ebchost

July 17, 2011

i try to build counter with mysql

mysql_query("INSERT INTO `rss_visitor` VALUE ('','$ip','$id_kor','$date1', '$date2','$opcija') ") or die(mysql_error())

format of date1, 2 is

$date1 = date('Y-m-d');

$date2 = date('H:m:s');

this code is ok and work.

Q:

How to present with mysql all data from today ? or before 2 day, .... etc...

SELECT * FROM `rss_visitor` WHERE ? // what go hiar

June 21, 2011

$datum = date('Y-m-d H:m:s');

June 21, 2011

i try tu insert current date and time, but, value inserted in table is 0000-00-00 00:00:00 all the time (wth no value)

[syntax=php]$sql = "INSERT INTO `rss_vesti` SET

`SAJT_ID` = '$rss_izvor_id',

`NASLOV` = '$title1',

`SLIKA` = '$image2',

`SADRZAJ` = '$descritpion_trim1',

`IZVOR` = '$link1',

`DATUM` = 'CURDATE()'";

mysql_query($sql)or die(mysql_error());[/syntax]

wer is the probelm?

June 20, 2011

is posible to integrate this 2 example and how ?

$podaci2 = mysql_query("SELECT * FROM rss_vesti LIMIT $granica") or die(mysql_error());						
while($row = mysql_fetch_assoc( $podaci2 ))
{
	$id		= $row['ID'];
	$naslov		= $row['TITLE'];
	$sadrzaj	= $row['DESCRIPTION'];
	$slika		= $row['IMAGE'];
       }

<?php

$sTemplate = <<<XML
<div class="imageElement">
    <h3>{title}</h3>
    <p>{description}</p>
    <a href="#" title="open image" class="open"></a>
    <img src="{fileurl}" class="full" />
    <img src="{fileurl}" class="thumbnail" />
</div>
XML;

$aUnits = array(
    'image 1' => array('$row', 'pic1.jpg'),
    'image 2' => array('Image 2 description', 'pic2.jpg'),
    'image 3' => array('Image 3 description', 'pic3.jpg'),
    'image 4' => array('Image 4 description', 'pic4.jpg'),
    'image 5' => array('Image 5 description', 'pic5.jpg')
);

$sGalleryObjects = '';
$sFolder = 'data_images/';
foreach ($aUnits as $sTitle => $aUnitInfo) {
    list($sDescription, $sFileName) = $aUnitInfo;
    $sFilePath = $sFolder . $sFileName;
    $sGalleryObjects .= strtr($sTemplate, array('{fileurl}' => $sFilePath, '{title}' => $sTitle, '{description}' => $sDescription));
}

echo <<<EOF
<link rel="stylesheet" href="css/main.css" type="text/css" />
<link rel="stylesheet" href="css/jd.gallery.css" type="text/css" />
<script src="js/mootools-1.2.1-core-yc.js" type="text/javascript"></script>
<script src="js/mootools-1.2-more.js" type="text/javascript"></script>
<script src="js/jd.gallery.js" type="text/javascript"></script>

<div class="example">
    <h3><a href="#">SmoothGallery examples. First - single gallery, second - multiple galleries</a></h3>
    <div>
        <div id="single_gallery">
            {$sGalleryObjects}
        </div>

        <script type="text/javascript">
            function startGallery() {
                var myGal1 = new gallery($('single_gallery'), {
                    timed: true,
                    delay: 5000,
                    embedLinks: false
                });
            }
            window.addEvent('domready',startGallery);
        </script>
    </div>
</div>
EOF;

?>

June 19, 2011

Can you pinpoint your problem for us please

this is code for echo rss value

$items = $rss_link->channel->item;

				foreach($items as $item)
					{
						$title 			= $item->title;
						echo "<h2>LOGO:</h2>$title</br>";
						$link 			= $item->link;
						echo "<h2>LINK :</h2>$link</br>";
						$published_on 		= $item->pubDate;
						echo "<h2>Published_on:</h2> $published_on</br>";

