jungle Posted August 23, 2014 Share Posted August 23, 2014 I can not get the values from the javascript add row to go dynamically as a row into MySql only the form values show up as the form below as one row. I made it as an array, but no such luck, I have tried this code around a multitude of ways. I don't know what I am doing wrong, kindly write out the correct way. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Dynamic Fields js/php to MySql need to submit dynamically to the database</title> <?php require ('database.php'); ?> <script type="text/javascript"> var counter = 1; var collector = ""; function addfields(indx) { var tbl = document.getElementById('table_id'); var newtr = document.createElement('tr'); counter = counter + indx; newtr.setAttribute('id','tr'+counter); newtr.innerHTML = '<td><input type="checkbox" name="checkb'+counter+'" id="checkb'+counter+'" value="'+counter+'" onclick="checkme('+counter+')"></td><td><input type="text" name="text1[]"></td><td><textarea name="textarea1[]"></textarea></td>'; tbl.appendChild(newtr); } function checkme(dx) { collector += dx+","; } function deletetherow(indx) { var col = collector.split(","); for (var i = 0; i < col.length; i++) { var remvelem = document.getElementById('tr'+col); var chckbx = document.getElementById("checkb"+col); if(remvelem && chckbx.checked) { var tbl = document.getElementById('table_id'); tbl.removeChild(remvelem); } } } </script> </head> <body> <form enctype="multipart/form-data" id="1" style="background-color:#ffffff;" action="<?php echo $_SERVER['PHP_SELF']; ?>"></form> <table id="table_id" > <tr id="tr1" class="trmain"> <td> </td> <td> <input type="text" name="text1[]"> </td> <td> <textarea name="textarea1[]"></textarea> </td> </tr> </table> <input type="button" value="Add" onClick="addfields(1);" /> <input type="button" value="Delete" onClick="deletetherow()" /> <input type="submit" value="Send" id="submit" name="submit"/> <?php if(isset($_POST['submit'])) { for ($i=0; $i < count($_POST['text1']); $i++ ) { $ced = stripslashes($_POST['text1'][$i]); $erg = stripslashes($_POST['textarea1'][$i]); } $bnt = mysql_query("INSERT INTO tablename (first, second) VALUES ('$ced', '$erg')")or die('Error: '. mysql_error() ); $result = mysql_query($bnt); } ?> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/290621-dynamic-fields-jsphp-to-mysql-need-to-submit-dynamically-to-the-database/ Share on other sites More sharing options...
CroNiX Posted August 23, 2014 Share Posted August 23, 2014 Because you only tell it to insert 2 values: $bnt = mysql_query("INSERT INTO tablename (first, second) VALUES ('$ced', '$erg')")or die('Error: '. mysql_error() ); $result = mysql_query($bnt); That should probably be in your loop where you are retrieving $ced and $erg from $_POST. Quote Link to comment https://forums.phpfreaks.com/topic/290621-dynamic-fields-jsphp-to-mysql-need-to-submit-dynamically-to-the-database/#findComment-1488746 Share on other sites More sharing options...
CroNiX Posted August 23, 2014 Share Posted August 23, 2014 PS in the future please post code between [ code] //code here [ /code] tags Quote Link to comment https://forums.phpfreaks.com/topic/290621-dynamic-fields-jsphp-to-mysql-need-to-submit-dynamically-to-the-database/#findComment-1488747 Share on other sites More sharing options...
ginerjm Posted August 24, 2014 Share Posted August 24, 2014 Since this is a multi-post I'll let you get my answer from your other site posting. Same problem - same answer. Quote Link to comment https://forums.phpfreaks.com/topic/290621-dynamic-fields-jsphp-to-mysql-need-to-submit-dynamically-to-the-database/#findComment-1488784 Share on other sites More sharing options...
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