No Database Selected

system · February 9, 2012, 5:05am

Hello, everyone. I’m quite a beginner at php, and I may be mixing legacy code, so do please point out all the problems in my code.

I have this problem of not getting my database. I am trying to create my own blog. However, I am unable to pull up the blog posts, for I get the “No database selected” error.

I’ve searched the internet, but most fixes came from adding the mysql_select_db(), which I already have. I added in mysql_error() after every function call to see where the error came from. The exact error is: “Error in query (3): No database selected.” This comes from the else{} function in GetBlogPosts{}, which is near the end. I’m at a total loss on how to resolve this. I’ve also checked and rechecked my privileges for the account. I’ve deleted and remade them, too.

[php]

<?php include('blogpost.php'); /* adds the database connection mysql_connect("localhost_name", "username", "password") */ $con = mysql_connect("127.0.0.1", "root", ""); if (!$con) { die ('Could not connect to server: ' . mysql_error()); } mysql_select_db('bento-blog', $con) or die ('Could not connect to database: ' . mysql_error()); //now trying to retrieve the blog posts from the database function GetBlogPosts($inId=null, $inTagId=null) { //asking if each argument is empty or null if (!empty($inId)) { //give blog ID, want just this single blog post $sql = "SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC"; $query = mysql_query($sql); if (!$query) { die('Error in query (1) : ' . mysql_error()); } } else if (!empty($inTagId)) { //give tag ID, want all posts with this tag $sql = "SELECT blog_posts.* FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =" . $tagID . " ORDER BY blog_posts.id DESC"; $query = mysql_query($sql); if (!$query) { die('Error in query (2) : ' . mysql_error()); } } else { //otherwise want all blog posts $sql = "SELECT * FROM blog_posts ORDER BY id DESC"; $query = mysql_query($sql); if (!$query) { die('Error in query (3) : ' . mysql_error()); } } //process data returned from query, create object, and create an array $postArray = array(); while ($row = mysql_fetch_assoc($query)) { $myPost = new BlogPost($row["id"], $row['title'], $row['post'], $row['author'], $row['datePosted']); array_push($postArray, $myPost); } return $postArray; } mysql_close($con); ?>

[/php]

Thank you for any assistance you can offer.

sarthakpatel · February 9, 2012, 7:15am

Hi,

Can you tell me from where this function ‘GetBlogPosts’ called?

Or

can you post the “blogpost.php” file here.

sarthakpatel · February 9, 2012, 7:18am

Hi

you ca try this

[php]

/*

adds the database connection

mysql_connect(“localhost_name”, “username”, “password”) */
$con = mysql_connect(“127.0.0.1”, “root”, “”);

if (!$con)

{

die ('Could not connect to server: ’ . mysql_error());

}
mysql_select_db(‘bento-blog’, $con) or die ('Could not connect to database: ’ . mysql_error());

//now trying to retrieve the blog posts from the database
function GetBlogPosts($inId=null, $inTagId=null)
{

//asking if each argument is empty or null

if (!empty($inId))

{

//give blog ID, want just this single blog post

$sql = “SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC”;

$query = mysql_query($sql);

if (!$query)

{

die('Error in query (1) : ’ . mysql_error());

}

else if (!empty($inTagId))

{

//give tag ID, want all posts with this tag

$sql = “SELECT blog_posts.* FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =” . $tagID . " ORDER BY blog_posts.id DESC";

$query = mysql_query($sql);

if (!$query)

{

die('Error in query (2) : ’ . mysql_error());

}

else

{

//otherwise want all blog posts

$sql = “SELECT * FROM blog_posts ORDER BY id DESC”;

$query = mysql_query($sql);

if (!$query)

{

die('Error in query (3) : ’ . mysql_error());

}

//process data returned from query, create object, and create an array

$postArray = array();

while ($row = mysql_fetch_assoc($query))

{

$myPost = new BlogPost($row[“id”], $row[‘title’], $row[‘post’], $row[‘author’], $row[‘datePosted’]);

array_push($postArray, $myPost);

}

return $postArray;
}
include(‘blogpost.php’);
mysql_close($con);

[/php]

lydiapurple · February 9, 2012, 8:35pm

The only difference I could tell between the code you gave me from my original code is that the

[php]include(‘blogpost.php’)[/php]

was at the bottom before the mysql_close() function.

I put in your code, and got a slew of errors.