						$description 		= $item->description;
						$descritpion_strip	= strip_tags($description);
						$description_trim 	= trim($descritpion_strip);
						echo "<h2>description_trim:</h2> $description_trim</br><hr>";

						$image_object = preg_match_all('/\"(.*?)\"/', $description, $matches);
						$image = str_replace( '"','',$matches[0][0]);


                           			}

this part of code colecting url of image

$image_object = preg_match_all('/\"(.*?)\"/', $description, $matches);
$image = str_replace( '"','',$matches[0][0]);

and sometimes, show bad value

for example:

$image = "width: 180px; float: right; margin:3px; text-align:center;"

How can I avoid this and get the correct value image url?

thanks

June 19, 2011

i think we (i) need update for this problem

How to correct this (upgrade)?

for some image, i have url value like this

$image = "width: 180px; float: right; margin:3px; text-align:center;"

you can check it hiar: http://moravaprom.com/ebchost/index.php

(as you can see, i make progres in my work, but i still have litle off beginig problem)

thanks in forvard

June 18, 2011

Pgination value for $X = 1 is:

first, -2, -1, 0, [1], 2, 3, 4, last

Pagination value for $x = $total_pages

first, 34, 35, 36, [37], 38, 39, 40, last

Note: I have 37 pages.

<?php
$range = 3;

for($x=($page - $range); $x < (($page + $range) + 1); $x++)
           {
   ?>
           <div style=' text-align: center; float:left; font-family: verdana; width: 20px; padding: 5px; margin: 1px; border-style:solid; border-width: 1px; border-color:#C8C8C8;'>
  <?php
	    echo ($x == $page) ? '<strong><a href="index.php?page='.$x.'"> '.$x.' </a></strong>' : '<a href="index.php?page='.$x.'"> '.$x.' </a> ';
  ?>
</div>
<?php
   }
?>

How to solve this problem?

June 15, 2011

Why this?

How to fixed?

June 15, 2011

if(empty($image)) {

$image = "http://test@test.com/test";

}

ok, thanks.

if you remember example of geting images url, from text (rss feed) for input in data base, you help me in that problem?

i $image string i have value somtimes, but is not corect url adress, in that case also, i want to giv new valu for $image

June 15, 2011

how to write this code?

If somthyng (string $image) is not image, or have no value, then $image = "http://www.somelink...";

June 15, 2011

There is probably a sql special character as part of the data. You should echo the $sql variable as part of the die(...) statement so what we could see exactly what the query is, right before the Bild point where the syntax failed.

Ultimately, you will probably find that the error is because you must escape ALL string data values being put into a query statement so that any special sql characters in the data don't break the sql syntax. See this link - mysql_real_escape_string

i think that is ok now

i will add lot`s of rss source, and then i will see

thankse

June 15, 2011

Its maby UTF-8 problem ?

June 15, 2011

Everything works perfectly when I work without inserting rss data in the database.

Wheb I've started entering data into the database, in the database is write about 200 lines in the table, and then show an error (after about 200 inserted record in database)

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 'Bild'.

', 'http://www.vesti-online.com/Sport/Fudbal/144014/Tejgo-Zelim-' at line 1

before these errors, you can see a news item to be inserted into the base hiar: http://moravaprom.com/ebchost/rss_reader.php

hiar is code of : rss_reader.php

<?php

header('Content-Type: text/html; charset=utf-8');

echo "<h1>RSS vesti</h1><hr>";

$serverurl = "";
$username = "";
$password = "";

$database = "";
$rss_sajtovi = "";

mysql_connect($serverurl, $username , $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

$konekcija = mysql_connect("localhost", "morava_rss", "brodolom") or die ("Parametri za povezivanje sa bazom nisu dobri, ili baza ne postoji!");

$sajtovi = mysql_query("SELECT * FROM rss_sajtovi") or die(mysql_error());

