arpeggio Posted December 12, 2017 Share Posted December 12, 2017 (edited) Hi everyone! New to the group and also PHP coding. What I have so far is 2 radio buttons and when or the other is ticked, it then displays its hidden drop down box to choose values. The radio buttons work correctly and show the proper drop down boxes but do not write correct values to the database! I am trying to have the chosen radio box called "INQUIRY" write a value of "1" to "IncidentType" column. Or when the radio box called "COMPLAINT" is chosen, write a value of "2" to the same column. It seems to always write "1" to the DB column regardless which radio box is ticked. Also when the drop down box is chosen (which are populated from another DB column), I am looking for it to write the "complaint.ComplaintID" or the "inquiry.InquiryID" to the column "CaseTypeSubCategory" as the ID integer, which it does not. I'm really stuck on solving this and would appreciate any help someone would be able to offer. I think my insert.php page has the correct values but if anyone would like to see it I will be happy to post the code. Thank you in advance for replies! Here is my code so far for the page: <html> <head> <style> #ca { display: none; } </style> <script src="[url=https://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js]https://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js[/url]" type="text/javascript"></script> <script> $(document).ready(function() { $("input:radio").click(function() { $('.searchSelect').hide(); $('#' + $(this).attr("value")).show(); }); }); </script> </head> <body> <form action="insert.php" method="post"> <?php //Connect to our MySQL database using the PDO extension. $pdo = new PDO('mysql:host=10.150.100.194:3308;dbname=publiccommenttracking', '[color=#0000FF]Iputmyusernameandpassword[/color]', '[color=#0000FF]inapublicpostontheinternet[/color]'); //Our select statement. This will retrieve the data that we want. $sql = "SELECT complaint.ComplaintID, complaint.Description as Des, inquiry.InquiryID,Inquiry.Description as InquiryDescription FROM complaint left join inquiry on complaint.ComplaintID = inquiry.InquiryID"; //Prepare the select statement. $stmt = $pdo->prepare($sql); //Execute the statement. $stmt->execute(); //Retrieve the rows using fetchAll. $users = $stmt->fetchAll(); ?> <ul id="countrylist"> <li> <label for="US-country"> <input name="IncidentType" type="radio" id="US-country" value="com" checked /> INQUIRY</label><input type="hidden" name="IncidentType" value="1"> </li> <li> <label for="CA-country"> <input name="IncidentType" id="CA-country" type="radio" value="ca"/> COMPLAINT</label><input type="hidden" name="IncidentType" value="2"> </li> </ul> <select name="CaseTypeSubCategory" id="com" class="searchSelect" title="Search in"> <?php foreach($users as $user): ?> <option value="<?= $user['inquiry.InquiryID']; ?>"><?= $user['InquiryDescription']; ?></option> <?php endforeach; ?> </select> <select name="CaseTypeSubCategory" id="ca" class="searchSelect" title="Search in"> <?php foreach($users as $user): ?> <option value="<?= $user['complaint.ComplaintID']; ?>"><?= $user['Des']; ?></option> <?php endforeach; ?> </select> <p><input type="submit" value="Submit"><br> </form> </body> </html> Edited December 15, 2017 by Zane Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/ Share on other sites More sharing options...
requinix Posted December 12, 2017 Share Posted December 12, 2017 If you have a problem storing stuff in the database then you need to post the code that stores stuff in the database. At least the fields seem named appropriately. Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554645 Share on other sites More sharing options...
Barand Posted December 12, 2017 Share Posted December 12, 2017 If you want 1 or 2 depending on which radio button is selected then why don't you give them values of 1 and 2? Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554647 Share on other sites More sharing options...
arpeggio Posted December 12, 2017 Author Share Posted December 12, 2017 Sorry forgot to post my insert.php page: <html> <body> <?php $con = mysql_connect("xxxx","xxxx","xxxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("publiccommenttracking", $con); $sql="INSERT INTO publiccommenttracking.reportpartydetail (IncidentType, CaseTypeSubCategory ) VALUES ('$_POST[incidentType]','$_POST[CaseTypeSubCategory]'"; ?> <?php $query = "SELECT InquiryID,Description FROM inquiry ORDER BY InquiryID "; $result = mysql_query($query) or die(mysql_error()."[".$query."]"); if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554651 Share on other sites More sharing options...
Barand Posted December 12, 2017 Share Posted December 12, 2017 <option value="<?= $user['inquiry.InquiryID']; ?>"> The column names in the result set will NOT include the table name prefixes. Should be <option value="<?= $user['InquiryID']; ?>"> Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554652 Share on other sites More sharing options...
arpeggio Posted December 12, 2017 Author Share Posted December 12, 2017 Thank you Barand I'll give it a try! Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554653 Share on other sites More sharing options...
Barand Posted December 12, 2017 Share Posted December 12, 2017 At least the fields seem named appropriately. Only if you think it's a good idea to have two hidden field with the same name (and the values 1 and 2) as the radio buttons. Regardless of which button is selected, both these will be posted and the value of one of them will overwrite the other. Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554654 Share on other sites More sharing options...
arpeggio Posted December 12, 2017 Author Share Posted December 12, 2017 Barand I do see that is what's happening! Being new to PHP, do I take out the <input type="hidden" name="IncidentType" value="2"> completely or is there a way to make the radio's post either a "1" for one radio and a "2" for the other radio to the database depending on which one is chosen? Thanks again! Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554656 Share on other sites More sharing options...
Barand Posted December 12, 2017 Share Posted December 12, 2017 The hidden fields are wrong and shouldn't be there. As I said in my first reply, give the radio buttons values 1 and 2. Whichever is selected is the one that will be posted. If neither is selected then you won't get any value, so it may be useful to set one of them as "checked" as a default. Quote Link to comment https://forums.phpfreaks.com/topic/305905-form-radio-buttons-and-drop-down-boxes-not-writing-to-database/#findComment-1554658 Share on other sites More sharing options...
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