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kamaboko

Chapter 12: Script 12.5 Errors

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Hello,
 
I'm working through exercise 12.5 and had been getting the following errors:
 
To double check, I downloaded the exercise code and got the same error messages. 
 
1.mysql_close() expects parameter 1 to be resource
mysql_close($dbc);
 
2. Warning: mysql_select_db() expects parameter 2 to be resource

mysql_select_db('myblog', $dbc);
 

 

Any suggestions?

 

Thanks,

K

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Hello Hartley,

 

I tried your suggestion.  In fact, I've been messing with this for quite some time now.  I've pasted Larry's code, along with how the MySQL database is set up.  I'm using PHP 5.5.12 and MySQL 5.6.17.   I get the following error:

 

Could not add the entry because:

The query being run was: INSERT INTO entries (entry_id, title, entry, date_entered) VALUES (0, 'test title','test text', NOW())

 

 

Database:

entry_id INT UNSIGNED NOT NULL AUTO_INCREMENT

title VARCHAR(100) NOT NULL

entry TEXT NOT NULL

date_entered DATETIME

 

 

<!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" xml:lang="en" lang="en">
<head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8" />
    <title>Add a Blog Entry</title>
</head>
<body>
<h1>Add a Blog Entry</h1>
<?php // Script 12.5 - add_entry.php
/* This script adds a blog entry to the database. */

if ($_SERVER['REQUEST_METHOD'] == 'POST') { // Handle the form.

    // Connect and select:
    $dbc = mysqli_connect('localhost', '**', 'password');
    mysqli_select_db('myblog', $dbc);
    
    // Validate the form data:
    $problem = FALSE;
    if (!empty($_POST['title']) && !empty($_POST['entry'])) {
        $title = trim(strip_tags($_POST['title']));
        $entry = trim(strip_tags($_POST['entry']));
    } else {
        print '<p style="color: red;">Please submit both a title and an entry.</p>';
        $problem = TRUE;
    }

    if (!$problem) {

        // Define the query:
        $query = "INSERT INTO entries (entry_id, title, entry, date_entered) VALUES (0, '$title', '$entry', NOW())";
        
        // Execute the query:
        if (@mysqli_query($query, $dbc)) {
            print '<p>The blog entry has been added!</p>';
        } else {
            print '<p style="color: red;">Could not add the entry because:<br />' . mysqli_error($dbc) . '.</p><p>The query being run was: ' . $query . '</p>';
        }
    
    } // No problem!

    mysqli_close($dbc); // Close the connection.
    
} // End of form submission IF.

// Display the form:
?>
<form action="add_entry.php" method="post">
    <p>Entry Title: <input type="text" name="title" size="40" maxsize="100" /></p>
    <p>Entry Text: <textarea name="entry" cols="40" rows="5"></textarea></p>
    <input type="submit" name="submit" value="Post This Entry!" />
</form>
</body>
</html>

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SOLVED.

 

OK...in the book, on page 354, line 35 of the code it is written: if(@mysql_query($query, $dbc)).

 

It should be if(@mysqli_query($dbc, $query)).  I made this change after reading more about mysqli_query.  Apparently the syntax is mysqli_query(connection,query,resultmode).

 

Thanks for the assistance though.

K

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