PHP :: Displaying A URL Stored In A Table As An Image On The Page?

Determine Image Information On Image Stored In MySQL BLOB
I store images in my DB as BLOB. When I want to place them in an html
table, I want to determine the width of the image in order to asign the
correct width to the column inside table.
For JPEGs this is working fine, this is my code:

$image = @imagecreatefromstring($r["foto"]);
return @imagesx($image);

where $r["foto"] is the field selected from the db which contains the image.

For GIFs however, its getting really on my nerves !!
I know imagecreatefromstring is not supported in gd2 and I tried several
thinks. None of them worked ! :
-
$temp = tmpfile();
list($width, $height, $type, $attr) = getimagesize($temp);
echo $attr;

gives me :Warning: getimagesize: Unable to open 'Resource id #6' for reading
-
$handle = fopen("/tmp/tmp.dat", "w+");
fwrite($handle, $r["foto"]);
list($width, $height, $type, $attr) = getimagesize($handle);
echo $attr;

gives me : Warning: getimagesize: Unable to open 'Resource id #6' for
reading
-
Doing this with imagecreatefromgif($temp); wont work either !

Can somebody help me out please? by telling me what I'm doing wrong or what
I should do.

Downloading File Stored In Mysql Table Through Ssl
I have a table where I store some binary data. To retrieve this
information as a downloadable file, I use the following code:

header('Content-Type: $mimeType');
header("Content-length: $size");

// for filename, i store the filename in the table, so this does not
actually
// point to a physical file name on the file system
header("Content-Disposition: attachment; filename=$fileName");
header("Content-Description: PHP Generated Data");
echo $data;
return true;

If I go through a normal port (http), the file downloads no problem.
When I try this through ssl (https), I get the following error:

'Internet Explorer cannot download / from 192.168.0.9.'

Through port 80, the filename is set correctly. Through ssl, the
filename doesn't seem to be getting set. Is there a problem with
using 'Content-Disposition: attachment' through ssl?

Image Upload Problem - New Image Not Displaying
I have a consistent problem that needs knocking on the head. I have an image upload script that uploads an image and replaces the old image. The image that is displayed is sometimes that old image and the page must be manually refreshed to get the new one. I need the new image to be displayed with out have to manually refresh the page.

So far I have identified the following:

•   Ive been testing in ie and firefox – the problem only happens in firefox.

•   If the upload script is activated more than once it works as desired – but not the first time. I need it to work the first time.

•   I have set the page to automatically refresh. – this works in ie, but firefox still needs a manual refresh.

•   I’ve included the following which makes no difference:

•   header("Expires: Mon, 26 Jul 1997 05:00:00 GMT");
•   header("Cache-Control: no-cache");
•   header("Pragma: no-cache");

Howto Insert Picture Stored In Directory With Url In Db Onto Web Page?
If I read one tutorial more on how to store a image inside a mysql database I will blow my mind.

I read everywhere on the forums to store the image on the harddrive and use the url in the database and then pull the image out of the directory and onto a web page.

I am new to PHP and mysql but have looked now for two weeks on how to code a page to do this. It really has to be simple, so simple no one has taken the time to write a tutorial on it. There are lots, many, plenty, please no more tutorials on using BLOB to store an image inside a database.

What is the code to pull a image from a dir. using a url in the db and put it on a web page, I have been trying my own ideas and not working....NO I don't want to use BLOB.

darn thing with php, you can't see the code that does the trick...all I can see is the std <img scr="blab,blab,blab">

Displaying Parts Of A Page While The Page Is Not Complete Yet
A script I'm writing outputs a HTML table with many rows. Each row is the result of a query which takes about one minute to perform. When I run the script in a browser, the page takes several minutes to load and then displays the entire table. Is it possible (using PHP) to display the table as it gets constructed, one row at a time?

