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我希望我的页面在不刷新页面的情况下从 PHP 代码返回结果，但它会将我重定向到页面。一切对我来说似乎都很好，但我显然错过了一些东西。

HTML 代码：

<html>
  <body>
  <script src="http://code.jquery.com/jquery-1.10.1.min.js" type="text/javascript">    </script>
  <script src="my_script.js" type="text/javascript"></script>
      <form action="bet_process.php" method="POST" id="myForm">
  <h2 class="title">Betting Settings<span class="line"></span></h2>
  <div class="form-box">
 <label for="bet">Bet Amount</label>
    <input type="text" name="bet" id="bet" class="text" placeholder="Amount">
</div>
<div class="form-box">
<label for="pay">Multiplier </label>
<input type="text" class="text" name="pay" id="pay">
</div>
<div class="form-box last">
<label for="profit">Profit </label>
<input type="text" name="profit" id="profit" class="text" placeholder="Profit">
  </div><!-- End Box -->
<div class="clearfix"></div>
<div class="form-box">
<label for="chance">Win Chance (%)</label><input type="text" name="chance" id="chance" class="text" value="50" placeholder="Win % - 50.5% Default">
</div>

            <p>Slide to choose win chance or enter it in the input!</p><br><input type="range" id="chanceslider" class="vHorizon" step="0.01" min="0.01" max="98" style="background-color: #00aec8; width: 50%;">
</div>





 <div class="form-box">
    <input type="submit" name="start1" class="button medium color" value="Roll Dice" id="sub">
   </div><!-- End Box -->
<span id="result"></span>
</div>


</form>
</body>
</html>

PHP代码：

<?php

require 'db.php';
$uid = $_SESSION['uid'];
$rand = rand(100, 10000)/100;
$select_gg_amt = mysql_query("SELECT * FROM `users` WHERE `username` = '$uid'");
$select_gg_row = mysql_fetch_array($select_gg_amt);
$balance = $select_gg_row['balance'];
$amount1 = $_POST['profit'];
$amount2 = $_POST['bet'];
$time_ago = date("F j, Y, g:i a");
if(isset($_POST['start1'])) {

if(isset($_POST['bet'], $_POST['pay'], $_POST['profit'], $_POST['chance'])) {

   if($balance > 0 && $_POST['bet'] > 0) {

      if($_POST['bet'] <= $balance) {

   if($rand < $_POST['chance']) {

 echo '<h3>You rolled a <strong>' .$rand. ' </strong> out of 100 on the percentile dice!  You won!</h3>';
 $result = 'Win';

 mysql_query("UPDATE `users` SET `balance` = `balance` + '$amount1' WHERE `username` = '$uid'");
 mysql_query("INSERT INTO `bets`(`amount`, `time_ago`, `username`, `multiplier`, `roll`, `result`) VALUES ('$amount2', '$time_ago', '$uid', '{$_POST['pay']}', '$rand', '$result')");

 }

else if($rand > $_POST['chance']) { 
    echo '<h3>You rolled a <strong>' .$rand. '</strong> out of 100 on the percentile dice!  You lost...</h3>';
   $result =  'Loss';
   mysql_query("UPDATE `users` SET `balance` = `balance` - '$amount2' WHERE `username` = '$uid'");
   mysql_query("INSERT INTO `bets`(`amount`, `time_ago`, `username`, `multiplier`, `roll`, `result`) VALUES ('$amount2', '$time_ago', '$uid', '{$_POST['pay']}', '$rand', '$result')");

 }

}
else { echo '<h3>You can only bet an amount you\'re capable of paying for!</h3>'; }
}
else if($amount2 > 0) { echo '<h3>You need to have a <strong>balance</strong> greater than 0 to bet.  Sorry!</h3>'; }
}

} 

?>

JavaScript 代码：

$("#sub").click( function() {
$.post( $("#myForm").attr("action"), 
     $("#myForm :input").serializeArray(), 
     function(info){ $("#result").html(info); 
});
clearInput();
});

$("#myForm").submit( function() {
 return false;    
});

function clearInput() {
$("#myForm :input").each( function() {
   $(this).val('');
});
}
4

1 回答 1

1

你真的需要正确缩进你的代码，否则它几乎是不可读的。

将您的 javascript 替换为：

$(function() {
    $("#myForm").on('submit', function(e) {
        e.preventDefault();
        $.ajax({
            type: 'POST',
            url : $(this).attr('action'),
            data: $(this).serialize()
        }).done(function(info) {
            $('#result').html(info);
        }).fail(function(a,b,c) {
            console.log('an error occured');
            console.log(a);
            console.log(b);
            console.log(c);
        });
        $(":input", this).not('#sub').val('');
    });
});

表单的提交事件将是正确的事件，并且当单击提交按钮时触发它，您不需要按钮的事件处理程序。

我添加了一个fail()方法，因为我们很难测试你的 PHP 和数据库的东西，但至少如果它失败了你会收到一条错误消息？

编辑：

如果那是您的标记，就像在问题中一样，您缺少整个 HEAD 部分，并且脚本被加载到 BODY 部分。

您至少需要添加 HEAD 标签和标题。

于 2013-07-06T01:41:39.990 回答