williamh69
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Everything posted by williamh69
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Hi guys i am populating products from db, but the images does not showing sequentially, there are gaps between images ,,how i can fix this this is my code: <?php $query = "SELECT * FROM products"; $select_products = mysqli_query($connection, $query); while ($row = mysqli_fetch_assoc($select_products)){ $product_id = $row['product_id']; $product_title = $row['product_title']; $product_image = $row['product_image']; $product_description = $row['product_description']; $product_quantity = $row['product_quantity']; $product_price = $row['product_price']; $short_desc = $row['short_desc']; ?> <div class="col-sm-4 col-lg-4 col-md-4"> <div class="thumbnail"> <img src="./images/<?php echo $product_image ?>" width="478" height="1034" alt=""> <div class="caption"> <h4 class="pull-right">$<?php echo $product_price ?></h4> <h4><a href="#"><?php echo $product_title ?></a> </h4> <p><?php echo $product_description ?>.</p> </div> <div class="ratings"> <p class="pull-right">18 reviews</p> <p> <span class="glyphicon glyphicon-star"></span> <span class="glyphicon glyphicon-star"></span> <span class="glyphicon glyphicon-star"></span> <span class="glyphicon glyphicon-star"></span> <span class="glyphicon glyphicon-star-empty"></span> </p> </div> </div> </div> <?php } ?>
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Hi guys i have this error, Parse error: syntax error, unexpected end of file in C:\wamp64\www\nigthclub\videos.php on line 103 but i know is probably easy to solve but for some reason i dont see the error. please help me <?php session_start(); ?> <?php error_reporting (E_ALL ^ E_NOTICE); ?> <!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"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href="style.css" rel="stylesheet" type="text/css" /> <link href="tablefiestas.css" rel="stylesheet" type="text/css" /> <title>La Taverna de Juan</title> </head> <body> <div id="container"> <div id="header"> <!--MENU BAR--> <?php include("includes/db.php"); ?> </div> <!--VIDEOS--> <?php $query = ("SELECT * from videos"); $result = mysqli_query($connection,$query); $num_per_page =05; ?> <table border="1" width="100%"> <tr> <td>Video Id</td> <td>Titulo</td> <td>Video</td> </tr> <tr> <?php while ($row=mysqli_fetch_assoc($result)) { $videoid = $row['videoid']; $titulo = $row['titulo']; $url_video = $row['url_video']; ?> <td><?php echo $videoid ?></td> <td><?php echo $titulo ?></td> <td><?php echo $url_video ?></td> </tr> <?php}?> </table> <?php $query = "SELECT * from videos"; $pr_result = mysqli_query($connection,$query); $totalrecord = mysqli_num_rows($pr_result); echo $totalrecord; ?> <br> <div id="footer"> <!--FOOTER--> <?php include("includes/footer.inc.php"); ?> </div> </div> </body> </html>
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ok thanks
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hi trying to make a grid of two columns , data from db using mysqli and php, but it does not work...can you guys help me thanks in advance <?php $limit=2; $count=0; echo"<table border='0' align='center' cellpadding='2' cellspacing='2' width='70%'>"; $query = "SELECT * FROM posts"; $select_all_categories_query = mysqli_query($connection, $query); while ($row = mysqli_fetch_assoc($select_all_categories_query)){ $post_image = $row['post_image']; $post_title = $row['post_title']; $post_content = $row['post_content']; $post_user = $row['post_user']; $post_date = $row['post_date']; if ($count < $limit) { if($count ==0) { echo"<tr>"; } echo"<div class='post-img'>"; echo"<div class='img'>"; echo"<img src='assets/img/blog/$post_image' alt='Blog'>"; echo"</div>"; echo"<div class='tag'>"; echo"<a href='#'><span class='icon'><i class='fas fa-tags'></i></span> Business</a>"; echo"</div>"; echo"</div>"; echo"<div class='cont'>"; echo"<h6><a href='blog-single.html'>$post_title</a></h6>"; echo"<p>$post_content</p>"; echo"<div class='info'>"; echo"<a href='#'><span class='author'><img src='assets/img/blog/4.png' alt='Post'></span>$post_user</a>"; echo"<a href='#' class='right'><span class='icon'><i class='fas fa-clock'></i></span> $post_date</a>"; echo"</div>"; echo"</div>"; }else { $count=0; echo"<div class='post-img'>"; echo"<div class='img'>"; echo"<img src='assets/img/blog/$post_image' alt='Blog'>"; echo"</div>"; echo"<div class='tag'>"; echo"<a href='#'><span class='icon'><i class='fas fa-tags'></i></span> Business</a>"; echo"</div>"; echo"</div>"; echo"<div class='cont'>"; echo"<h6><a href='blog-single.html'>$post_title</a></h6>"; echo"<p>$post_content</p>"; echo"<div class='info'>"; echo"<a href='#'><span class='author'><img src='assets/img/blog/4.png' alt='Post'></span>$post_user</a>"; echo"<a href='#' class='right'><span class='icon'><i class='fas fa-clock'></i></span> $post_date</a>"; echo"</div>"; echo"</div>"; } $count++; } echo"</tr></table>"; ?> this is the result i want this kind of result thank you
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Got it... my mistake.. calling wrong function syntax
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Hi guys I AM NEW TO PHP AND MYSQLI... TRYING TO CALL FUNCTION BUT I GOT ERROR CANT FIGURE OUT WITH IT IS, THANKS FOR HELP ME functions.php <?php function Updatepost(){ global $db; $query = "UPDATE posts SET title='Yo me quiero casar con gloria', body='Me gusta el 69' WHERE id=3"; $result = mysqli_query($db, $query); if(!$result){ echo mysqli_error($db); } else{ echo "updated row sucessfully"; } } ?> Index.php <?php include("includes/db.php"); include("functions/functions.php"); ?> <?php function Updatepost(); ?> ERROR Parse error: syntax error, unexpected ';', expecting '{' in C:\wamp64\www\gloria-blog\index.php on line 9 THANKS
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hi guys i cant edit my data... what is my error <?php // get value of id that sent from address bar $menu_id=$_GET['menu_id']; // Retrieve data from database $sql="SELECT * FROM menus WHERE menu_id='$menu_id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result) ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><strong>Update data in mysql</strong> </td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><strong>Menu name</strong></td> </tr> <tr> <td> </td> <td align="center"> <input name="menu_name" type="text" id="menu_name" value="<? echo $rows['menu_name']; ?>"> </td> <td> </td> <td align="center"> <input type="submit" name="Submit" value="Submit"> </td> <td> </td> </tr> </table> </td> </form> </tr> </table> <?php // close connection mysql_close(); ?> error said:_ notice: Undefined index: menu_id in C:\wamp\www\sparklenshine2\admin\edit.inc.php on line 4 the tablet is: menu_id | menu_name
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i found this class to use it <?php class Cleaner { var $stopwords = array(" find ", " about ", " me ", " ever ", " each ", " update ", " delete ", " add ", " insert ", " where ", " i ", " a ", " my ");//you need to extend this big time. var $symbols = array('/','\\','\'','"',',','.','<','>','?',';',':','[',']','{','}','|','=','+', '-','_',')','(','*','&','^','%','$','#','@','!','~','`'); function parseString($string) { $string = ' '.$string.' '; $string = $this->removeStopwords($string); $string = $this->removeSymbols($string); return $string; } function removeStopwords($string) { for ($i = 0; $i < sizeof($this->stopwords); $i++) { $string = str_replace($this->stopwords[$i],' ',$string); } return trim($string); } function removeSymbols($string) { for ($i = 0; $i < sizeof($this->symbols); $i++) { $string = str_replace($this->symbols[$i],' ',$string); } return trim($string); } }
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no,,, look for the stop words in the text, echo them back, (so like this you can delete manually or using any function to erase it) before enter to database.
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Some search engines don't record extremely common words in order to save space or to speed up searches. These are known as "stop words."
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Hi guys, thank so much for all your help. I have this question: I have a two tables one is a content text, and the other one there are the stop words. I would like to match my content with those stop words and echo them, before submit to db. content stopwords content_id words content any suggestion thank you
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thank you guys
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Hi guys... please review my website. I appreciate your comments. www.sparklenshinecs.com
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ok thank you very much, i really appreciate your help. thank you thank you
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thank you, i re-wrote the code....this is what i have now <?php $menu_id = $_GET['cat']; $sql = mysql_query("SELECT * from paginas WHERE menu_id=$menu_id"); if(mysql_num_rows($sql)>0) { while($row = mysql_fetch_array($sql)) { $title = $row['title']; $sub_title = $row['sub_title']; $content = $row['content']; $title_tag= str_replace(' . ',' | ',$row['title_tag']); $keywords= $row['keywords']; $description= $row['description']; echo "<title>$title_tag</title>"; echo"<meta name='keywords' content='$keywords'>"; echo"<meta name='description' content='$description'></br>"; } echo"<h1>$title</h1>"; echo"<h2>$sub_title</h2>"; echo"<p>$content</p>"; } else { echo "No results found!"; } ?> what you think?
