Question

我找到了一个自动提交表单数据的教程，但我想要做的就是添加一个提交按钮来将数据传递给ajax。

我的目标是拥有一个包含多个输入的表单，当用户单击提交按钮时，它会通过ajax发送它并更新页面而不重新加载页面。此外，另一个关键部分是它将所有输入发布到数组中的方式，以便在运行更新脚本时，输入字段中的名称属性与数据库中的列匹配。

我想我很亲密。我搜索过，但没有找到我的确切解决方案。提前谢谢。

<script type="text/javascript" src="/js/update.js"></script>

<form method="POST" action="#" id="myform">

    <!-- start id-form -->
    <table border="0" cellpadding="0" cellspacing="0"  id="id-form">
    <tr>
        <th valign="top">Business Name:</th>
        <td><input type="text" name="company_name" class="inp-form" /></td>
        <td></td>
    </tr>
    <tr>
        <th valign="top">Address 1:</th>
        <td><input type="text" name="address_1" class="inp-form" /></td>
        <td></td>
    </tr>
    <tr>
        <th valign="top">Address 2:</th>
        <td><input type="text" name="address_2" class="inp-form" /></td>
        <td></td>
    </tr>



<tr>
    <th>&nbsp;</th>
    <td valign="top">
            <input id="where" type="hidden" name="customer_id" value="1" />
            <button id="myBtn">Save</button>

<div id="alert">    
    </td>
    <td></td>
</tr>
</table>
<!-- end id-form  -->
</form>

update.js

var myBtn = document.getElementById('myBtn'); 
myBtn.addEventListener('click', function(event) {

updateform('form1'); }); 

function updateform(id){
        var data = $('#'+id).serialize();
       // alert(data);
         $.ajax({
            type: 'POST',
            url: "/ajax/update_company_info.php",
            data: data,
             success: function(data) {
                 $('#id').html(data);


                 $('#alert').text('Updated');
                 $('#alert').fadeOut().fadeIn();

              },
              error: function(data) { // if error occured
                    alert("Error occured, please try again");
                },
                        }); }

update_customer_info.php

<?php

include($_SERVER['DOCUMENT_ROOT'] . '/load.php');

// FORM: Variables were posted
if (count($_POST))
{
$data=unserialize($_POST['data']);
// Prepare form variables for database
foreach($data as $column => $value)
    ${$column} = clean($value);

// Perform MySQL UPDATE
$result = mysql_query("UPDATE customers SET ".$column."='".$value."'
    WHERE ".$w_col."='".$w_val."'")
    or die ('Error: Unable to update.');
}


?>

Answer 1

结束搞清楚。感谢大家的帮助。

<p id="alert"></p>    
<form id="form" method="post" action="/ajax/update_company_info.php">

    <!-- start id-form -->
    <table border="0" cellpadding="0" cellspacing="0"  id="id-form">
    <tr>
        <th valign="top">Business Name:</th>
        <td><input type="text" name="company_name" class="inp-form" /></td>
        <td></td>
    </tr>
    <tr>
        <th valign="top">Address 1:</th>
        <td><input type="text" name="address_1" class="inp-form" /></td>
        <td></td>
    </tr>
    <tr>
        <th valign="top">Address 2:</th>
        <td><input type="text" name="address_2" class="inp-form" /></td>
        <td></td>
    </tr>



<tr>
    <th>&nbsp;</th>
    <td valign="top">
            <input id="where" type="hidden" name="customer_id" value="1" />
            <input type="submit" value="Save" id="submit">

    </td>
    <td></td>
</tr>
</table>
<!-- end id-form  -->
</form>

update.js

$(document).ready(function() {

$('form').submit(function(evt) {
  evt.preventDefault();

   $.each(this, function() {
            // VARIABLES: Input-specific
            var input = $(this);
            var value = input.val();
            var column = input.attr('name');

            // VARIABLES: Form-specific
            var form = input.parents('form');
            //var method = form.attr('method');
            //var action = form.attr('action');

