How to create a table and style it in PHP

Question

I have managed to take information from my database of the website, however I have no idea how to present the results in a table and how to style it. I tried putting <table> outside of the whole PHP, clearly did not work. I tried echoing a <table> tag before the result echo and a closing </table> tag after it, but that did not do it. This is the code I am working with:

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "onlib";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    //Takes all the results from the table with genre 5.
    $sql = "SELECT name, description, content FROM books WHERE genre='5'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            echo "<span style='color:white;'>"."<br> Name: ".$row["name"]."<br> Description: ".$row["description"]."<br> Content: ".$row["content"] ."<br>"."</p>";
        }
    } else {
        echo "0 results";
    }

    $conn->close();
?>

I am still new in PHP, trying to understand how the whole thing works. Thanks in advance!

Show us what you actually tried, instead of just telling us vague stories about what didn’t work … _“I am still new in PHP”_ - this has rather little to do with the specific language, and is more or less just a general issue of logical thinking … What is the structure I _want_ to create, and _where_ do the necessary tags have to be output for that to happen. — CBroe, Jun 08 '18 at 12:01
Do you get any output at all from the script as it is currently? — fightstarr20, Jun 08 '18 at 12:02
I searched for a few things, could not get it to work. The code as it is right now shows the information from the database as it is supposed to. Name: Alice in Wonderland Description: A nice book Content: This is the whole content. If I add another book, it will show it under this one with a line between them. — Todor Yakov, Jun 08 '18 at 12:05
if ($result->num_rows > 0) { // output data of each row echo ""; while($row = $result->fetch_assoc()) { echo ""."
Name: ".$row["name"]."
Description: ".$row["description"]."
Content: ".$row["content"] ."
".""; echo "
"; } } else { echo "0 results"; } This is the whole if I tried to do, but a table is not created. — Todor Yakov, Jun 08 '18 at 12:08
Basically PHP just creates plain HTML. I'd suggest to create a pure HTML without PHP and then replace the dynamic part (which should get repeated gor each roe, normally the ... part) with the PHP code. — Herbert Scheffknecht, Jun 08 '18 at 12:09

score 0 · Answer 1 · answered Jun 08 '18 at 12:13

Try this:

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "onlib";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    //Takes all the results from the table with genre 5.
    $sql = "SELECT name, description, content FROM books WHERE genre='5'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        echo "<table>";
        echo "<thead>
          <tr>
             <th>Name</th>
             <th>Description</th>
          </tr>
         </thead>";
        echo "<tbody>";
        while($row = $result->fetch_assoc()) {
            echo "<tr>";
            echo "<td>".$row["name"]."</td>";
            echo "<td>".$row["description"]."</td>";
            echo "</tr>";            
        }
        echo "</table>";
    } else {
        echo "0 results";
    }

    $conn->close();
?>

Thank you. This does style it in a table! I have forgotten to include the information about each row. Thank you. — Todor Yakov, Jun 08 '18 at 12:26

Nic3500 · Answer 2 · 2018-06-08T15:24:04.943

Try this approach:

[... some HTML header and code ...]

<table>
    <thead>
        <tr><th>Name</th><th>Description</th><th>Content</th></tr>
    </thead>
    <tbody>
    <?php
        [ ... extract something from the database ...]

        while($row = $result->fetch_assoc()) 
        {
            echo "<tr><td>$row[name]</td><td>$row[description]</td><td>Content</td></tr>\n";
        }
    }
    ?>
    </tbody>
</table>

[... some HTML footer ...]

Obviously, replace the [...] sections with your own code. The idea is here to use PHP to output the dynamic part of your HTML code, and is the power of PHP combined with HTML.

I try to avoid "echoing" to much HTML in my PHP, but it does work (reference other answers).

Detail: when setting up a <table> you should setup a header section <thead> <tfoot> (if applicable) and body section <tbody>. Ref What is the purpose for HTML's tbody?

