[英]Making a secure connection to a SQL database in php
我試圖確保數據庫憑據的安全,但是它們始終顯示在Firebug控制台中。
我只想將表單信息保存到數據庫中。
我的表格(form.html)
<form id="Form" method="POST" >
<table>
<tr>
<td>Name:</td>
<td><input id="name" name="name" type="text" /></td>
</tr>
<tr>
<td>Address:</td>
<td><input id="address" name="address" type="text" /></td>
</tr>
<tr>
<td>Telephone:</td>
<td><input id="telephone" name="telephone" type="text" /></td>
</tr>
<tr>
<td>Email:</td>
<td><input id="email" name="email" type="email" /></td>
</tr>
</table>
<div class="SubmitButton">
<input type="submit" id="submit" name="submit" value="Submit" />
</div>
</form>
<div id="Response"></div>
</div>
<!-- javascript to handle form data -->
<script type="text/javascript">
$("document").ready(function(){
var frm = $('#Form');
frm.submit(function (ev) {
ev.preventDefault();
var email = $('#email').val();
if(email)
{
$.ajax({
type: "POST",
dataType:"json",
url: "submit.php",
data: frm.serialize(),
success: function () {
$('#Response').empty();
$('#Response').wrapInner("<span class='SuccessMessage'>Your information has been submitted successfully!</span>")
return;
},
error: function () {
$('#Response').empty();
$('#Response').wrapInner("<span class='ErrorMessage'>Your information was not submitted successfully. Please try again.</span>")
}
});
} else {
$('#Response').empty();
$('#Response').wrapInner("<span class='EmailBlankMessage'>Email address is a required field.</span>")
return;
}
});
});
</script>
這是我的submit.php
<?php
if($_POST)
{
require(dirname(__FILE__)."/../config.php");
$name = $_POST["name"];
$address = $_POST["address"];
$telephone = $_POST["telephone"];
$email = $_POST["email"];
$name = mysql_real_escape_string($name);
$address = mysql_real_escape_string($address);
$telephone = mysql_real_escape_string($telephone);
$email = mysql_real_escape_string($email);
mysql_query("INSERT INTO XXXXX(name, address, telephone, email) VALUES('$name', '$address', '$address', '$email')");
}else {
header("Location: /Form.html");
die();
}
?>
這是我的config.php
<?
$hostname = "XXXXXX";
$database = "XXXXXX";
$username = "XXXXXX";
$password = "XXXXXX";
mysql_connect($hostname, $username , $password) or die (mysql_error());
mysql_select_db($database ) or die (mysql_error());
配置文件位於根目錄外部。 但是我遇到的2個主要問題是,它不會將表單數據插入表中,而且config.php中的數據庫憑據也會顯示在firebug控制台的響應中。 我不明白我的代碼有什么問題。
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