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Hi guys, thank you for all your help i really appreciated. I have my website, www.sparklenshinecs.com, and i am trying to do a SEO on it. However when I run a seo report appears two different lines of code as follow http://www.sparklenshinecs.com/index.php?content=paginas&cat=7 and http://www.sparklenshinecs.com/index.php?content=paginas&cat=7 if I run the first link it gives me the following error on the page: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/content/10/9601510/html/paginas.inc.php on line 7 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/content/10/9601510/html/paginas.inc.php on line 38 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/content/10/9601510/html/paginas.inc.php on line 56 and if I run the second link: it gives me the normal page. Someone can please explain me whats going on, and how I avoid those warnings.
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<?php $menu_id = $_GET['cat']; $query="SELECT * from paginas WHERE menu_id=$menu_id"; $result=mysql_query($query); while($row=mysql_fetch_array($result,MYSQL_ASSOC)) { $page_id = $row['page_id']; $title = $row['title']; $sub_title = $row['sub_title']; $content = $row['content']; echo "<h1>$title</h1>"; echo"<br>"; echo"<h2>$sub_title</h2>"; echo"<br>"; echo"<p>$content</p>"; } ?> here it is
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adam i used your code but gives me also the same result
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hi cyber i tried but t it gives me the same result
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hi guys i have this two db's menu paginas menu_id page_id name menu_id title content I populate the menu with the following code, which working fine. <?php function query($parent_id) { //function to run a query $query = mysql_query ( "SELECT * FROM menus" ); return $query; } function fetch_menu($query) { while ( $result = mysql_fetch_array ( $query ) ) { $menu_id = $result ['menu_id']; $menu_name = $result ['menu_name']; $menu_link = $result ['menu_link']; echo "<li class='has-sub '><a href='index.php?content=paginas&cat=$menu_id'><span>$menu_name</span></a>"; echo "</li>"; } } fetch_menu (query(0)); //call this function with 0 parent id ?> then I link every menu in my list with each page with this code, but appears the first entry in all the menus, I used the following code: <?php $menu_id = $_GET['cat']; $query="SELECT page_id, title, sub_title, content from paginas where menu_id=$page_id"; $result=mysql_query($query); while($row=mysql_fetch_array($result,MYSQL_ASSOC)) { $page_id = $row['page_id']; $title = $row['title']; $sub_title = $row['sub_title']; $content = $row['content']; echo "<h1>$title</h1>"; echo"<br>"; echo"<h2>$sub_title</h2>"; echo"<br>"; echo"<p>$content</p>"; } ?> what i am doing wrong..... thank you for ur help
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<?php function login() { $con = mysql_connect("localhost", "xxxx", "xxxx") or die('Could not connect to server'); mysql_select_db("xxxx", $con) or die('Could not connect to database'); } ?> this is the loging function.... from file called mylibrary
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hi guys i have the following db table called topnavigation id name url title this is the code to get the navigation from mysql <?php include("mylibrary/login.php"); $sql = mysql_query("SELECT * FROM topnavigation"); while($row = mysql_fetch_array($sql)){ $id = $row['id']; $name = $row['name']; $url = $row['url']; $title = $row['title']; echo"<li>"; echo"<a href='$url' title='$title'>$name</a>"; echo"</li>"; } ?> But i have this error: atal error: Cannot redeclare login() (previously declared in C:\wamp\www\sparklenshine\mylibrary\login.php:3) in C:\wamp\www\sparklenshine\mylibrary\login.php on line 7 thank your for help
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Thank you very much........
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hi guys, i change all the code: <?php $dbhost = 'localhost'; $dbuser = 'whbweb'; $dbpass = 'andres69'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if(! $conn ) { die('Could not connect: ' . mysql_error()); } if (!isset($page_id)) $page_id=1; $sql = 'SELECT keywords,title,description FROM meta_tags where id=$page_id'; mysql_select_db('whb'); $retval = mysql_query( $sql, $conn ); if(! $retval ) { die('Could not get data: ' . mysql_error()); } while($row = mysql_fetch_assoc($retval)) { $title= $row['title']; $keywords= $row['keywords']; $description= $row['description']; echo"<meta name='title' content='$title'>"; echo"<p>"; echo"<meta name='keywords' content='$keywords'>"; echo"<p>"; echo"<meta name='description' content='$description'></br>"; } mysql_close($conn); ?> Now I have the following error: Could not get data: Unknown column '$page_id' in 'where clause'
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hi guys, i have this database table called meta_tags id keywords description title and I have the following code to retrieve the data <?php $id = 1; $query=("SELECT title, keywords, description FROM meta_tags WHERE id =$id"); $result=mysql_query($query); while($row=mysql_fetch_array($result,MYSQL_ASSOC)) { $title= $row['title']; $keywords= $row['keywords']; $description= $row['description']; echo"<meta name='title' content='$title'>"; echo"<meta name='keywords' content='$keywords'>"; echo"<meta name='description' content='$description'>"; } ?> but I have this error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\WHB\includes\metags.inc.php on line 10 can you guys help me please.... thank you