Displaying # Of Records In A Table
Can anyone throw me a link to some prefab code, or a tutorial on how I can get a page to display the number of records that meet a certain criteria.

example:
I have a page that stores information on audio FX. I'd like to include a feature that would let the users know how many items are in a particular section.

Player (10)
Items (5)
Weapons (0)
etc.

All help appreciated,

Displaying A HTML Table With PHP
I'm new to PHP and I'm trying to get my head around databases at the moment. So far I have a page that kinda takes the mickey and takes a few details from you, it then creates a cookie that can be removed later. It also prompts you to add the data you gave to a database.

All this works perfectly, but when it comes to displaying the results of the database, all I can show is the first entry in the table. Can someone please tell me how I'm supposed to see all of them?

I've read other articles but they don't make much sense to me.

BTW, the database details are as follows:

Database name: visualanteorg
Table name: people
Fields: id, name, shoe_size, job

As you will see if you choose to look at the pages, getting the information out is not a problem, it's just formatting it how I want it, so each entry appears under the other in the table.

Displaying Table In Php Code
I have a PHP script that has a drop down to select a record, then it should display some text from a database and then display a table of information below it. But, the table is displaying at the top instead. Code:

Grabbing Data From A Table And Displaying It
I'm grabbing data from a table and displaying it but I want to to spand out into multiple pages...I got it all setup but when I click on the next link nothing happens... I got it setup at "3" right now just for testing..I have 5 entries in the database.. anyways could someone help me out? here is the code:

<?php

// Get connection info
include "../connect.php";
// End connection info
if(!$rowstart) $rowstart=0;
// Start database select
$result = mysql_query(
"SELECT * FROM links WHERE notactive='$notactive' order by date desc limit $rowstart,3"); //Select Database
$result2 = mysql_query("select * from links");
if (!$result) {
echo("<P>Error performing query: " .
mysql_error() . "</P>"); exit(); }
// Display the text of each host
while ( $row = mysql_fetch_array($result) ) {
if ($notactive = 1) {
echo ("<table width=500 height=57 border=0 cellpadding=0 cellspacing=0>");
echo ("<tr>");
echo ("<td align=left valign=top><strong>" . $row["name"] . "</strong></td>");
echo ("</tr>");
echo ("<tr>");
echo ("<td align=left valign=top>" . $row["desc"] . "</td>");
echo ("</tr>");
echo ("<tr>");
echo ("<td align=left valign=top><a href=" . $row["url"] . ">". $row["url"] ."</a></td>"); }
echo ("</tr>");
echo ("<br><br>");
echo ("</table>");
}
if ($rowstart>$numrows)
{
echo "<A HREF=$php_self?rowstart=rowstart-3>";
echo "< Previous Page</A>";
}
echo "|";
$numrows=mysql_num_rows($result2);
if($rowstart+3<$numrows)
{
echo "<A HREF='$php_self?$rowstart=rowstart+3'>Next Page></A>";
}
?>

Displaying Information In A 1 Row/2 Column Table
I'm very new to php scripting and haven't quite figured out what I need to do here.

I need to display information in a 1 row/2 column table. I have the MySQL query working fine, but the scripting part in php is killing me!

In the first column a graphic about 70x70 will be shown (plus some additional padding). In the second column will be the description information. The second column will be around 400 pixels wide.

How do I set this up in php? It's easy enough in HTML, but I've about killed myself trying to figure out how to properly lay this out in php!

Displaying Array Results In Table
I'd like to display the results of an array in a table format with the Alias, Alias Destination & Username as column headings & the results underneath. PHP Code:

Displaying Records In A Split Table?
trying to split up results of a query into a split table. example: PHP Code:

Displaying Data In Html Table.
I wondered if anyone could offer some guidance, I trying to write a
php script to connect to a database, and display the records in a
table.

I found the code here in a php4 text, and when I run this directly
through the php intrupeter, the line Successfully connected is display
and 4 as the number of rows.