You have non-constant assumption issues with the variables in that echo. — IncredibleHat, Jun 08 '18 at 12:19
If you would add the explanation that HTML tables require more than `
` including `` and `` this would be a very good answer. — Elin, Jun 08 '18 at 12:21

score 0 · Answer 3 · answered Jun 08 '18 at 12:13

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "onlib";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
//Takes all the results from the table with genre 5.
$sql = "SELECT name, description, content FROM books WHERE genre='5'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    echo '<table>';
    echo '<tr><td>Name</td><td>Description</td><td>Content</td></tr>'
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<tr><td>".$row["name"]."</td><td>".$row["description"]."</td><td>".$row["content"] ."</td></tr>";
    }
    echo '</table>';
} else {
    echo "0 results";
}

$conn->close();

?>

Thank you. This does style it in a table! I have forgotten to include the information about each row. Thank you. — Todor Yakov, Jun 08 '18 at 12:25

score 0 · Answer 4 · answered Jun 08 '18 at 12:16

0

The basic table format is this

for example:

<table>
    <tr>
        <td>row 1 item 1</td>
        <td>row 1 item 2</td>
    </tr>
    <tr>
        <td>row 2 item 1</td>
        <td>row 2 item 2</td>
    </tr>
</table>

so try:

if ($result->num_rows > 0) {
    echo "<table>";
    while($row = $result->fetch_assoc()) {
        echo "<tr>";
        echo "<td>".$row["name"]."</td>";
        echo "<td>".$row["description"]."</td>";
        echo "<td>".$row["content"]."</td>";
        echo "</tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}

answered Jun 08 '18 at 12:16

John Gkouziotis

59
6

I just receive the information on one line when I try this. – Todor Yakov Jun 08 '18 at 12:20
@TodorYakov Are you sure you did it exactly like this ? Because i've tested it and it works fine for me – John Gkouziotis Jun 08 '18 at 12:24
Apologies about that, it does indeed do it in a table, I just have the background to black and the letters to grey. Thank you! – Todor Yakov Jun 08 '18 at 12:28

Anuga · Accepted Answer · 2018-06-08T13:07:37.973

<?php

function tableV1 ($row) {
    echo '<tr>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['description'] . '</td>';
    echo '<td>' . $row['content'] . '</td>';
    echo '</tr>';
}

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "onlib";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } 
?>

Always do Database Connection first, before Outputting anything, that way, you can create custom error message to show instead of a failed database conntection or no content at all.

<style type="text/css">
table {}
tbody {}
td {}
th {}
thead {}
tr {}
</style>

Style is used inside the <head></head> to style the table, CSS it's called.

<table>
    <thead>
        <th>Name</th>
        <th>Description</th>
        <th>Content</th>
    </thead>
    <tbody>
<?php
// Takes all the results from the table with genre 5.
$sql = "SELECT name, description, content FROM books WHERE genre='5'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    // Output data of each row
    while($row = $result->fetch_assoc()) {
        tableV1($row);
    }

} else {
    echo '<tr><td colspan="3">0 results</td></tr>';
}
?>
    </tbody>
</table>

Output contents from database.

<?php

$conn->close();

?>

Close database connection in the end. All together:

<?php

function tableV1 ($row) {
    echo '<tr>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['description'] . '</td>';
    echo '<td>' . $row['content'] . '</td>';
    echo '</tr>';
}

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "onlib";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } 
?>
<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8" />
        <meta http-equiv="X-UA-Compatible" content="IE=edge">
        <title>Page Title</title>
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <style type="text/css">
        table {}
        tbody {}
        td {}
        th {}
        thead {}
        tr {}
        </style>
    </head>
    <body>
        <table>
            <thead>
                <th>Name</th>
                <th>Description</th>
                <th>Content</th>
            </thead>
            <tbody>
<?php
// Takes all the results from the table with genre 5.
$sql = "SELECT name, description, content FROM books WHERE genre='5'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    // Output data of each row
    while($row = $result->fetch_assoc()) {
        tableV1($row);
    }

} else {
    echo '<tr><td colspan="3">0 results</td></tr>';
}
?>
            </tbody>
        </table>
    </body>
</html>
<?php

$conn->close();

?>

How to create a table and style it in PHP

5 Answers5