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POAdding form data to mysql database PHP and AJAX
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  1. POAdding form data to mysql database PHP and AJAX
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    copied!<p>I am wanting to have a form that will let me add names to my database without refreshing the form. I have been working on it but it does not seem to work. I am quite new at this so any help is appreciated. </p> <p>For my index.html I have:</p> <pre><code> &lt;script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"&gt; &lt;/script&gt; &lt;script type="text/javascript" &gt; $(function() { $(".submit").click(function() { var name = $("#firstname").val(); var username = $("#lastname").val(); var dataString = 'firstname='+ firstname + 'lastname=' + lastname if(firstname=='' || lastname=='') { $('.success').fadeOut(200).hide(); $('.error').fadeOut(200).show(); } else { $.ajax({ type: "POST", url: "join.php", data: dataString, success: function(){ $('.success').fadeIn(200).show(); $('.error').fadeOut(200).hide(); } }); } return false; }); }); &lt;/script&gt; &lt;body&gt; &lt;form method="post" name="form"&gt; &lt;ul&gt;&lt;li&gt; &lt;input id="firstname" name="firstname" type="text" /&gt; &lt;/li&gt;&lt;li&gt; &lt;input id="lastname" name="lastname" type="text" /&gt; &lt;/li&gt;&lt;/ul&gt; &lt;div &gt; &lt;input type="submit" value="Submit" class="submit"/&gt; &lt;span class="error" style="display:none"&gt; Please Enter Valid Data&lt;/span&gt; &lt;span class="success" style="display:none"&gt; Registration Successfully&lt;/span&gt; &lt;/div&gt;&lt;/form&gt; </code></pre> <p>For join.php:</p> <pre><code>&lt;?php include("db.php"); if($_POST) { $firstname=$_POST['firstname']; $lastname=$_POST['username']; mysql_query("INSERT INTO persons (firstname,lastname) VALUES('$firstname','$lastname')"); } ?&gt; </code></pre> <p>and db.php:</p> <pre><code>&lt;?php $mysql_hostname = "localhost"; $mysql_user = "root"; $mysql_password = ""; $mysql_database = "test"; $bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Could not connect database"); mysql_select_db($mysql_database, $bd) or die("Could not select database"); ?&gt; </code></pre>
 

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