This is blogpost.php

[php]<?php

class BlogPost
{
var $id;
var $title;
var $author;
var $tags;
var $datePosted;

function __construct($inId=null, $inTitle=null, $inPost=null, $inAuthorId=null, $inDatePosted=null)
{
	//all the if statements to make sure not pass nothing to server, which produces error
	if (!empty($inId))
	{
		$this->id = $inId;
	}
	if (!empty($inTitle))
	{
		$this->title = $inTitle;
	}
	if (!empty($inPost))
	{
		$this->post = $inPost;
	}
	if (!empty($inDatePosted))
	{
		//to get date in dd/mm/yyyy format
		// explode() is remove the components of the date
		$splitDate = explode("-", $inDatePosted);
		$this->datePosted = $splitDate[2] . "/" . $splitDate[1] . "/" . $splitDate[0];
	}
	if (!empty($inAuthorId))
	{
		//to get the author's name using a query
		$query = mysql_query("SELECT poster_name FROM People WHERE id = " . $inAuthorId);
		$row = mysql_fetch_assoc($query);
		$this->author = $row["poster_name"];
	}
	
	//setting default, will be overwritten if tag database not empty
	$postTags = "No Tags";
	if (!empty($inId))
	{
		//performing a LEFT JOIN
		//this selects data from another table but only when it matches the data from the "left" (the other selected data) 
		$sql = "SELECT tags.* LEFT JOIN (tags) ON (blog_post_tags.tag_id = tags.id) WHERE blog_post_tags.blog_post_id = " . $inId;
		$query = mysql_query($sql);
		$tagArray = array();
		$tagIDArray = array();
		while($row = mysql_fetch_assoc($query))
		{
			array_push($tagArray, $row["name"]);
			array_push($tagIDArray, $row["id"]);
		}
		//check to make sure array has length>0 with if-else
		if (sizeof($tagArray) > 0)
		{
			foreach ($tagArray as $tag)
			{
				if ($postTags == "No Tags")
				{
					$postTags = $tag;
				}
				else
				{
					$postTags = $postTags . ", " . $tag;
				}
			}
		}
	}	
	$this->tags = $postTags;
}

}

?>[/php]

I’m calling it from index.php from the content div.

[code]

<?php
include (“includes/includes.php”);

	$blogPosts = GetBlogPosts($inId=null, $inTagId=null);
	
	foreach ($blogPosts as $post)
	{
		echo "<div class='post'>";
		echo "<h2>" . $post->title . "</h2>";
		echo "<p>" . $post->post . "</p";
		echo "<span class='footer'>Posted By: " . $post->author . " Posted On: " . $post->datePosted . " Tags: " . $post->tags . "</span";
		echo "</div>";
	}
?>
</div>

[/code]

sarthakpatel · February 10, 2012, 6:04am

Hi,

here is he solution.

[php]include(‘blogpost.php’);
/*
adds the database connection
mysql_connect(“localhost_name”, “username”, “password”) /
$con = mysql_connect(“127.0.0.1”, “root”, “”);
if (!$con)
{
die ('Could not connect to server: ’ . mysql_error());
}
mysql_select_db(‘bento-blog’, $con) or die ('Could not connect to database: ’ . mysql_error());
//now trying to retrieve the blog posts from the database
function GetBlogPosts($inId=null, $inTagId=null)
{
//asking if each argument is empty or null
if (!empty($inId))
{
//give blog ID, want just this single blog post
$sql = “SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC”;
$query = mysql_query($sql);
if (!$query)
{
die('Error in query (1) : ’ . mysql_error());
}
}
else if (!empty($inTagId))
{
//give tag ID, want all posts with this tag
$sql = "SELECT blog_posts. FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =" . $tagID . " ORDER BY blog_posts.id DESC";
$query = mysql_query($sql);
if (!$query)
{

die('Error in query (2) : ’ . mysql_error());
}
}
else
{
//otherwise want all blog posts
$sql = “SELECT * FROM blog_posts ORDER BY id DESC”;
$query = mysql_query($sql);
if (!$query)
{
die('Error in query (3) : ’ . mysql_error());
}
}
//process data returned from query, create object, and create an array
$postArray = array();
while ($row = mysql_fetch_assoc($query))
{
$myPost = new BlogPost($row[“id”], $row[‘title’], $row[‘post’], $row[‘author’], $row[‘datePosted’]);
array_push($postArray, $myPost);
}
return $postArray;
}
mysql_close($con);
[/php]

And

[php]

<?php include ("includes/includes.php"); $blogPosts = GetBlogPosts($inId, $inTagId); foreach ($blogPosts as $post) { echo "

"; echo "

" . $post->title . "

"; echo "

" . $post->post . "</p"; echo "Posted By: " . $post->author . " Posted On: " . $post->datePosted . " Tags: " . $post->tags . "</span"; echo "

"; } ?>

[/php]

You are passing null value when calling function ‘GetBlogPosts’.