            // VARIABLES: Where to update in database
            var where_val = form.find('#where').val();
            var where_col = form.find('#where').attr('name');

  $.ajax({
      url: "/ajax/update_company_info.php",
      data: {
                        val: value,
                        col: column,
                        w_col: where_col,
                        w_val: where_val
      },
      type: "POST",
      success: function(data) {         

      $('#alert').html("<p>Sent Successfully!</p>");

                        }

  }); // end post
  });// end each input value
}); // end submit
}); // end ready

update_customer_info.php

    <?php

include($_SERVER['DOCUMENT_ROOT'] . '/load.php');

function clean($value)
{
    return mysql_real_escape_string($value);
}
// FORM: Variables were posted
if (count($_POST))
{

    // Prepare form variables for database
    foreach($_POST as $column => $value)
        ${$column} = clean($value);

    // Perform MySQL UPDATE
    $result = mysql_query("UPDATE customers SET ".$col."='".$val."'
        WHERE ".$w_col."='".$w_val."'")
        or die ('Error: Unable to update.');
}
?>

Answer 2

我认为你想在提交时更新表格。你应该这样做使用下面给出的按钮删除提交。
```
<button id="myBtn">Save</button>.
```

您应该在ur js文件中添加以下给定的代码。

var myBtn = document.getElementById('myBtn'); 
myBtn.addEventListener('click', function(event){ 
    Updateform('give id of the form'); 
}); 
function updateform(id){
    var data = $('#'+id).serialize();
    // alert(data);
    $.ajax({
        type: 'POST',
        url: "/ajax/update_company_info.php",
        data: data,
        success: function(data) {
            $('#id').html(data);
            // alert(data);
            //alert(data);
        },
        error: function(data) { // if error occured
            alert("Error occured, please try again");
        },
    });

您可以使用unserialize（）检索php代码中的输入值作为一个数组。所以你可以保存数据数据库和你想要的任何东西。我希望你得到答案。因此，你的代码将成为

   <form method="POST" action="#" id="form1">
    <!-- start id-form -->
   <table border="0" cellpadding="0" cellspacing="0"  id="id-form">
   <tr>
    <th valign="top">Business Name:</th>
    <td><input type="text" name="company_name" class="inp-form" /></td>
    <td></td>
</tr>
<tr>
    <th valign="top">Address 1:</th>
    <td><input type="text" name="address_1" class="inp-form" /></td>
    <td></td>
</tr>
<tr>
    <th valign="top">Address 2:</th>
    <td><input type="text" name="address_2" class="inp-form" /></td>
    <td></td>
</tr>
   <tr>
<th>&nbsp;</th>
<td valign="top">
        <input id="where" type="hidden" name="customer_id" value="1" />
    <button id="myBtn">Save</button>
</td>
<td></td> </tr> </table> <!-- end id-form  --> </form>

您的js代码变为

 var myBtn = document.getElementById('myBtn'); 
 myBtn.addEventListener('click', function(event)
{   Updateform('form1'); }); 
function updateform(id){
            var data = $('#'+id).serialize();
           // alert(data);
             $.ajax({
                type: 'POST',
                url: "/ajax/update_company_info.php",
                data: data,
                 success: function(data) {
                     $('#id').html(data);
               // alert(data);
                 //alert(data);
                  },
                  error: function(data) { // if error occured
                        alert("Error occured, please try again");
                    },
                            }); }

update_company_info.php将成为

       $data=unserialize($_POST['data']);
        // you can retrieve all values from data array and save all .

＆GT;

Answer 3

而不是：

$(".submit").click(function() {

为您的表单提供类似于＆＃39; myform＆＃39;：<form method="POST" action="#" id="myform">

的ID

并使用它来防止默认提交表单：

$("#myform").submit(function(e) {
 e.preventDefault();
 //your code
}

使用Ajax，PHP，MYSQL更新表单

3 个答案: