Jump to content

williamh69

Members
 View Profile  See their activity

  • Posts

    25

  • Joined

  • Last visited

 Content Type 

Posts posted by williamh69

    populating from db cause problem displaying images

    in CSS Help

    Posted

    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 } ?>

    capture-20200112-200626.png

    Parse error: syntax error, unexpected end of file in ... on line 103

    in PHP Coding Help

    Posted · Edited by Psycho
    Added code tags

    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>

     

    display grid from mysqli and php

    in PHP Coding Help

    Posted

    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

    OnPaste.20191110-215034.png.b3062a8bdf1e9722a9fde5c0beacb0c9.png

     

    i want this kind of result

    OnPaste.20191110-215329.png.df9bc64588c58221e90ae714d026bbf8.png

    thank you

    Call functions

    in PHP Coding Help

    Posted

    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

     

    update problem

    in PHP Coding Help

    Posted

    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

     

    ECHO STOP WORDS BEFORE ENTER TO DB

    in PHP Coding Help

    Posted

    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);
    }
}

    ECHO STOP WORDS BEFORE ENTER TO DB

    in PHP Coding Help

    Posted

    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

    php and mysql question

    in PHP Coding Help

    Posted

    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?

    php and mysql question

    in PHP Coding Help

    Posted

    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&amp;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.

     

     

     

    can retrieve the correct data

    in PHP Coding Help

    Posted 

    <?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

    can retrieve the correct data

    in PHP Coding Help

    Posted

    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

    dynamic navigation with php from mysql

    in PHP Coding Help

    Posted

    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

    dynamic metatags with php and mysql

    in PHP Coding Help

    Posted

    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'

    dynamic metatags with php and mysql

    in PHP Coding Help

    Posted

    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

     

     

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.