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POHow to show data from database without refreshing the page using jquery, PHP
 

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  1. POHow to show data from database without refreshing the page using jquery, PHP
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    <p>Is there anyway to load the newly inserted rows to the database without refreshing the page using jQuery? I am able to send the data to the database without refreshing using jquery but I am stuck at showing that data back without refreshing the page. How do I display the data from the table without refreshing page and make it appear under the table in index.php which I have to display the retrieved data? Thanks </p> <p>This is my index.php</p> <pre><code>&lt;html&gt; &lt;head&gt; &lt;script src="jquery.js" type="text/javascript"&gt;&lt;/script&gt; &lt;script type="text/javascript"&gt; $(document).ready(function(){ $('#submit').click(function(){ $.ajax({ type: "POST", url: 'send.php', data: "user=" + $('#user').val() + "&amp;comment=" + $('#comment').val(), success: function(data){ $('input[type=text]').val('') $('#status').html(data); } }); }); }); &lt;/script&gt; &lt;/head&gt; &lt;body&gt; &lt;div&gt; &lt;input type="text" name="user" id="user"&gt; &lt;input type="text" name="comment" id="comment"&gt; &lt;input name="submit" type="submit" id="submit"&gt; &lt;div id="status"&gt;&lt;/diV&gt; &lt;/div&gt; &lt;br&gt; &lt;?php $con=mysqli_connect("localhost","root","","user"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM comments"); echo "&lt;table width='640' &gt; &lt;tr&gt; &lt;/tr&gt;"; while($row = mysqli_fetch_array($result)) { echo "&lt;tr &gt;"; echo "&lt;td style='vertical-align:text-top'&gt;" . $row['user'] . "&lt;/td&gt;"; echo "&lt;td&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;" . wordwrap($row['comment'], 90, '&lt;br&gt;',true) . "&lt;/td&gt;"; echo "&lt;/tr&gt;"; } echo "&lt;/table&gt;"; mysqli_close($con); ?&gt; </code></pre> <p>And here is my send.php which submits the data to the table.</p> <pre><code>&lt;?php $mysqli = new mysqli("localhost", "root", "", "user"); if ($mysqli-&gt;connect_errno) { echo "Failed to connect to MySQL: (" . $mysqli-&gt;connect_errno . ") " . $mysqli-&gt;connect_error; } $username = $_POST["user"]; $comment = nl2br($_POST['comment']); $stmt = $mysqli-&gt;prepare("INSERT INTO comments (user, comment) VALUES (?,?)"); $stmt-&gt;bind_param('ss', $username, $comment); $stmt-&gt;execute(); echo"comment posted successfully"; ?&gt; </code></pre>
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    2. COCheck the "related" sidebar -- there are many ajax examples. Your best bet is to understand how they work so that you can apply the same principles to your project.
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