joshm101
Members-
Posts
16 -
Joined
-
Last visited
Recent Profile Visitors
The recent visitors block is disabled and is not being shown to other users.
joshm101's Achievements
Member (2/5)
0
Reputation
-
Can anyone help me re-write this? Thanks
-
The events table is separate. It is connected to the Calendar which is for just adding notes to the calendar, nothing special.
-
Please see the files I am using. https://www.dropbox.com/s/5kk18az68ic402v/Bowls Program.zip?dl=0 There is an exported MySQL file in there to use in phpmyadmin. Database will need to be created in phpmyadmin called bowls_program The navigation isn't properly coded as of yet. It just has dummy data. If you goto the bowlersinfo folder, my scripts are in there to use. My aim: To create a form that can input records into the database and can create, view, edit and delete data. I want an automated function for the Membership Status. I don't have the exact date as of yet so just anything in the future will do for testing purposes. When the date gets to e.g. 2019-07-20 the membership will reset back to non member and then once the bowlers purchase the new membership, we can update as we go. I would ideally like the SQL query to be automated to the same date every year. Hope this helps.
-
-
It resets for all members back to non member on the renewal date. There is no counting of days.
-
My background is 1 year study of PHP, MySQL and I am fairly new with the date functions. I am building a free project for a Lawn Bowling Club. I have a Table that shows the input data from the form fields and this is the area of the paid membership status. So when the date gets to (Sometime next year, just a dummy date was used) the date provided, I would like the status value to go back to Non Member, instead of still being financial in the database.
-
Here is where I am currently at: if ($dateTime > new DateTime('2019-06-26')) { if($stmt->rowCount() > 0) { $sql = "UPDATE bowlers_info SET status = 'Non Member'"; $stmt = $this->db->prepare($sql); $result = $stmt->execute(); } print "<option>Full Member</option>"; print "<option>Social Member</option>"; print "<option selected>Non Member</option>"; }
-
Sorry about that. I want the select input option changed in the database when the Date is equal to the date given.
-
I am a PHP newbie and need some help. I have a select input that requires some php date coding. Any help would be much appreciated. PLUS, I am in Australia so I'm not sure if there is any confliction when writing the date. <select style="height:100px;" name="status[]" id="status[]" multiple> <?php $statusstring = implode(",", $status); $status = $_SESSION['status']; if ($dateTime = new DateTime('2019-07-19')) { print "<option>Full Member</option>"; print "<option>Social Member</option>"; print "<option selected>Non Member</option>"; } else if ($status == 'Full Member') { print "<option selected>Full Member</option>"; print "<option>Social Member</option>"; print "<option>Non Member</option>"; } else if ($status == 'Social Member') { print "<option>Full Member</option>"; print "<option selected>Social Member</option>"; print "<option>Non Member</option>"; } else if ($status == 'Non Member') { print "<option>Full Member</option>"; print "<option>Social Member</option>"; print "<option selected>Non Member</option>"; } else { print "<option>Full Member</option>"; print "<option>Social Member</option>"; print "<option>Non Member</option>"; } ?> </select>
-
This is what I have: <select name="message_date" id="message_date"> <option value="0">Select a Date</option> <?php date_default_timezone_set('Australia/Melbourne'); $sunday= date("N"); $sunday=7 - $sunday; $sunday_increment = new DateInterval('P'.$sunday .'D'); $sevendays_increment = new DateInterval('P7D'); $d0 =new DateTime(); $d0->add($sunday_increment); for($i=date('W');$i<53;$i++) { $sundate= $d0->format("Y-m-d"); $d0->add($sevendays_increment); print "<option value='$sundate'>$sundate</option>"; } ?> </select>
-
Ok I am pretty new with databases, I am currently writing a PHP/MySQL program that: 2 tables, a form with date input from a select. I need data from both tables. Here is my SQL Query: $sql = "SELECT sd.sponsor_id, sd.sp_name FROM sponsor_details as sd INNER JOIN sponsor_message as sm on sd.sponsor_id = sm.sponsor_id WHERE message_date = 'Whatever the date from the form input'"; This is where the problem is. I don't know how to get a form data match with an SQL record. Any help would be much appreciated thanks.
-
Thanks maxxd, I wasn't knocking your logic I just didn't give you enough information.(My fault). However I do have another question. Please see my scripts attached. Every time I seem to access the database through xampp when running a script, it logs me out of my logged in session. I can't find the problem. Any help would be much appreciated thanks. https://www.dropbox.com/sh/gabmonzk0rbawhm/AAAZPJIFJPV9aM-yUGMPfitoa?dl=0
-
the data is coming from a different php script. So therefore I thought to get the data from another page of a form. Using SESSION works for me.
-
I have a College Project which is to create a backend system that can manipulate the logged in users information. I wanted to have update functionality as an admin to manipulate the users details. Therefore the hashed password was giving me troubles. If I update the users details when the password field is empty, I don't want it to run the SQL wiith the password field. But if there is something in that field, I would like to execute the sql statement with the hashed password field.
-
Sorry guys, found the error. I needed to add: if(empty($_SESSION['password'])). I am new here, please be nice