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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.

<scriptsrc="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><metacharset="UTF-8"/><scriptsrc="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script><scriptsrc="select.js"type="text/javascript"></script></head><body><buttonid="button"> Mostrar </button><br><inputtype="text"id="id" /><divid="content"></div></body></html>

select.js

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

            functionshow_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, $query) ordie ("error");
    echo"<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";
    while ($row = mysqli_fetch_array($show)) {

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

Outputenter 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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