$vesti = mysql_query("SELECT * FROM rss_vesti") or die(mysql_error());


while($row = mysql_fetch_assoc( $sajtovi ))
	{
		$id_sajta 	          = $row['ID_SAJTA'];
		$naziv 		  = $row['NAZIV'];
		$url_sajta 	  = $row['URL'];
		$rss_izvor 	  = $row['RSS'];
		$rss_kategorija = $row['kategorija'];
		$rss_odobren 	  = $row['odobren'];
		$korisnik_id 	  = $row['korisnik_id'];
		$kontrola 	  = $row['kontrola'];		

		$url_rss 	= $rss_izvor;

		$rss_link 	= simplexml_load_file($url_rss);

		if($rss_link)
			{
				echo '<h1>'.$rss_link->channel->title.'</h1>';

				$channelimg = $rss_link->channel->image->url;
				echo $channelimg;

				echo "<img src='$channelimg' alt='$rss_link->channel->title' width='400'>";
				echo "<hr>";

				$items = $rss_link->channel->item;

				foreach($items as $item)
					{
						$title                    = $item->title;
						$link                     = $item->link;
						$published_on     = $item->pubDate;
						$description         = $item->description;
						$description_trim = trim($description);
						$image_object     = preg_match_all('/\"(.*?)\"/', $description, $matches);
						$image                 = str_replace( '"','',$matches[0][0]);

?>
							<div style='overflow:hidden; width:420px; padding: 10px; border-style:solid; border-width: 1px; margin: 5px; '>
								<?php
									echo '<h2><a href="'.$link.'">'.$title.'</a></h2>';
								?>	
									<div style='overflow:hidden;margin:5px; float: left;'><?php echo "<img src='$image' alt='$rss_link->channel->title' width='200'>";?></div>
									<div><?php echo strip_tags($description_trim);?></div>
								<?php
									echo '<span>('.$published_on.')</span>';
								?>
							</div>
<?php
								$sql= "INSERT INTO `rss_vesti` VALUES('', '$id_sajta', '$title', '$image', '$description', '$link', '$published_on')";
                           						mysql_query($sql,$konekcija)or die(mysql_error());
					}
			}
}
?>

June 14, 2011

your trying to get a field called post.id. but you stated your table name is postos. Is that a typo on the forum or a typo on your table name?

typo mistake an the forum

thanks,

now is corect

and,
in my DB structure i have 2 tables:

1. categories - id, name,

2. posts - id, cat_id, title, contents, date_posted

June 14, 2011

I write first time INER JOIN and i have error

Unknown column 'posts.id' in 'field list'

this is the query

<?php
$query = "SELECT posts.id AS post_id, categories.id AS category_id,
			title, contents, date_posted, categories.name
		  FROM posts
		  INER JOIN categories ON categories.id = posts.cat_id";
	mysql_query($query)or die(mysql_error());

?>

and,

in my DB structure i have 2 tables:

1. categories - id, name,

2. postos - id, cat_id, title, contents, date_posted

wer is the problem?

thanks

June 13, 2011

update:

stop with out die("alredy exists");

i want to show message "alredy exists" with form, with out adding new record.

thanks

June 13, 2011

How to stop code for add new record in this situation if i have "test" categories alredy in table "categoryes" wen i want to ad new category with name "test".

file: add_category.php

(note: this file include_once("init.php"); for connection with database, who include_once "blog.php"; with function add_category and category_exist,

<?php
if ($submit)
{
	if($name)
		{			
			if(strlen($name)>24)
				{
					$error = "Naziv kateogrije ne moze biti duzi od 24 karaktera";
				}
				else
					if (category_exist($name))
						{
							$error = "Kategorija postoji";						
						}
						else		
							$upisano = "kategorija: $name je uspesno upisana u bazu";
							add_category($name);


		}
}
?>
<html>

<body>
<h1>Add Category</h1>
<hr>
<?php

if($error)
	{
		echo "<p>$error</p>\n";
	}
	else
		echo "</br> $upisano </br>\n";
?>

