, DiscNo
1 5
1 7
2 1
2 2
2 3

Then your select queries will need to change to pull back the users selection…

In order to help you, I needed to know how you’re storing the data. Without knowing that I can’t point you in the right direction.

#9

I get where, you are coming from !!! I can see the logic behind that !!

Not realising how important this was i have actually changed my membership script, which creates different tables etc … Would it make sense to provide you with a full copy of my database ?

There’s nothing sensitive in it …

#10

No, I don’t need your database…

I’m not coding it for you, but just trying to help you along where you get stuck, break things into mini-steps, accomplish those mini-steps and move forward. If you have a problem on a mini-step ask for help. People are going to offer more help on a small issue vs a big issue.

Re-designing your database was the first big step, now that you’re done with that, it’s just code.

#11

No i wouldn’t expect you to code it for me at all …I greatly appreciate you have gone over and above already !!! I understand your advise on the database and have redesigned it accordingly, i now know what to do and available documentation is easily enough understood to work this out, but the part i can’t get my head around was the intial bit where i capture the DiscNos from the passed URL after hitting the submit button.

[php]http://mydomain.com/post.php?SF155=on&SF163=on&SF179=on&submit=Submit+[/php]

#12

This

[php]echo “

<input type=“checkbox” name=”. $row[“DiscNo”] ." id=“checkbox”>";
[/php]

Should be set to an array using the []

[php]echo “

<input type=“checkbox” name=”. $row[“DiscNo”] ." id=“checkbox[]”>";[/php]

Then cycle through the results and add them to you database.

[php]<?php

foreach($_POST[‘checkbox’] as $discno) {
echo “discno= $discno
”;
//Do you insert statement here.
}

?>[/php]

#13

Thanks again for your help, i have added the third piece of code to post.php but i haven’t as yet written the INSERT statement as it stands post.php displays this error :

Warning: Invalid argument supplied for foreach() in /home/kslchost/public_html/post.php on line 2

Now maybe i’m just being stupid in thinking that i can run that code on it’s own, i know it won’t do anything without an Insert statement but should it show that error as it is ?

Thanks again

#14

Disregard my last … Having spent sometime checking through my code i found a space which was causing the problem on line 20 just after post before "

[php]echo "<form action=“post.php” method=“post " />”;[/php]

No i’m getting a list of discs on post.php like so

discno= SF176
discno= SF177
discno= SF178
discno= SF179

Is this what i should expect ?

#15

[php]echo “

<input type=“checkbox” name=“checkbox[]” value=”. $row[“DiscNo”] ." id=“checkbox[]”>";[/php]

PHP cares about the input name not the id. And you probably want the value to show what was selected.

#16

Hmmm that conflicts with what Topcoder advised … However i do appreciate your response. The direction which Topcoder provided is now working for me perfectly with the following code …

discs.php
[php] $result = mysqli_query($con,“SELECT DISTINCT DiscNo, Description FROM cdg ORDER BY DiscNo”);
echo “

<input type=“checkbox” onClick=“toggle(this)” /> Toggle All
”;
echo “<form action=“post.php” method=“post” />”;
echo "";

while($row = mysqli_fetch_array($result))
{
echo “

”;
echo “";
echo “”;
echo “”;
echo “”;
echo “”;
echo “”;
}
echo “
Tag Description DiscNo View Songs
<input type=“checkbox” value=”. $row[“DiscNo”] ." name=“checkbox[]”>” . $row[‘Description’] . “” . $row[‘DiscNo’] . “” . $row[‘TrackNo’] . “<a href=“tracks.php?DiscNo=” . $row[“DiscNo”] . “”>View Tracks
”;
echo "<input type=“submit” name=“submit” value=“Submit " />”;
echo “”;

mysqli_close($con);[/php]

post.php

[php]<?php
include(“include/session.php”);
$con=mysqli_connect(“localhost”,“kslc_admin”,“password”,“kslc_cdg”);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST[‘checkbox’]))
{
foreach($_POST[‘checkbox’] as $discno) {
echo “discno= $discno
”;

$result = mysqli_query($con,“INSERT INTO members_discno_selection (DiscNo, userid) VALUES (’$discno’, ‘$session->username’)”);
if($result){
echo “Successfully Added
”;
}
else{
echo “Query failed”;
}
}
}
?>[/php]

The only thing i need to do is prevent it from creating duplicate entries … Should i open a new thread ?

#17

No you should just change your INSERT statement, notice the IGNORE Keyword? This will prevent the duplicates from happening.

Change:

[php]$result = mysqli_query($con,“INSERT INTO members_discno_selection (DiscNo, userid) VALUES (’$discno’, ‘$session->username’)”);
[/php]

To:

[php]$result = mysqli_query($con,“INSERT IGNORE INTO members_discno_selection (DiscNo, userid) VALUES (’$discno’, ‘$session->username’)”);[/php]

#18

You truly are awesome !! Thank you once again !!

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