The database table we connect to, has three field username, firstname,
surname.

When I tried a while loop, to display the data in a table. No values
were displayed.

Could anyone possibly demonstrate how to display the username,
firstname, surname values in a basic table?

<?php
function db_connect(){
global $MYSQL_ERRNO, $MYSQL_ERROR;

$link_id = mysql_connect("localhost","user04", "password04");

if(!$link_id){
$MYSQL_ERROR = "Connection failed";
return 0;
}
else if(empty($dbname) && !mysql_select_db("database04")){
$MYSQL_ERROR = mysql_error();
return 0;
}
else return $link_id;
}

!$link_id = db_connect();
if(!$link_id) die(error_message('Error connecting'));

$result = mysql_query("SELECT * FROM users");
if(!$result) error_message('Error in selection');
else echo "Successfully connected. ";

$query_data = mysql_fetch_row($result);
$total_records = $query_data[0];

if(!$total_records) error_message('No records');
else echo "$total_records.";
?>

Displaying Table Field List
I got a function that displays a list of the fields in a certain mysql table. I got it off of this site. The problem is that it only displays the word Array and I'm not sure why. This is the code for the function: Code:

Displaying A Table Using ID Number As Hyperlink
i am able to insert and display data to the end user, but i am trying to let the end user edit a row. I am unsure how this is usually done, but i am have the following code to display by data

echo "<table border='1'><thead><tr>";
for($i = 0;$i < mysql_num_fields($sql_result);$i++)
{
echo "<th>".mysql_field_name($sql_result,$i).
"</th>";
}
echo "</tr></thead>
<tbody>";
for ($i=0;$i < mysql_num_rows($sql_result);$i++)
{
echo "<tr>";
$row = mysql_fetch_row($sql_result);
foreach($row as $value)
{
echo "<td>".$value."</td>";
}
echo "</tr>";
}
echo "</tbody></table>";

The user have a jump menu (combo box) to pick the table they need to show information on, so the columns displayed are going to vary, but i have a column called ID which is the same in each tables which is the index (auto increment) field.

I was thinking about using this number comnbined with a hyperlink, so the user can click it to take them to a page to edit that row.... But i have no idea how to do this, or if there is an easier way which i just do not know; or how do most people achieve what i require...

Combining Data From Two Fields, Displaying With Another Table
What I am trying to do is pull data from two tables and display it on one page. The code below doesn't display any errors, but all it prints is the results from the first query. Ideas? Thanks.

(I also need to somehow put a variable in for the warranty table data so it is tied to the product_status table data. You can see there is a WHERE in the first query of pid, but no WHERE in the second query. That is a seperate question, though. It would make this way too long, so I will post it separately). PHP Code:

Displaying SELECT Results In A Formatted Table
I was wondering how to show a table with a max limit of 20 results per page. I have various data being pulled and its putting it in a table, but its displayed all in one row. I want to format it so its a 4 column/5 row table. With that I then want the page to generate a numbered list at the bottom that calculates how many pages it takes to show 20 items on each. Code:

Displaying SQL Query Results In 3 Column Table
I would like to be able to display the results of an sql query in an html table, but the results need to be displayed in three columns, for example:

Column 1 Column 2 Column 3
Result 1 Result 2 Result 3
Result 4 Result 5 Result 6
Result 7 Result 8 Result 9

I've tried to find the answers on the internet but i can't find anything that suits my needs!

Image Displaying
It looks as if you are creating the second image, but not saving it anywhere.

This generates an image: imagejpeg($image_p, null, 100)."
";

But what about $image? You do a copyresampled then destroy it. If you want
to save the image, pass the second argument to imagejpeg.

Image Not Displaying?
My code doesn't seem to want to display the images in the location provided. I get renderings but floorplans are not working, any suggestions in what i am doing wrong?

I checked permissions, i made sure the file exists, and thats what my code is suppose to be doing is if it exists than print the file, which i t does

<?php
  print "<h1><img src='http://homes.pacificscene.com/images/browse/logos/" . $row[0] . ".jpg' alt='" . $row[1] . "' /></h1>";
  print "<p>" . $row[6];
$linkresult = mysql_query("SELECT * from psh_commlinks where commID = '" . $row[0] . "'");
while ($linkrow = mysql_fetch_row($linkresult)) {
print "<br />For more information about " . $linkrow[1];
print ", please visit:<br /><a href='" . $linkrow[2] . "'>" . $linkrow[2] . "</a>.";
}
print "</p>";
if (file_exists("http://homes.pacificscene.com/browse/features/" . $row[10])) {
print "<p><a href='features/" . $row[10] . "' target='_blank'>View Features List</a></p>";
}
if (file_exists("http://homes.pacificscene.com/browse/brochures/" . $row[8])) {
print "<p><a href='brochures/" . $row[8] . "' target='_blank'>Download PDF brochure</a></p>";
}
$query = "SELECT * from psh_siteplans WHERE communityID = '$ID' ORDER BY orderID";
   $result = mysql_query($query);
while ($row = mysql_fetch_row($result)) {
print "<form action=" . $_SERVER['PHP_SELF'] . "?ID=" . $ID . "&siteID=" . $row[0] . " method='POST'>";
print "<table border='0' cellspacing='0' cellpadding='5'><tr>";
                    $window = "'http://homes.pacificscene.com/popup.php?type=renderings&ID=" . $row[0] . "','popUp','width=775,height=550'";

print "<td width='206' rowspan='8' valign='top'>";
print '<a href="javascript:;"><img border="0" src="http://homes.pacificscene.com/images/browse/renderings/' . $row[0] . '-TH.jpg" onclick="MM_openBrWindow(' . $window . ')" /></a>';
print "</td>";
print "<td width='135'><strong>Model</strong></td>";
                  print "<td>" . $row[1] . "</td></tr>";
  print "<tr><td><strong>Appx Sq. Ft</strong>.</td>";
                  print "<td>" . $row[2] . "</td></tr>";
print "<tr><td><strong>Stories</strong>.</td>";
                  print "<td>" . $row[3] . "</td></tr>";
print "<tr><td><strong>Bedrooms</strong>.</td>";
                  print "<td>" . $row[4] . "</td></tr>";
print "<tr><td><strong>Baths</strong>.</td>";
                  print "<td>" . $row[5] . "</td></tr>";
  print "<tr><td><strong>Garage</strong>.</td>";
                  print "<td>" . $row[6] . "</td></tr>";

$window = "'http://homes.pacificscene.com/popup.php?type=floorplans&ID=" . $row[0] . "','popUp','width=825,height=625,scrollbars=yes'";
if (file_exists("http://homes.pacificscene.com/images/browse/floorplans/" . $row[0] . ".jpg")) {
print '<tr><td><a href="javascript:;" onclick="MM_openBrWindow(' . $window . ')">Floorplan</td></a>';
print "<td> </td></tr>";
}
$window = "'http://homes.pacificscene.com/popup.php?type=renderings&ID=" . $row[0] . "','popUp','width=775,height=550'";
if (file_exists("http://homes.pacificscene.com/images/browse/renderings/" . $row[0] . "-TH.jpg")) {
print '<tr><td><a href="javascript:;" onclick="MM_openBrWindow(' . $window . ')">Enlarged Rendering                         </td></a>';
print "<td> </td></tr>";
}
    print "</table><br />";
}
  ?>

Displaying Multiple Queries To Same Table Excluding Different Rows!
i've got a table with 224 town listings and their ID's in it from 6 counties, and I want it to displayin a 6 columned table, each county getting their own column. Anyway, here's what I've got and what's happening: PHP Code:

Displaying An Image From A MYSQL DB
Can someone point me in the right direction as to display an image in PHP that is being stored in MySQL? Basically I know how to store the image in MySQL, but getting it to display on a web page is what I am having trouble with.

