Sending php variable to another php page

Question

I need a little help with some coding. I have a page named index.php and a picture.php.

index.php

$pid = 0;

for($ctr = 1; $ctr <= 2; $ctr++)
{
    echo '<tr>';
    $pid++;

    echo '<td>';
    echo '<a id="various3" href="picture.php" title="try" idno="5"><img src="picture.php" width="150" height="210">';
    //echo '<br>'.$pagenm;
    echo '</td>';

    echo '</td>';
    echo '</tr>';
}

After $pid++ is executed, I want its value to be sent to picture.php, so that it can be used for a retrieving query.

picture.php

$host = 'localhost';
$user = 'root';
$pw = '';
$db = 'thepillar';

mysql_connect($host, $user, $pw);
mysql_select_db($db);
$sql = "select pics, ext from infopics where id='5'";
// mysql_real_escape_string($sql, mysql_connect);
$result = mysql_query($sql) or die('Bad query at 12!' . mysql_error());
while ( $row = mysql_fetch_array($result, MYSQL_ASSOC) ) {
    $db_img = $row['pics'];
    $type = $row['ext'];
}
$img = base64_decode($db_img); // print_r($db_img );
$img = imagecreatefromstring($img);
header("Content-Type: image/jpeg");
imagejpeg($img);
imagedestroy($img);

I have only assigned 5 as the id to be used to retrieve the image from the database. This is because I don't know how to fetch $pid from index.php and assign it to a variable which I can use in my SQL query.

Any help would be much appreciated. Thanks in advance.

Are you grabbing picture.php with ajax? or just loading the page as a full regular page? — Spidy
– Spidy, Commented Mar 4, 2011 at 3:57

Your Common Sense · Accepted Answer · 2011-03-04 04:00:54Z

1

index.php

for($ctr=1; $ctr<=2; $ctr++)
{
     echo '<tr><td><a id="various3" href="picture.php?id=', 
     // I use $ctr insteaf of $pid because is the same... starts with 1
     $ctr,'" title="try" idno="5"><img src="picture.php" width="150" height="210">',
     '</td></td></tr>';
}

picture.php

<?php
...
$sql = sprintf("select pics, ext from infopics where id='%d'", $_GET['id']);
...
?>

edited Mar 4, 2011 at 4:00

Your Common Sense

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answered Mar 4, 2011 at 3:58

Joseadrian

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2 Comments

zerey Over a year ago

.dude if i use $ctr instead of $pid the count will return to zero upon reaching the condition $ctr<=2. but A BIG THANKS TO YOU! this works FINE with me! specially the $sql = sprintf("select pics, ext from infopics where id='%d'", $_GET['id']);

Joseadrian Over a year ago

why it will return to zero? i'm just printing it. you're welcome jaja

Long Ears · Accepted Answer · 2011-03-04 03:57:54Z

1

You could include it in the query string of picture.php, e.g.:

echo '<a id="various3" href="picture.php?id=', $pid, '" ... ';

Then in picture.php:

if (!isset($_GET['id']) || !ctype_digit($_GET['id'])) {
    // todo: some proper error handling
    exit;
}
$sql = "select pics, ext from infopics where id=" . $_GET['id'];

answered Mar 4, 2011 at 3:57

Long Ears

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Comments

Your Common Sense · Accepted Answer · 2011-03-04 04:10:11Z

0

Don't store pictures in database, but save names only.
You won't need to pass anything anywhere and you'll avoid other silly mistakes from your code.

make your main page like this

<?php
$host = 'localhost';
$user = 'root';
$pw = '';
$db = 'thepillar';

mysql_connect($host,$user,$pw); 
mysql_select_db($db); 
$sql = "select pic from infopics limit 2";
$result = mysql_query($sql) or trigger_error(mysql_error().$sql); 
while($row = mysql_fetch_array($result,MYSQL_ASSOC))
{
  echo '<tr>';
  echo '<td>';
  echo '<a id="various3" href="/pictures/'.$row['pic'].'" title="try" idno="5">
        <img src="/pictures/thumbnails/'.$row['pic'].'" width="150" height="210">';
  echo '</td>';
  echo '</tr>';
} 
?>

that's all

answered Mar 4, 2011 at 4:10

Your Common Sense

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