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Editorial Team
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Asked: 2026-06-14T19:40:19+00:00

I’ve created a table (notebook) where I keep user’s text. The table is like

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I’ve created a table (notebook) where I keep user’s text. The table is like this one:

+------+------------+-------+------+--------+------------------------+
|  id  |  Username  | title | html | public |  create_date           |
|--------------------------------------------------------------------|
|   1  | sorge13248 | Test  | ...  |  n     |2012-11-24 9:00:00      |
+------+------------+-------+------+--------+------------------------+

id = INT(25) A_I NOT NULL
Username = VARCHAR(65) NOT NULL
title = VARCHAR(26) NOT NULL
html = LONGTEXT NOT NULL
public = VARCHAR(11) NOT NULL
create_date = TIMESTAMP CURRENT_TIMESTAMP NOT NULL

So, I’ve tried to use this code to list all notebook’s record for a user:

$link = mysql_connect("localhost", "mysql_user", "password");
mysql_select_db("notebook", $link);

$result = mysql_query("SELECT * FROM notebook WHERE Username = '"$username."'", $link);

if(0 == mysql_num_rows($result)) {
    echo 'No notebook was created for this user';
}
else { 
    while(row = mysql_fetch_array($result)) {
       // do whatever you want with the data here
    }
}

The code has been taken from: How to list rows for a query or display 'no records' using a single query and I’ve edited it for use it with my table.

Now, I’m confused, what I must write in “// do whatever you want with the data here” string to list the records?

EDIT: Now, I have this code, but Chrome says “HTTP 500 Error, Internal Server Error”:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>Notebook</title>
</head>

<body>
<h1>Notebook</h1><br/>
<?=
$link = mysql_connect("localhost", "database_username", "password");
mysql_select_db("database_name", $link);

$result = mysql_query("SELECT * FROM notebook", $link);

// If if result set contains rows
if(0 == mysql_num_rows($result)) {
    echo 'No notebook was created for this user';
}
else { 
    while(row = mysql_fetch_array($result)) {
   printf( '%d: &quot;%s&quot; by %s<br />', $row['id'], 
                 htmlspecialchars($row['title']), 
                 htmlspecialchars($row['Username']) );
}
}
?>
</body>

</html>

Where is the error? The MySQL credentials are provide correctly…

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1 Answer

  1. Editorial Team

    Form HTML with whatever you want to display, so your while loop would look like, i.e.:

    while(row = mysql_fetch_array($result)) {
   printf( '%d: &quot;%s&quot; by %s<br />', $row['id'], 
                 htmlspecialchars($row['title']), 
                 htmlspecialchars($row['Username']) );
}

    which would list all items, like

    1: "Test" by sorge13248

    PS: you should always use the same naming convention to avoid problems. in your DB all columns are lowercased, except Username).

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