Displaying An Image From Directory
I am new to php and having some troubles with something that should be relatively simple. I am trying to read an image file from a directory and display it in an html table. The catch is, the directory is not a set variable because it depends on the user who is logged in . this directory is called "$imagedirectory" below and is passed through a SESSION.
for example $imagedirectory = "/htdocs/page/members/johnny_tracks/images/imagefile.jpg"

the area of concern is highlighted below <img src="$imagedirectory"/>

I have checked that the directory is correct AND that the file exists. I have no clue how to show the image, every I run it the output is either nothing or "/htdocs/page/members/username_tracks/images/imagefile.jpg". Code:

Displaying A Binary Image
I would like to know if anyone knows how I could go about sending an image to the browser that is binary. Here is what I am doing right now...but it's not working:

header("Content-Type: image/jpeg");
header('Content-Disposition: inline; filename="image.jpg"');
$theImage = $thePage->GetAs("image.jpg", 350, 270, 60);
imagejpeg($theImage);

I have an ASP script that I am going off of to pull this together, and it uses "Response.BinaryWrite", is there an equivalent in PHP?

Problem Displaying Image From Database
I used the code below to pull out an image from a mysql database
but nothing shows on the page and there was no error message to
indicate any problem.What could be wrong.The image is actually in the
database with all the details:

<?php
if(isset($_GET[loginid]))
{
$con=mysql_connect("localhost","name","pass");
if(!$con)
{
die('Could not connect to database:'.mysql_error());
}
mysql_select_db("datab",$con);
$id = $_GET[loginid];
$query = "SELECT name, type, size, content " .
"FROM image WHERE id= '$id'";
$result = mysql_query($query) or die('Error, query failed');
$file=mysql_fetch_array($result);
$name=$file['name'];
$type=$file['type'];
$content=$file['content'];
header('Content-length: '.strlen($content));
header("Content-type: $type");
header("Content-Disposition: inline; filename=$name");
echo $content;
}
mysql_close($con)
?>

Displaying Image Stop The Script?
im using the gd function for displying an image, but after:

imagepng($image);
imagedestroy($image);

seems the script stop without execute all the rest. why that?

Image Display With 1 Displaying Every 10 Seconds...
I'm using:

<?php
$images = array("1.jpg","2.jpg","3.jpg");
srand(time());
for ($i=0;$i<3;++$i)
echo "<img src='$images[$i]'>";
?>

but it displays all the images. I need it to display one every 10 seconds.

Displaying An Image Based On URL Origination-URL Redirect
I have several domains that use splash pages that point to my main
site:
i.e. http://www.mysite.com/index.php.

How do I replace an image, named "Logo.gif" on the main page
"http://www.mysite.com/index.php" with the image named "logo.gif"
from the URL splashpage, " http://www.mysite.com/SanDiego/"

Both images are named logo.gif because the PHP script references
logo.gif through the script in too many places...It is just easier to
keep the name logo.gif

Problem:
If a user is on the San Diego page and clicks to the main page I want
the user to feel that the site is all about SanDiego.
If a user is on the Los Angeles page and clicks to the main page I want
the user to feel that the site is all about Los Angeles.

Hence the site and database will still be the same..I just want them to
feel the experience is personalized for their city and this will be
determined by changing the logo.gif.

I am using an image named logo.gif that will display SanDiego for the
San Diego page
I am using an image named logo.gif that will display Los Angeles for
the Los Angeles page

I have seen some PHP code that uses php echo $picture I am not sure if
this will work....

Displaying A Page At A Specified Position
I have a form about half way down a web page, which posts criteria for the
WHERE part of an sql which then returns the results at the bottom of the
page.

Trouble is when the page reloads it goes to the top of the page again so the
results are not visible and the user is sometimes unaware that anything has
happened.

I could get a message to display at the top to tell user to scroll to
bottom but i would prefer the page to display at the start of the records(in
a html table).

Displaying Just A Blank Page??
I have commented out a lot of my code and have realised that the code below is what the problem is. I have placed ini_set('display_errors', 'on');
error_reporting(E_ALL); at the very top of the page. No errors are displaying. However I can't see what the problem could be. Code:

Displaying Mysql On Web Page
i am archiving data in mysql, each has an id number from smallest (oldest) to largest (newest), by the week. on my website, i would like to display only the newest week and the week before it. (or the weeks with the largest ids). how can i do that?

Displaying X Number Of Results Per Page
What I need to do is this. I query a MySQL database and, for example, 400 results are returned. I can get it to display the number of results, but I would like to set a value to display, say 20, results per page and generate numbered links with and offset to browse through and next/previous buttons -not unlike this page. Can anyone point me in the direction of a good article or if you have it, post some code here?

Displaying Page Render Time
Some sites (some forum software etc) have a line at the bottom of the page that says Page rendered in 0.0002414 seconds or something of that ilk.

My question is - how would one code this? The only way I could think would be to take the time at the top of the page and the time at the bottom of the page, take one from the other and display that - am I on the right track?

Displaying Text While Page Is Loading
i'm trying to display text while a page is loading using a method
similar to the following:

<?
ob_end_flush();
echo 'AAA<br>'
flush();
sleep(10);
echo 'BBB'
?>

in this script, AAA and BBB appear at the same time - when the page has
fully loaded - which is not what i want (i want AAA to appear and then
10 seconds later, BBB to appear). pursuant to the suggestions on
php.net's entry for flush, i've also tried the following to no avail:

<?
echo 'AAA<br>'
ob_flush();
flush();
sleep(10);
echo 'BBB'
?>

Displaying Last Visitor On Profile Page
I'm in the process of malking a php/mysql website where each registered member has a profile page. This is what I'm trying to achieve:

-When some member visits someone else, I want that someone to see the recent visitor's username on his/her profile page. Can someone give me some hints on how this could be done?

Security Code Image In PHPNuke Login Isnt Displaying
i installed the PHP Nuke in a remote web space bought from prackhost.Problem is that iam not able to get the security code to login to the site as admin or normal user. All the remaining images like links and logo are comming except that security code. Iam not able to do any furthur action.

Displaying X Number Of MySQL Results Per Page
I'm writing a search on a MySQL DB, and everything so far has just gone swimmingly. I've got a "results per page" drop down box in my form, with a few different values.

At the moment, I'm just using:

while($myrow = mysql_fetch_array($result)

to output results to screen.

I'm not too sure how to go about it, but I did have two thoughts:

1) Suck the results into a multi-dimensional array, and pass the array to a display results script (not sure if you can pass arrays to other scripts???

2) Display x number of results, then have the "Next" (or 2, 3... ) links call the search again, but displaying results x+1 - x+x.

1 sounds like too much trouble, and 2 sounds inefficient. Not too sure how else to go about it. Any info / thoughts you guys have would be much appreciated.

On a side note, how do you look at a specific row in the result?

Displaying Html Code On PHP Generated Page
I am trying to create a link page in my php application. this is a page that will display a block of html code that a visitor can copy and paste into their web site to create a link back to mine. My PHP page has a line in it like this:

define('TEXT_INFORMATION', '<html page code is here>');

I have added the html code to create the page as noted above. the page contains a same of the banner image file and the html code that goes alon with it. My problem is that I haven't been able to "quote" the html code so that ir will not be executed.

every time I try to run the application from my web server my page will display two copies of the banner iimage not on banner image and text block like it should. Any suggestions?

Displaying A URL Stored In A Table As An Image On The Page?

In a table in my database I have a column for storing the URL of an image of a product. What PHP code would I use to display the image? PHP Code: