PHP - Fetch Categories And Get Their Last Article In Same Query

I have a database in which i store articles that contain all type of special characters including slashes, single quotes, double quote etc.

Previously it was working fine, but now after the updating the xampp, things are not going as usual.

I insert are articles using following code

$query = "INSERT INTO `articles` (`source`, `heading`, `description`, `catagory`, `detail`, `date`, `timestamp`)
          values ('$source', '$heading', '$desc', '$catagory', '$detail', '$date', '$timestamp')";

Error I am getting is

Error Running query 1You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 's economy and emphasized that international community should provide its complet' at line 2

Can anybody tell me whats wrong with my script?

I have manage to create my own script that add news article, however, I would like to add a comment form below the news article. How can I relate the news and the comments? also, when I delete the articles, it should also delete the comments in mysql.

Your help is highly appreciated!

Hello,

can you help me on my news article?
I would like to be able to add photo on the article. Below is my codes

Quote

=======
html
=======
<?php
include_once ('post_news.php');
?>

<form Action="add_news.php" Method="post">
   <input name="postdate" type="text" id="postdate">
   <input name="title" type="text" id="title">
   <textarea name="newstext" cols="60" rows="15" id="newstext"></textarea>
<input type="submit" name="submit" value="Submit">
<input type="reset" name="reset" value="Reset">
</form>

==============
php - post_news.php
==============
<?php

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);

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

{

   require_once ('inc/dbConfig.php');
   
// Define the query.
$query = "INSERT INTO news (id, postdate, title, newstext) VALUES
(0,'{$_POST['postdate']}', '{$_POST['title']}','{$_POST['newstext']}')";

// Execute the query.
if (@mysql_query ($query)) {
   print "<p>Data has been added.</p>";
} else {
   print "<p>Could add the entry because: <b>" . mysql_error() . "</b>. The query was $query.</p>";
}

 mysql_close();

}

?>

=====
db structure
======
 id int(10) unsigned NOT NULL auto_increment,
   postdate timestamp(14),
   title varchar(50) NOT NULL,
   newstext text NOT NULL,
   PRIMARY KEY (id),
   KEY postdate (postdate)



Hope you guys can help me.

Thanks!

I'm building an article system, what im trying to do is when a user choose the article to be published tomorrow to able to.  Write today the article but the system will show it from tomorrow

Code: [Select]
        $gettoday = date("Y:m:d");
$query = "SELECT * FROM tblnews WHERE MainArticle = 1 AND NewsDate = '".$gettoday."'  ORDER BY `Id` DESC LIMIT 1";
$result = mysql_query($query);

This is my code, but when i use this and the article is not posted for today is not showing anything.. any suggestions please?

Thank you

I'm feeling a little overwhelmed/burned-out/confused...   

I have a page "article_index.php" that contains a summary of each article plus a link to it, e.g.

Code: [Select]
<a href="<?php echo WEB_ROOT; ?>articles/postage-meters-save-you-money">
When you click on a link, it goes to "article.php" and uses a mod_rewrite to transform the "pretty URL" to regular URL that "article.php" can use to query the correct article from my database.

-----

Here is what I need help with...

When a user is on a given article page, I want to sto

- ReturnToPath
- ArticleID

in the SESSION.

I am drawing a blank on how to get the "ArticleID" when a user is on a give page?!   

Hope you guys follow me?!


Debbie

Hello,

I am trying to create article directory for learning PHP MySql and i want to write a script for recording number of views for an article and store it in Database. Can anyone please direct me to a nice simple tutorial or help me with the steps or algorithm?

Suppose the link for an article is http://www.domain.com/article.php?id=123
Do i need to put extra variables and create a tracking PHP file for it?

hi there,

basically in my database i have  DATETIME for when an article was posted, using this, I'd like to make it possible for viewers of the article to see when it was posted in a "Posted Today", "Posted Yesterday" or Posted 23 hours ago" format. If it was posted longer ago than yesterday, then I would like it to just be the normal date.

I know how to work out if it was posted today, just through:

$date = date('j F', strtotime($rows['date']));
$today=date('j F');

but not sure how to do the hours.

thanks!

This topic has been moved to Apache HTTP Server.

http://www.phpfreaks.com/forums/index.php?topic=322454.0

Hi all -

I'm setting up a custom PHP blog. It pulls the data from a MySQL database which includes HTML tags (<p><div><span> etc...).

I would like to display only up to 50 words per post on the blog page, which users can then read and click a link to then see the entire post.

I've developed some code which does this, however, it seems to be stripping my HTML tags... Very sad!

Would be very very grateful if one (or more) of you kind lot would have a look at my code and let me know if there is a easier (I'm all for easy) and proper way of implementing this so that it works without stripping my HTML tags.

Cheers!!!

Code: [Select]
// Counts number of blog words in the content
$blog_content_words = str_word_count($blog["content"], 1);

// Sets the blog_words variable to 0
$blog_words = 0;

// Prints title on page as a permalink to a post
echo '<h1><a href="blog.php?id='.$blog["id"].'">'.$blog["title"].'</a></h1><p>';

// Loops while blog_word is under 50
while($blog_words < 50) {

// Prints a word from blog_content_words array of blog post and adds a space afterwards
echo $blog_content_words["$blog_words"]." ";

// Adds 1 to the blog_words counter
++$blog_words;
}

// Adds a read more link to the post which links to full blog post
echo '... <a href="blog.php?id='.$blog["id"].'">[read more]</a></p>';

i need to submit 1 article (3 input fields: title, category, description), to multiple sites. it must be done with 1 click.

one more thing... each site to which i have to submit has different input names, but are the same 3, also 1 is for title 1 for category and 1 for description.

I have two tables, one called Categories and one called SubCategories. Each subcategory needs to be displayed under its parent category in a HTML table.

So I have this code, but I don't know how to list each Sub-Category under each Category in a HTML table.

<?php

require ('includes/initialize.php');

require 'classes/database.class.php';


$sql = "SELECT * FROM Categories, SubCategories WHERE Categories.ID = SubCategories.CatID";

$Result = $database->Query($sql);
 
$Data = $database->Fetch($Result);

?>

Hello dear friends,

If i've the following ( categories - gifts )

<?PHP
$sql ="select *  from  categories";
$result= mysql_query($sql);
while($line= mysql_fetch_array($result)){
$qryrow="select * from gifts where gifts='$line[1]'";
}
?>

it means select category and its gifts

now my question how to view it all in table for example if i add

Code: [Select]
<TABLE cellSpacing=1 cellPadding=1 width="100%">
<TBODY>
<TR>
<TD>
<?php echo "". $line[1].""; ?>
</TD>
</TR>
</TBODY>
</TABLE>
it will show me like this
 

How can i make it like this



my problem how to make table like this and where to put the php variable  .. 

thanks

How can i make unlimited sub categories when i have this mysql fields: id, name, parent


Code: [Select]
$query = "SELECT * FROM cats";
$result = mysql_query($query);
while ($var = mysql_fetch_array($result)) {
  echo $var['name'];
  // now i need here some code for sub cats, but sub cat in subcat :(
}
Can someone help me?

Hi there, I have a piece of code that takes all posts with the category of 'model' and puts it on the models page. It then displays the post as a gallery using fancy box. The problem is I want all posts with the category of 'faces' to go on the faces page, but at the moment they are appearing on both pages. Here is the code I have. Any help will guarantee a donation as I am at my wits end trying to figure this out.

Code: [Select]
<?php
  global $post;

  $args = array(
    'numberposts'     => 8,
    'offset'          => 0,
    'category'        => 'models',
    'orderby'         => 'post_date',
    'order'           => 'DESC',
    'post_type'       => 'post',
    'post_status'     => 'publish' );

  $posts_array = get_posts( $args );
  $x = 1; // track model
  foreach( $posts_array as $post ) {
    setup_postdata($post); ?>
    <div id="model-thumb">
<?php
    $arg2 = array( 'post_type' => 'attachment', 'numberposts' => -1, 'order' => 'ASC', 'post_status' => null, 'post_parent' => $post->ID );
    $attachments = get_posts($arg2);
    if ($attachments) {
        $i = 1; // track images
        foreach ( $attachments as $attachment ) {
            $image_attributes = wp_get_attachment_image_src( $attachment->ID, 'full' );
            if ($i == 1) { // Get thumbnail info to display with second image
                $img1src = $image_attributes[0];
                $img1w = $image_attributes[1];
                $img1h = $image_attributes[2];
            } elseif ($i == 2) { // Display thumbnail with link to second image
?>
           <a class="allModels" rel="gallery<?php echo $x; ?>" href="<?php echo $image_attributes[0]; ?>" 
?><img src="<?php echo $img1src; ?>" width="<?php echo $img1w; ?>" height="<?php echo $img1h; ?>"/></a><br /><p><?php the_content();?></p></div>

<?php
            } else { // Other images
?>
                <div style="display:none;"><a class="allModels" rel="gallery<?php echo $x; ?>" href="<?php echo $image_attributes[0]; ?>"><img src="<?php echo $image_attributes[0]; ?>" width="<?php echo $image_attributes[1]; ?>" height="<?php echo $image_attributes[2]; ?>"/></a></div>

<?php
            }
            $i++; // Increase image count
        }
    }

    $x++; // Increase model count
  }

?>