Putting SQL data into HTML table

php html sql html-table

26,434

Solution 1

First of all, the most important thing to keep in mind is:

You are using deprecated and unsecure code

The mysql_ functions are strongly discouraged, for various reasons:

Are deprecated and will be removed in future versions of PHP,
Are insecure leading to possible SQL injections,
Lack many features present in more current versions of PHP

See the linked question for much more in-depth explanations.

Now, to the code itself:

You are not using `mysql_connect` to connect to the server

You should use mysql_connect to specify the server, the username and the password that will be used to access the data in the database. From your code, it seems that it was supposed to be present, because there's a $db variable used in the mysql_connect function, but not properly initialized nor used again anywhere else.

You should use mysql_connect in a way similar to this:

$db = mysql_connect('localhost', $user, '$password');
if (!$db) {
    die('Not connected : ' . mysql_error());
}

(Don't forget to set your username and password!)

You are using `mysql_select_db` twice in a row:

    $db_select = mysql_select_db($dbname,$db);
        if (!db_select) {
            die("Database selection also failed miserably: " . mysql_error());
        }

followed by

mysql_select_db("databaseiheko");

Note the $dbname and $db variables, you don't have them on your code, this function won't work like this.
The second mysql_select_db overwrites the first, but you don't specify a server connection to be used.

You should use the first version, but you should use mysql_connect before it.

You have typos in your code

if (!db_select) { should be if (!$db_select) {
echo &row['DatumFuif'] should be echo $row['DatumFuif']

Solution 2

mysql_ functions are deprecated, but if you want to use them, i suggest these corrections:

You can correct your code this way: the mysql connect is needed:

 <?php
 //connect to your database
 mysql_connect("serverIpAddress","userName","password");
 //specify database
 mysql_select_db("yourDatabaseName") or die;
 //Build SQL Query
 $query = "select * from tblfuiven";
 $queryResult=mysql_query($query);
 $numrows=mysql_num_rows($queryResult);

numrows will contain the number of found records in the db. add an echo for the number of rows and let us know if the number of rows is still one. Then use mysql_fetch_assoc to get rows:

  while($row = mysql_fetch_assoc($queryResult)) {
        ?>
            <tr>
                <td><?php echo $row['NaamFuif']?></td>
                <td><?php echo $row['GemeenteFuif']?></td>
                <td><?php echo &row['DatumFuif']?></td>
            </tr>
        <?php

        }
        ?>

EDIT: You can test the code and let us know the number of rows that you obtain, using this code(write your real user name and password and db name:

         <?php
         mysql_connect("localhost","root","root");     
         mysql_select_db("databaseName") or die;
            $results = mysql_query("SELECT * FROM tblfuiven");
            $numrows=mysql_num_rows($queryResult);
           while($row = mysql_fetch_assoc($queryResult)) {
            ?>
                <tr>
                    <td><?php echo $numrows ?></td>
                    <td><?php echo $row['NaamFuif']?></td>
                    <td><?php echo $row['GemeenteFuif']?></td>
                    <td><?php echo $row['DatumFuif']?></td>
                </tr>

            <?php
           }
           ?>

26,434

Author by

Matt

Just an avid coder

Updated on September 13, 2020

Comments

Matt over 3 years

I'm trying to get the data from my mySQL database and put them into a HTML table. After searching a lot on the internet, but I coudn't find code that worked for me.

Currently, I have this code

<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title></title>
</head>
<body>
    <table>
        <thead>
            <tr>
                <td>Naam</td>
                <td>Gemeente</td>
                <td>Datum</td>
            </tr>
        </thead>
        <tbody>
       <?php
          $db_select = mysql_select_db($dbname,$db);
            if (!db_select) {
                die("Database selection also failed miserably: " . mysql_error());
            }
            mysql_select_db("databaseiheko");
            $results = mysql_query("SELECT NaamFuif, GemeenteFuif, DatumFuif FROM tblfuiven");
            while($row = mysql_fetch_array($results)) {
            ?>
                <tr>
                    <td><?php echo $row['NaamFuif']?></td>
                    <td><?php echo $row['GemeenteFuif']?></td>
                    <td><?php echo &row['DatumFuif']?></td>
                </tr>

            <?php
            }
            ?>   
            </tbody>
            </table>
</body>
</html>

The only thing that I get is the first row of my table (Naam-Gemeente-Datum). Am I doing something wrong or did I forgot something?

Matt over 11 years

After adding your code, it still doesn't work. Current code --> pastebin.com/p7J4rZek
Gaucho over 11 years

@Matt in my answer there is a question. Could i know the value of $numrows ?
Gaucho over 11 years

@Matt in mysql_connet you used $dbhost etc., within quotes. Write directly the string within quotes, for example "localhost" instead of "$dbhost"
Matt over 11 years

When I echo something, I don't get an output (example: echo 'Hello World"). Even after directly writing the string, I get the same results.
That Brazilian Guy over 11 years

@Matt Then probably your php.ini settings are different from the usual
Gaucho over 11 years

@Matt, add the row <td><?php echo $numrows ?></td> near the other 3 td and watch the output on the webpage.
Gaucho over 11 years

@Matt use this pastebin.com/JmuecfGh and write the correct user name, password and database name. Why you use variables that you didn't declared?