stevochegeoj Posted January 29, 2021 Share Posted January 29, 2021 Quote I have this code here. i wanted to know how i can be able to make multiple selections from the returned data from the database and how to save them Quote <select name="troom" multiple class="form-control" required> <option selected="true" disabled="disabled">--Select a room--</option> $records =mysqli_query($con, "select * from room where TRoom not in (select TRoom from roombook where cin <= '$checkout' and cout >= '$checkin');"); while ($data = mysqli_fetch_array($records)) { echo "<option value='" . $data['TRoom'] . "'>" . $data['TRoom'] . "</option>"; //displaying data in option menu } } ?> </select> Quote Link to comment Share on other sites More sharing options...
Barand Posted January 29, 2021 Share Posted January 29, 2021 Name the select "troom[]" so the options are posted as an array Then foreach ($_POST['troom'] as $room) { // save $room } Quote Link to comment Share on other sites More sharing options...
stevochegeoj Posted January 30, 2021 Author Share Posted January 30, 2021 i don't why its returning an empty list. or how will it display? i am learning php Quote Link to comment Share on other sites More sharing options...
Barand Posted January 30, 2021 Share Posted January 30, 2021 Example... <?php if ($_SERVER['REQUEST_METHOD']=='POST') { echo "<b>You selected</b><ul>"; foreach ($_POST['fruit'] as $fruit) { echo "<li>$fruit</li>"; // update database here } echo "</ul><hr>"; } ?> <!DOCTYPE html> <html lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <title>Example</title> </head> <body> <h1>Example</h1> <hr> <form method="POST"> Fruit: <br> <select name="fruit[]" multiple size="10"> <option>Apple</option> <option>Banana</option> <option>Date</option> <option>Kiwi</option> <option>Lemon</option> <option>Melon</option> <option>Orange</option> <option>Pear</option> <option>Quince</option> </select> <br> <input type="submit" name="btnSubmit" value="Submit"> </form> </body> </html> Quote Link to comment Share on other sites More sharing options...
stevochegeoj Posted January 30, 2021 Author Share Posted January 30, 2021 26 minutes ago, Barand said: Example... <?php if ($_SERVER['REQUEST_METHOD']=='POST') { echo "<b>You selected</b><ul>"; foreach ($_POST['fruit'] as $fruit) { echo "<li>$fruit</li>"; // update database here } echo "</ul><hr>"; } ?> <!DOCTYPE html> <html lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <title>Example</title> </head> <body> <h1>Example</h1> <hr> <form method="POST"> Fruit: <br> <select name="fruit[]" multiple size="10"> <option>Apple</option> <option>Banana</option> <option>Date</option> <option>Kiwi</option> <option>Lemon</option> <option>Melon</option> <option>Orange</option> <option>Pear</option> <option>Quince</option> </select> <br> <input type="submit" name="btnSubmit" value="Submit"> </form> </body> </html> my problem with this is, the option list is populated from the database. Do you mind implementing the code you posted in the quoted code? <select name="troom" multiple class="form-control" required> <option selected="true" disabled="disabled">--Select a room--</option> $records =mysqli_query($con, "select * from room where TRoom not in (select TRoom from roombook where cin <= '$checkout' and cout >= '$checkin');"); while ($data = mysqli_fetch_array($records)) { echo "<option value='" . $data['TRoom'] . "'>" . $data['TRoom'] . "</option>"; //displaying data in option menu } } ?> </select> Quote Link to comment Share on other sites More sharing options...
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