我找到了一个自动提交表单数据的教程,但我只想添加一个提交按钮以将数据传递给 ajax。
我的目标是拥有一个具有多个输入的表单,当用户单击提交按钮时,它通过 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相匹配。
我想我很接近。我已经搜索过,但没有找到我的确切解决方案。提前致谢。
<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> </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>
更新.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.');
}
?>
最佳答案
终于搞清楚了。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> </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>
更新.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.');
}
?>
关于javascript - 使用 Ajax、PHP、MYSQL 更新表单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24858331/