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Showing Mysql Rows Content In Html Table Using Php

I'm new here and I have a question, I looked for the solution everywhere and I still cant manage to solve this. I want to show the results of the SELECT statement (in php) in a tab

Solution 1:

In this line

echo "<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";

The final </td> is missing a >


Solution 2:

Everything is perfect in your script if you are sure about your mysql details. Please check your js files getting included in script.

<script src="http://code.jquery.com/jquery-2.1.4.min.js"></script>

Solution 3:

I removed jquery.js and inserted another script at head. It worked for me. I connected it to localhost. check it.

main.html

<html>
    <head>
        <meta charset="UTF-8"/>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
        <script src="select.js" type="text/javascript"></script>
    </head>
    <body>
    <button id="button"> Mostrar </button>
    <br>
    <input type="text" id="id" />
    <div id="content"></div>
    </body>
</html>

select.js

 $(document).ready(function(){
        $("#button").click(function () {

            function show_all() {
                $.ajax({
                    type: "POST",
                    url: "select.php",
                    data:{action:"showroom"},
                    success: function (data) {
                        $("#id").hide();
                        $("#content").html(data);
                    }
                });
            }

            show_all();
        });
    });

select.php

<?php
$link=mysqli_connect("localhost", "root","","manishYii");

if (mysqli_connect_errno() )
    echo "Fallo en la conexion con mysql" .mysqli_connect_error();

$action=$_POST["action"];
if ($action=="showroom") {
    $query = "SELECT FirstName, LastName, EmailID from members";
    $show = mysqli_query($link, $query) or die ("error");
    echo "<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";
    while ($row = mysqli_fetch_array($show)) {

        echo "<tr><td>" .$row['FirstName']."</td><td>".$row['LastName']."</td><td>".$row['EmailID']."</td></tr>";
    }
    echo "</table>";
}
?>

Output enter image description here


Solution 4:

try this

while ($row = mysqli_fetch_assoc($show)) {

    echo "<tr><td>" .$row['cod']."</td><td>".$row['nmbre']."</td><td>".$row['drccn']."</td></tr>";
}

or

while ($row = mysqli_fetch_assoc($show)) {

    echo "<tr><td>" .$row['0']."</td><td>".$row['1']."</td><td>".$row['2']."</td></tr>";
}

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