Question

大家好，我是一名新的网络开发学习者。

用户单击提交按钮后，我有一个带有表单的页面，然后按钮将表单内容提交到下一页执行并保存到数据库中。

这样，我的页面将转到另一个页面执行并返回到登录页面。

例如：index.php和exec.php

的index.php：

<form name="g-form" action="gbtn-exec.php" method="post" class="goat-vote" onsubmit="return validategForm()">
<input type="text" name="g-product" placeholder="Brand / Product Name" style="-moz-border-radius: 5px; border-radius: 5px; padding-left:20px; opacity:.5; border:none; margin-left:110px; width:440px; height:38px; font-family:'Proxima Nova Rg';color:#000; font:1.6em;" />


<p class="g-question">Why you love it?</p>

<textarea name="g-reason" style="-moz-border-radius: 5px; border-radius: 5px; padding:5px; opacity:.5; border:none; margin-left:110px; width:450px; height:150px; font-family:'Proxima Nova Rg';color:#333; font-size:1em;"></textarea>

<input name="g-btn" class="vote-btn" type="submit" value="vote" style="margin-left:470px; cursor:pointer;"></form>

exec.php

if ($_POST["g-product"] && $_POST["g-reason"] != "" )
{
$gproduct = $_POST["g-product"];
$greason =  $_POST["g-reason"];

$insert ="INSERT INTO jovine.vote (vote_id ,product_name ,reason ,type) VALUES (NULL, '$gproduct', '$greason', 'goat')";
$result = mysql_query($insert,$con);
echo "<script>";
echo "alert('Thank you. Your vote has been recorded.');";
echo "window.location.href='index.php';";
echo "</script>";
}

我的问题是，我如何使用ajax在后台运行exec.php？谢谢！

Answer 1

当您标记jQuery时 - 只需向exec.php文件发送HTTP请求：

$('.vote-btn').on('click', function() {
    $.ajax({
      url: "exec.php",
      data: { g-product: $('#g-product').val(), g-reason: $('#g-reasons').val() }
    }).done(function() {
      alert('Thank you. Your vote has been recorded.');
      window.location.href='index.php';
    });
});

如何使用ajax在后台运行执行页面？

1 个答案: