inserting Data into MySQL help

seephp · February 14, 2007

I am writing A simple variatino of a blogging engine and it I keep getting the error:

Could not select the database because: Column count doesn't match value count at row 1The query was INSERT INTO blog_entries (blogid, title, entry, date_entered) VALUES (0, 'I\'m trying this again!' 'This time I fixed a simple HTML problem!', NOW()).

The code is:

<!DOCTYPE html
PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<head>
<title>Creat A sticky note</title>
</head>
<body>
<?php
ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);
if (isset ($_POST['submit'])) {
if ($dbc = mysql_connect
('localhost', 'root', 'vertrigo'))
{		

		if (!@mysql_select_db ('mysticky'))
		{	
			die ('<p>Could not select the database because: <b>' .
			mysql_error() . '</b></p>');
		}

} else {
	die ('<p>Could not select the database because: <b>' .
			mysql_error() . '</b></p>');

}

$query = "INSERT INTO blog_entries
(blog_id, title, entry, date_entered)
VALUES (0, '{$_POST['title']}'
'{$_POST['entry']}', NOW())";

		if (mysql_query ($query)) {
			print '<p>Your StickyWeb Sticky has been stuck.</p>';

		} else {
			print "<p>Could not select the database because: <b>" .
			mysql_error() . "</b>The query was $query.</p>";

		}

		mysql_close();
}
?>
<form action="addNote.php" method="post">
<p>Create your Websticky Sticky note!</p>
<p>Enter your title:<input type="text" name="title" size="40" maxsize="100" /></p>
<p>Enter your note:<textarea name="entry" columns="40" rows="5"></textarea></p>
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>

ataria · February 14, 2007

INSERT INTO blog_entries

(blog_id, title, entry, date_entered)

SHOULD BE.

INSERT INTO `blog_entries`

(`blog_id`, `title`, `entry`, `date_entered`)

Jenk · February 14, 2007

You are missing out on a comma between two of your values.

seephp · February 14, 2007

Now I am getting this error:

Could not select the database because: 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 ''blog_id', 'title', 'entry', 'date_entered') VALUES (0, 'Trying this AGAIN!' ' at line 2The query was INSERT INTO blog_entries ('blog_id', 'title', 'entry', 'date_entered') VALUES (0, 'Trying this AGAIN!' 'I hope this works!', NOW()).

I had changed the code to:

<!DOCTYPE html
PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<head>
<title>Creat A sticky note</title>
</head>
<body>
<?php
ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);
if (isset ($_POST['submit'])) {
if ($dbc = mysql_connect
('localhost', 'root', 'vertrigo'))
{		

		if (!@mysql_select_db ('mysticky'))
		{	
			die ('<p>Could not select the database because: <b>' .
			mysql_error() . '</b></p>');
		}

} else {
	die ('<p>Could not select the database because: <b>' .
			mysql_error() . '</b></p>');

}

$query = "INSERT INTO blog_entries
('blog_id', 'title', 'entry', 'date_entered')
VALUES (0, '{$_POST['title']}'
'{$_POST['entry']}', NOW())";

		if (mysql_query ($query)) {
			print '<p>Your StickyWeb Sticky has been stuck.</p>';

		} else {
			print "<p>Could not select the database because: <b>" .
			mysql_error() . "</b>The query was $query.</p>";

		}

		mysql_close();
}
?>
<form action="addNote.php" method="post">
<p>Create your Websticky Sticky note!</p>
<p>Enter your title:<input type="text" name="title" size="40" maxsize="100" /></p>
<p>Enter your note:<textarea name="entry" columns="40" rows="5"></textarea></p>
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>

Greaser9780 · February 14, 2007

As said previously, you forgot a comma. You have:

"INSERT INTO blog_entries

(blog_id, title, entry, date_entered)

VALUES (0, '{$_POST['title']}'

'{$_POST['entry']}', NOW())";

Change to :

"INSERT INTO blog_entries

(blog_id, title, entry, date_entered)

VALUES (0, '{$_POST['title']}',<--------note the added comma!

'{$_POST['entry']}', NOW())";

seephp · February 14, 2007

Thanks, I wasn't sure where that comma was supposed to go ;D.

Greaser9780 · February 14, 2007

Always b4 or after the single quote,',

seephp · February 14, 2007

I added the comma but I'm still getting:

Could not select the database because: 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 ''blog_id', 'title', 'entry', 'date_entered') VALUES (0, 'I reall hope this wor' at line 2The query was INSERT INTO blog_entries ('blog_id', 'title', 'entry', 'date_entered') VALUES (0, 'I reall hope this works this time!', 'If this works I can begin working on my project for Master Alley!', NOW()).

Greaser9780 · February 14, 2007

U don't need the single quotes around you row identifiers.Try changing:

('blog_id', 'title', 'entry', 'date_entered')

to:

(blog_id, title, entry, date_entered)

Greaser9780 · February 14, 2007

$query = "INSERT INTO `blog_entries`

(blog_id, title, entry, date_entered)

VALUES (0, '{$_POST['title']}',

'{$_POST['entry']}', NOW())";

seephp · February 14, 2007

YES! It finally works.

redarrow · February 14, 2007

<!DOCTYPE html
PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<head>
<title>Creat A sticky note</title>
</head>
<body>
<?php

//php.ini settings.

ini_set ('display_errors', 1);

//php.ini error settings.

error_reporting (E_ALL & ~E_NOTICE);

// if the form of title and entry are blank

if(($title=="none")||($entry=="none")){

// tell user to fill in all the form

echo "please fill in all the form cheers";

}
//database

$db=mysql_connecy("localhost","username","password");
mysql_select_db("databse",$db) or die (mysql_error());


//protect database

$blog_id=addslases($_POST['blog_id']);
$title=addslashes($_POST['title']);
$entry=addslashes($_POST['entry']);
$date_entered=addslashes($_POST['date_entered']);

//date for date entred.

$date_entered=date("d-m-y");


//if the user post an entry.

if (isset($_POST['submit'])) {

$query = "INSERT INTO blog_entries(blog_id,title,entry,date_entered)
VALUES('$blog_id', '$title', '$entry', '$date_entered')";

$result=mysql_query($query) or die(mysql_error());

// 1st message if correct.

echo '<p>Your StickyWeb Sticky has been stuck.</p>';

}else{

// 2nd message not correct.

echo "<p>Could not select the database because:";


}

?>
<form action="addNote.php" method="post">
<p>Create your Websticky Sticky note!</p>
<p>Enter your title:<input type="text" name="title" size="40" maxsize="100" /></p>
<p>Enter your note:<textarea name="entry" columns="40" rows="5"></textarea></p>
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>

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inserting Data into MySQL help

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