newbie dummy question (DIV and mysql queries)

brlissab · May 10, 2009

Hello friends
after tried some tables to fix displayed queries, seems it is much more flexible with css.
here some tries:

Code:

<?php
require_once ('mysqli_connect.php');
 $q="SELECT * FROM imagens";
 $result = @mysqli_query ($dbc, $q)
 or die("Error: ".mysqli_error($dbc));
while ($rows = mysqli_fetch_array($result))
{
extract ($rows);
$image_nome = $rows['imagem_nome'];
 $image_data = $rows['imagem_data']; 
  $image_id = $rows['imagem_id']; 
            }
?>
<html>
<head>
       <link rel="stylesheet" href="includes/css1.css" type="text/css" media="screen" />
     <meta http-equiv="content-type" content="text/html; charset=utf-8" />
     <title>Details</title>
 </head>
 <body>
 <div id="container">
<img src="<?php echo "../images/$rows['imagem_nome']" ?>" border='0'>
</div>
</body>
</html>

gives me an error
"Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in D:\Program Files\EasyPHP 3.0\

http://www\prod\testcss.php

on line 21"
seems that i have a little problem with double quotes? tried to chane into ' but doesn't have great impact on script....
If if a kind soul could advertise me i would be great.
Thnk u

feherke · May 10, 2009

Hi

Code:

<img src="<?php echo "../images/$rows[imagem_nome]" ?>" border='0'>

Feherke.

http://rootshell.be/~feherke/

brlissab · May 10, 2009

yes thank u!
ive replaced the code between the DIV tag for

Code:

<img src="<?php echo "a href='../images/{$rows['imagem_nome']}'
                           border='0'>
            <img src='../images/{$rows['imagem_nome']}' border='0'
              width='100' height='80'></a>" ?>

works, no error on executing the script but doesn't display correctly the image (only u know a little graphic....) and the request has a while loop who normally displays all the images on the db.....
What must i change on my code?
thank u

brlissab · May 19, 2009

tried this and have my columns displayed.

Code:

<?php
$connection = mysql_connect("localhost" , "user" , "pass") or die ("Can't connect to MySQL");
$db = mysql_select_db("db" , $connection) or die ("Can't select database.");

$cols = 2;
$size = (80/$cols) - 5;
$result = mysql_query("SELECT * FROM imagens order by imagem_id desc") or die (mysql_error());
if (mysql_num_rows($result) > 0){
    echo <<<CSS
<style>
html, body, div, img {padding:0px; margin:0px;}
.results { width: 30%; margin: 0 auto; clear:both; overflow:hidden; }
.cell {width: {$size}%; float:left;}
.result
{
clear:both; 
overflow:hidden;
margin: 0 auto;
position: absolute;
width: 30%;
}
</style>
<div class="result">    
CSS;
    while ($t = mysql_fetch_object($result)) {
        echo <<<HTML
<div class="cell">
<a href=../images/{$t->imagem_nome} border='0'>
<img src='../images/{$t->imagem_nome}' width='100' height='80' border='0'></a>
</div>
HTML;
}
echo "</div><!--close results div--!>";
} else {
    echo "no results";
}

?>

how can i reuse the query to retrive more results from it?
now i have the thumbnail on the left and i wnat to have infos from the query on the rigth like this
img | date
img | description
img | category

so the idea is to reuse the query to display other infos...how do i get that?

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newbie dummy question (DIV and mysql queries)

brlissab

Technical User

feherke

Programmer

brlissab

Technical User

brlissab

Technical User

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