Javascript 如何在单击单选按钮时提交此PHP民意测验表单？
我正在使用CSS技巧教程创建我自己的投票 我试图让投票在用户单击单选按钮选项时提交，而不是在用户单击“投票”按钮时提交 poll.php:Javascript 如何在单击单选按钮时提交此PHP民意测验表单？,javascript,php,jquery,html,forms,Javascript,Php,Jquery,Html,Forms,我正在使用CSS技巧教程创建我自己的投票 我试图让投票在用户单击单选按钮选项时提交，而不是在用户单击“投票”按钮时提交 poll.php: <?php require_once('Connections/conn_vote.php'); ?> <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "
<?php require_once('Connections/conn_vote.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($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")) {
$insertSQL = sprintf("INSERT INTO poll (id, question) VALUES (%s, %s)",
GetSQLValueString($_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);
$query_rs_vote = sprintf("SELECT * FROM poll WHERE id = %s", GetSQLValueString($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);
?>
<?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());
?>
<!DOCTYPE html>
<html>
<head>
<title>Poll</title>
</head>
<body>
<form id="form">
<label> Snoop Dogg
<input type="radio" name="Poll" value="snoop" id="Poll_0" />
</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>
</form>
<script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script>
<script src="http://malsup.github.io/min/jquery.form.min.js" type="text/javascript"></script>
<script type="text/javascript">
$('input[type=radio]').click(function() {
$("#form").ajaxSubmit({url: 'send.php', type: 'post'})
$("#form").replaceWith("<h2>Thank you for your vote!</h2>");
});
</script>
</body>
</html>
投票
嗅探犬
大人物
图帕克
$（'input[type=radio]'）。单击（函数（）{
$（“#表单”）.ajaxSubmit（{url:'send.php'，键入：'post'}）
$（“#表格”）。替换为（“感谢您的投票！”）；
});
SEND.PHP
<?php
require('connection.php');
// Radio
$poll = $_POST['Poll'];
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function pollQuery($value) {
$sql = "";
if ($value == "snoop") {
$sql = "UPDATE poll SET snoop = snoop + 1 WHERE id = 1";
return $sql;
} elseif ($value == "biggie") {
$sql = "UPDATE poll SET biggie = biggie + 1 WHERE id = 1";
return $sql;
} else {
$sql = "UPDATE poll SET tupac = tupac + 1 WHERE id = 1";
return $sql;
}
}
if ( isset($poll) ) {
// Run Query
$conn->query( pollQuery($poll) );
}
$conn->close();
?>
CONNECTION.PHP
<?php
// Connect info
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "poll";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
?>
使用
onClickonChange<input onChange="this.form.submit();" />
onChangeonClick<input type="radio" id="#tupacRadio" />
//get the value of the radio input
var tupacRadio = $('#tupacRadio').val();
不要试图将单选按钮变成它不是的东西，而是使用三个普通按钮。@zzzzBov你可以做到。@zzzzBov我实际上打算最终将单选输入样式设置为传统按钮，所以这是一个很好的建议。但是，我之所以将它们保留为单选按钮，是因为我想让它们具有语义——因为民意测验是一种“选择一个答案”的形式，所以我认为单选按钮是正确的选择。你觉得呢？如果它看起来像一个按钮，而且它的行为像一个按钮，那么它就是一个按钮。我同意Ajax是解决这个问题的方法。你可以做很多很酷的事情，比如在表单提交后隐藏表单，然后显示统计数据。@Rafael我现在什么都用AJAX；）这当然是web开发的未来，特别是随着NoSQL数据库的创建，比如MongoDBThanks，为您提供答案。我尝试了所有这些解决方案，但不幸的是，在我完整的代码环境中，它们都不起作用。单击“投票”将值提交到数据库表中，但单击单选按钮不会。单击单选按钮时没有任何更改。也许我应该详细说明我所说的“提交”是什么意思——我正在寻找一个解决方案，提交表单并将值输入数据库，这是民意测验的预期功能。我认为这是一个PHP问题，因为我已经尝试了所有建议的jQuery/JavaScript脚本，但都没有效果。这是我在上面的HTML中已有的jQuery:$（'input[type=radio]'）。单击（function（）{$（this）.closest（“form”）.submit（）；}）；不幸的是，将“$（this）.closest（“form”）.submit（）；”更改为“$（“form”）.submit（）；”没有更改任何内容。@AnnaBlabber我的示例正在我提供的链接中运行。请检查您的代码并重试。我看到它在那里工作，但在完整文件的上下文中它对我不起作用。它不会在我的数据库表中输入任何值，而单击“投票”则会输入值。在尝试了许多“在单选按钮点击时提交表单”脚本的变体之后，没有任何效果，这似乎不是JavaScript的问题，而是PHP的问题。确保在PHP代码中没有使用
if（isset（$\u POST['SubmitName']）{//insert values in db}$('input[type=radio]').click(function() {
$(this).closest("form").submit();
});
$('input').click(function(){
$.ajax({
type:'POST',
url:'results.php',
data:{radio:info}
}).done(function(data){
alert("show that ajax call was successful!");
});
});
<!DOCTYPE html>
<html>
<head>
<title>Poll</title>
</head>
<body>
<form id="form">
<label> Snoop Dogg
<input type="radio" name="Poll" value="snoop" id="Poll_0" />
</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>
</form>
<script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script>
<script src="http://malsup.github.io/min/jquery.form.min.js" type="text/javascript"></script>
<script type="text/javascript">
$('input[type=radio]').click(function() {
$("#form").ajaxSubmit({url: 'send.php', type: 'post'})
$("#form").replaceWith("<h2>Thank you for your vote!</h2>");
});
</script>
</body>
</html>
<?php
require('connection.php');
// Radio
$poll = $_POST['Poll'];
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function pollQuery($value) {
$sql = "";
if ($value == "snoop") {
$sql = "UPDATE poll SET snoop = snoop + 1 WHERE id = 1";
return $sql;
} elseif ($value == "biggie") {
$sql = "UPDATE poll SET biggie = biggie + 1 WHERE id = 1";
return $sql;
} else {
$sql = "UPDATE poll SET tupac = tupac + 1 WHERE id = 1";
return $sql;
}
}
if ( isset($poll) ) {
// Run Query
$conn->query( pollQuery($poll) );
}
$conn->close();
?>
<?php
// Connect info
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "poll";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
?>
<input onChange="this.form.submit();" />
$('input').click(function(){
$('form').submit();
});
$('input').click(function(){
$.ajax({
type:'POST',
url:'somePage.php',
data:{radio:info}
}).done(function(data){
//show that ajax call was successful!
});
});
<input type="radio" id="#tupacRadio" />
//get the value of the radio input
var tupacRadio = $('#tupacRadio').val();
$.ajax({
data:{value:tupacRadio}
})