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PHP

Posted by Richard Duffy
Richard Duffy
16,488 Points

While Loop Not Working!

while($row = mysqli_fetch_array($res_acct)) {
  $name = $row['account_title'];
  $id =  $row['id'];
}

This is only showing one row.

7 Answers

David Tonge
David Tonge
Courses Plus Student 45,640 Points

There's a lot of instances, in your code, of PHP being used within HTML that aren't enclosed in PHP tags.

David Tonge
David Tonge
Courses Plus Student 45,640 Points

I'm not saying that's the problem but it'd probably help.

David Tonge
PLUS
David Tonge
Courses Plus Student 45,640 Points
David Tonge
David Tonge
Courses Plus Student 45,640 Points

Hey Richard, I think you're reaching an error because "=" is an assignment operation and "==" is comparison. Right not you're assigning that function to the $row variable.

Richard Duffy
Richard Duffy
16,488 Points
Richard Duffy
Richard Duffy
16,488 Points

No, that has not seemed to work

<?php
include_once 'connect.php';
doDB();

$get_acct = "SELECT *, id, account_title FROM account_list  ORDER BY account_title DESC ";

$res_acct = mysqli_query($mysqli,$get_acct)or die(mysqli_error($mysqli));

if(mysqli_num_rows($res_acct) < 1) {
    $display_block = "<p><em>No accounts exist</em></p>";
} else {
    $display_block = <<<END_OF_TEXT
    <table>
    <tr>
    <th><h4>Account Name</h4></th>
    </tr>
END_OF_TEXT;
    while($row = mysqli_fetch_assoc($res_acct)) {
        $name = $row['account_title'];
        $id =  $row['id'];


    }

    $display_block .= <<<END_OF_TEXT
    <tr>
    <td><a href="show_account.php?account=$id">
    <strong> $name</strong>
    </a>
    &nbsp;
    <strong>$id</strong>

END_OF_TEXT;
}

    mysqli_free_result($res_acct);

    mysqli_close($mysqli);
    $display_block .= "</table>";
?>
<!DOCTYPE HTML>
<html>
    <head>
        <meta charset="UTF-8">
        <title>Account Lists</title>
    </head>
    <body>
    <?php echo $display_block;?>
    </body>
</html>
Richard Duffy
Richard Duffy
16,488 Points
Richard Duffy
Richard Duffy
16,488 Points

When I put == it gives me these errors

''' Notice: Undefined variable: row in /Applications/XAMPP/xamppfiles/htdocs/hosting list/inc/list.php on line 18

Notice: Undefined variable: id in /Applications/XAMPP/xamppfiles/htdocs/hosting list/inc/list.php on line 28

Notice: Undefined variable: name in /Applications/XAMPP/xamppfiles/htdocs/hosting list/inc/list.php on line 31

Notice: Undefined variable: id in /Applications/XAMPP/xamppfiles/htdocs/hosting list/inc/list.php on line 32 Account Name

'''

Richard Duffy
Richard Duffy
16,488 Points
Richard Duffy
Richard Duffy
16,488 Points

Also on my other document it is not displaying anything at all:

<?php
include_once 'connect.php';
doDB();

if(!isset($_GET['account'])) {
    echo("Error");
    exit;
}


$account_id = mysqli_real_escape_string($mysqli , $_GET['account']);

$yes        = "SELECT id FROM account_list
              WHERE id = '".$account_id."'";

$res        = mysqli_query($mysqli,$yes)or die(mysqli_error($mysqli));

if(mysqli_num_rows($res) < 1){
    $display_block = "<p><em>You have selected an invalid account. Please <a href='list.php'> try again</a></em></p>";
} else {
    while($row = mysqli_fetch_array($res)){
        $title = $row['id'];
    }

    $get_account_info = "SELECT  `account_info` ,`id`, `domain`, `ip`, `cgi`, `username`, `password`, `mode`, `meg`, `ns1`, `ns2`, `ns3`, `ns4`, `package`, `feature`, `language`,
                        FROM account_info WHERE id = '".$account_id."'";
    $res_acct         = mysqli_query($mysqli,$get_account_info)or die(mysqli_error($mysqli));

    while($row1 = mysqli_fetch_assoc($res_acct)){
        $domain  = $row['domain'];
        $ip      = $row['ip'];
        $cgi     = $row['cgi'];
        $user    = $row['username'];
        $pass    = $row['passowrd'];
        $mod     = $row['mode'];
        $neg     = $row['meg'];
        $ns1     = $row['ns1'];
        $ns2     = $row['ns2'];
        $ns3     = $row['ns3'];
        $ns4     = $row['ns4'];
        $pack    = $row['package'];
        $feat    = $row['feature'];
        $lang    = $row['language'];

        echo "Hello world";
        $display_block = "
        <h1>Hello World</h1>
        <table>
        <tr>
        <thead>
        <th>Hello</th>
        </thead>
        </tr>
        <tr>
        <td>$domain</td>
        <td>$ip</td>
        <td>$cgi</td>
        <td>$user</td>
        <td>$pass</td>
        <td>$mod</td>
        <td>$meg</td>
        <td>$ns1</td>
        <td>$ns2</td>
        <td>$ns3</td>
        <td>$ns4</td>
        <td>$pack</td>
        <td>$feat</td>
        <td>$lang</td>
        <tr>";

    }
}
    mysqli_free_result($yes);
    mysqli_free_result($get_account_info);
    mysqli_close($mysqli);


?>
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php echo $display_block;?>
Richard Duffy
Richard Duffy
16,488 Points
Richard Duffy
Richard Duffy
16,488 Points

And gives me this error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM account_info WHERE id = '1'' at line 2.

Richard Duffy
Richard Duffy
16,488 Points
Richard Duffy
Richard Duffy
16,488 Points

I have done what you said and now it has this error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM account_info WHERE id = '1'' at line 2.