javascript - How do I submit this PHP poll form on click of a radio button? -
i'm using css tricks' how design , create php powered poll tutorial create own poll.
i'm trying poll submit when user clicks 1 of radio button options, instead of submitting when click "vote" button.
poll.php:
<?php require_once('connections/conn_vote.php'); ?> <?php if (!function_exists("getsqlvaluestring")) { function getsqlvaluestring($thevalue, $thetype, $thedefinedvalue = "", $thenotdefinedvalue = "") { $thevalue = get_magic_quotes_gpc() ? stripslashes($thevalue) : $thevalue; $thevalue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($thevalue) : mysql_escape_string($thevalue); switch ($thetype) { case "text": $thevalue = ($thevalue != "") ? "'" . $thevalue . "'" : "null"; break; case "long": case "int": $thevalue = ($thevalue != "") ? intval($thevalue) : "null"; break; case "double": $thevalue = ($thevalue != "") ? "'" . doubleval($thevalue) . "'" : "null"; break; case "date": $thevalue = ($thevalue != "") ? "'" . $thevalue . "'" : "null"; break; case "defined": $thevalue = ($thevalue != "") ? $thedefinedvalue : $thenotdefinedvalue; break; } return $thevalue; } } $editformaction = $_server['php_self']; if (isset($_server['query_string'])) { $editformaction .= "?" . htmlentities($_server['query_string']); } if ((isset($_post["mm_insert"])) && ($_post["mm_insert"] == "form1")) { $insertsql = sprintf("insert poll (id, question) values (%s, %s)", getsqlvaluestring($_post['id'], "int"), getsqlvaluestring($_post['poll'], "text")); mysql_select_db($database_conn_vote, $conn_vote); $result1 = mysql_query($insertsql, $conn_vote) or die(mysql_error()); $insertgoto = "results.php"; if (isset($_server['query_string'])) { $insertgoto .= (strpos($insertgoto, '?')) ? "&" : "?"; $insertgoto .= $_server['query_string']; } header(sprintf("location: %s", $insertgoto)); } $colname_rs_vote = "-1"; if (isset($_get['recordid'])) { $colname_rs_vote = $_get['recordid']; } mysql_select_db($database_conn_vote, $conn_vote); $query_rs_vote = sprintf("select * poll id = %s", getsqlvaluestring($colname_rs_vote, "int")); $rs_vote = mysql_query($query_rs_vote, $conn_vote) or die(mysql_error()); $row_rs_vote = mysql_fetch_assoc($rs_vote); $totalrows_rs_vote = mysql_num_rows($rs_vote); ?> <!doctype html> <head> <title>poll</title> <script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script> </head> <body> <form action="<?php echo $editformaction; ?>" id="form1" name="form1" method="post"> <label> <input type="radio" name="poll" value="snoppdogg" id="poll_0" /> snoop dogg </label> <label> <input type="radio" name="poll" value="biggie" id="poll_1" /> biggie </label> <label> <input type="radio" name="poll" value="tupac" id="poll_2" /> tupac </label> <input type="submit" name="submit" id="submit" value="vote" /> <input type="hidden" name="id" value="form1" /> <input type="hidden" name="mm_insert" value="form1"> </form> <script type="text/javascript"> $('input[type=radio]').click(function() { $(this).closest("form").submit(); }); </script> </body> </html> <?php mysql_free_result($rs_vote); ?> conn_vote.php:
<?php # filename="connection_php_mysql.htm" # type="mysql" # http="true" $hostname_conn_vote = "localhost"; $database_conn_vote = "poll"; $username_conn_vote = "root"; $password_conn_vote = "root"; //$conn_vote = mysql_pconnect($hostname_conn_vote, $username_conn_vote, $password_conn_vote) or trigger_error(mysql_error(),e_user_error); $conn_vote = mysql_connect($hostname_conn_vote, $username_conn_vote, $password_conn_vote) or die('can\'t create connection: '.mysql_error()); mysql_select_db($database_conn_vote, $conn_vote) or die('can\'t access specified db: '.mysql_error()); ?> here's poll.php looks like:
this works great if click "vote", not if click radio button.
i tried adding onchange event each radio input, doesn't submit form either.
<input onchange="this.form.submit();" type="radio" name="poll" value="snoopdogg" id="poll_0" /> i have feeling has hidden inputs, can't figure out needs change.
any ideas on how submit form when user clicks radio button?
thanks in advance :)
edit:
i think it's php issue, not javascript issue. i've tried of these javascript solutions , none of them have worked:
<input onchange="this.form.submit();" type="radio" name="poll" value="snoopdogg" id="poll_0" /> and
<input onclick="this.form.submit();" type="radio" name="poll" value="snoopdogg" id="poll_0" /> and
$('input').click(function(){ $('form').submit(); }); and
$('input[type="radio"]').click(function() { $("form").submit(); }); and
$('input[type=radio]').click(function() { $(this).closest("form").submit(); }); and
$('input').click(function(){ $.ajax({ type:'post', url:'results.php', data:{radio:info} }).done(function(data){ alert("show ajax call successful!"); }); }); by "submit", mean want submit form, enter values database, , go results.php page, functionality of "vote" button.
use onclick instead of onchange.
<input onchange="this.form.submit();" /> you can use jquery submit form onclick:
$('input').click(function(){ $('form').submit(); }); and more slick, use ajax ;)
$('input').click(function(){ $.ajax({ type:'post', url:'somepage.php', data:{radio:info} }).done(function(data){ //show ajax call successful! }); }); edit: guess problem php code, since wasn't inserting values database. reason why form wasn't submitting though, because used onchange instead of onclick.
if want use ajax, need values input elements, in case radio inputs. not have submit form when use ajax.
for instance:
<input type="radio" id="#tupacradio" /> //get value of radio input var tupacradio = $('#tupacradio').val(); then once call ajax:
$.ajax({ data:{value:tupacradio} })
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