... rest of code

File: blog.php

<?php

function add_category($name)
{
	$name = mysql_real_escape_string($name);
	mysql_query("INSERT INTO categories SET name='{$name}'")or die(mysql_error());
}

function category_exist($name)
{

	$name = mysql_real_escape_string($name);
	$query = mysql_query("SELECT COUNT(1) FROM categories WHERE name = '{$name}'") or die(mysql_error());

	return (mysql_result($query,0)=='0')? false: true;
}

?>

June 11, 2011

$slika = str_replace( '"','',$matches[0][0]);

this is give resolt

Syntax

str_replace(find,replace,string,count)

thenks a lot

June 11, 2011

nothing new... :-\

June 11, 2011

$podaci = mysql_query("SELECT * FROM rss_vesti") or die(mysql_error());

		while($row = mysql_fetch_assoc( $podaci ))
			{
				$naslov = $row['NASLOV'];
				$sadrzaj = $row['SADRZAJ'];
				$izvor = $row['IZVOR'];
				$datum = $row['DATUM'];

				?>
				<table>
				<tr>
				<td>

				<?php 
					$garbage = 'just some bs <ul><li>text</li><li>text</li><li>text</li></ul>just some more bs';
					$test = preg_match_all('/\"(.*?)\"/', $sadrzaj, $matches);
					echo $matches[0][0];
					$slika = str_replace($matches, '"', '');
					echo $slika;
					echo "<img src='$slika' alt='' height='100' width='100'/>";

				?>

				</td>
				</tr>
				</table>
				<?php

			}

This code i use in this moment

resolt can be check it in http://www.moravaprom.com/ebchost/prikaz.php adres

June 11, 2011

now pictures have value "http://www.moravaprom.com/ebchost/prikaz.php"

June 11, 2011

http://www.moravaprom.com/ebchost/prikaz.php

you can see the problem hiar ?

I believe that the problem is in quotes "url" ?

June 11, 2011

he, i look wrong place for code

Sorry

June 11, 2011

$podaci = mysql_query("SELECT * FROM rss_vesti") or die(mysql_error());

		while($row = mysql_fetch_assoc( $podaci ))
			{
				$naslov = $row['NASLOV'];
				$sadrzaj = $row['SADRZAJ'];
				$izvor = $row['IZVOR'];
				$datum = $row['DATUM'];

				?>
				<table>
				<tr>
				<td>

				<?php 

				preg_match_all('/<ul>(.*?)<\/ul>/i', $sadrzaj, $matches);
				echo $matches[0][0]; ?>

				</td>
				</tr>
				</table>
				<?php

			}

i hava emty screen, no resolts?

But, in your original code, resolt egsisting.

June 11, 2011

For example

<img src="http://www.vesti-online.com/data/images/2011-06-05/156013_sampioni_kf.jpg?ver=1307647416"><br/>Upravo zavrÅ¡ena klupska sezona obeleÅ¾ena je potpunom dominacijom Partizana, s obzirom da su sekcije naÅ¡eg najbrojnijeg sportskog druÅ¡tva osvojile Äak 11 pehara u pet najpopularnijih kolektivnih sportova.<br/><br/>

this is part of som news article wich i have in my database in tabele.

How i can to separete imege url from rest part of text in 2 diferent variable

$image_url = " url image value from item";

$text = "rest of text value";

in this example final resolt wil be

$image_url = "http://www.vesti-online.com/data/images/2011-06-05/156013_sampioni_kf.jpg";

$text = "Upravo zavrÅ¡ena klupska sezona obeleÅ¾ena je potpunom dominacijom Partizana, s obzirom da su sekcije naÅ¡eg najbrojnijeg sportskog druÅ¡tva osvojile Äak 11 pehara u pet najpopularnijih kolektivnih sportova.";

thanks.

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ebchost

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Posts posted by ebchost

date query

date time current value

date time current value

integrate php, mysql and SmoothGallery

separate text from the image into two different variables

separate text from the image into two different variables

Pagination: $X > 0 and $X <= $totalpages

MySQL server has gone away

If $image is not picture or have no value then...

If $image is not picture or have no value then...

insert external rss data in database ater lots of record stop

insert external rss data in database ater lots of record stop

insert external rss data in database ater lots of record stop

INER JOIN first time writing

INER JOIN first time writing

die (""); problem

die (""); problem

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

separate text from the image into two different variables

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