PHP table error with mysql

Question

I have been programming a page to allow you to be able to store passwords within a database and I am having an issue where the table is being printed out 3 times but I have no idea why. I wanted the table to be on the far right of the screen and to have a list downwards of passwords that you have entered into the mysql database but it is not working

Here is the code:

<?php
     $tempUsername = $_SESSION['usernameBox'];
     mysql_connect("localhost", "username", "password");
     mysql_select_db("UserTables");
     $sql = "SELECT * FROM $tempUsername";
     $resultSet = mysql_query($sql);
     while($info = mysql_fetch_array($resultSet)) {
         $counter = count($info);
         for($i = 0; $i < $counter; $i++) {
?>

<table align="right">
     <th> Id </th <th> Website </th> <th> Password </th>
      <tr>
             <td align="center"> <?php echo $info['Id']; ?> </td>
             <td align="center"> <?php echo $info['Website']; ?> </td>
             <td align="center"> <?php echo $info['Password']; ?> </td>
      </tr>
</table>

<?php
        }
    }
?>

Please don't store passwords in plaintext. See: crackstation.net/hashing-security.htm — Chris Kitching
– Chris Kitching, Commented Mar 6, 2016 at 9:18
You output a full table inside your iteration loop over the entries. Off course you get one full table for each entry then... — arkascha
– arkascha, Commented Mar 6, 2016 at 9:19
In the mysql_connect that Is not my username or my password, however in the table I wanted them to see the passwords. — ImXari
– ImXari, Commented Mar 6, 2016 at 9:19

Death-is-the-real-truth · Accepted Answer · 2016-03-06 09:22:00Z

Remove the table code outside of for loop and also no need of for loop:-

<?php
     $tempUsername = $_SESSION['usernameBox'];
     mysql_connect("localhost", "username", "password");
     mysql_select_db("UserTables");
     $sql = "SELECT * FROM $tempUsername";
     $resultSet = mysql_query($sql);
?>
    <table align="right">
     <tr><th> Id </th <th> Website </th> <th> Password </th></tr>
<?php
    while($info = mysql_fetch_assoc($resultSet)) { // check that i changed array to assoc and remove for loop because no need
?>
    <tr>
        <td align="center"> <?php echo $info['Id']; ?> </td>
        <td align="center"> <?php echo $info['Website']; ?> </td>
        <td align="center"> <?php echo $info['Password']; ?> </td>
    </tr>
<?php } ?> 

</table>

Note:- Now it's time to move towards to mysqli_* or PDO, because mysql_* is a deprecated library now.

Professor Abronsius · Accepted Answer · 2016-03-06 09:18:54Z

You are not correctly closing the first php code block and there are errors in the html with unclosed tags ( th ) and missing table row tags.

<?php
     $tempUsername = $_SESSION['usernameBox'];
     mysql_connect("localhost", "username", "password");
     mysql_select_db("UserTables");
     $sql = "SELECT * FROM $tempUsername";
     $resultSet = mysql_query($sql);
     while($info = mysql_fetch_array($resultSet)) {
         $counter = count($info);
         for( $i = 0; $i < $counter; $i++ ) {
?>

    <table align="right">
        <tr>
            <th> Id </th>
            <th> Website </th>
            <th> Password </th>
        </tr>
        <tr>
            <td align="center"> <?php echo $info['Id']; ?> </td>
            <td align="center"> <?php echo $info['Website']; ?> </td>
            <td align="center"> <?php echo $info['Password']; ?> </td>
        </tr>
    </table>

<?php
        }
    }
?>

However, the approach above ( which follows your original ) will produce a new table on each iteration through the loop - I suspect you would prefer 1 table with many rows.? Perhaps more like:

<?php
     $tempUsername = $_SESSION['usernameBox'];
     mysql_connect("localhost", "username", "password");
     mysql_select_db("UserTables");
     $sql = "SELECT * FROM $tempUsername";
     $resultSet = mysql_query($sql);
?>
    <table align="right">
        <tr>
            <th> Id </th>
            <th> Website </th>
            <th> Password </th>
        </tr>

<?php
     while( $info = mysql_fetch_array( $resultSet ) ) {
         $counter = count($info);
?>
        <tr>
            <td align="center"> <?php echo $info['Id']; ?> </td>
            <td align="center"> <?php echo $info['Website']; ?> </td>
            <td align="center"> <?php echo $info['Password']; ?> </td>
        </tr>
<?php
    }
?> 
</table>

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