PHP Forum

**lothop** · 06-24-2004, 06:31 PM

How would I go about placing pictures after getting information back from the database?

For example.

I recall the Username and Email of each member. Which is placed in a table.
Then I recall their security level. And if their security level is "1" then display "1.gif" after their username and email.

Then it goes to the next entry but their security level is "2", so it now places 2.gif behind that users name and email.

Any help would be greatly appreciated.
Thanks,
Lothop

**Dark-Skys99** · 06-24-2004, 06:43 PM

This should do....

PHP Code:


			
<?php



function check($Security_level) { 

    if( $Security_level = 1 )  return '1.gif'; 

    if( $Security_level = 2 )  return '2.gif'; 

} 



$User_level = check($Security_level);



echo "$User_level";



?>

Hope this helps

If you need any more help just ask

-James

**Strike Hosting** · 06-24-2004, 06:43 PM

Edit: You beat me to it!

PHP Code:


			
<?php

$level = ($security_level == "2") ? "2.gif" : "1.gif";

echo "<img src='$level'>";

?>

**lothop** · 06-24-2004, 08:15 PM

Thanks for the quick responses

I'll give it a go and let you guys know whats up

Cher,
Lothop

**lothop** · 06-24-2004, 09:17 PM

I think I stuffed up somewhere lol. It returned text "1.gif"
Here is my php code in total.

PHP Code:


			
<? 

/* declare your DB variables */ 

$DBname = 'lothop_'; 

$DBuser = '*****'; 

$DBpass = '*****'; 

$DBhost = 'localhost'; 



/* Make The Connection */ 

mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable toconnect to database"); 

@mysql_select_db("lothop_") or die("Unable to select database $DBName"); 



/* SQL */ 

$sqlquery = "SELECT email,username FROM users"; 

$result   = mysql_query($sqlquery); 

$number   = mysql_num_rows($result); 



/* Display Results */ 

if($number < 1) 

    print "<CENTER><P>There Were No Results for Your Search</CENTER>"; 

else 

{ 

    while($data = mysql_fetch_array($result)) 

    { 

        $name = $data['email']; 

        $age  = $data['username']; 

        print "<p><b>Name:</b> $name<br><b>Age:</b> $age</p>";  

    } 

} 



function check($level_id) { 

    if( $level_id = 1 )  return '1.gif'; 

    if( $level_id = 10 )  return '2.gif'; 

} 



$User_level = check($level_id); 



echo "$User_level"; 





?>

Their Security levels are stored in the database as level_id in the table users.

Any help on this would be leg-end

Thanks,
Lothop

**Kyrnt** · 06-24-2004, 10:50 PM

lothop,

If you want to see the image, you need to put the text (which is the name of the image file) into an image tag.

PHP Code:


			
<?php

    $userLevelImg = check($level_id);

?>

<img src="yourImagesPath/<?= $userLevelImg ?>" />

**lothop** · 06-26-2004, 07:22 AM

I can't seem to get this to work.
I need it to be able to,

Display their username, display their email, then display their security picture.

I tried those codes, but I couldn't implement

PHP Code:


			
<?php 

    $userLevelImg = check($level_id); 

?> 

<img src="yourImagesPath/<?= $userLevelImg ?>" />

I get error "Fatal error: Call to undefined function: check() in /home/lothop/public_html/testing/memberstest.php on line 126"

And those other codes just seem to always display the same picture.

Is it possible to, check to see what $level_id they are off the database, then if their $level_id is 1, display 1.gif, if $level_id is 2, display 2.gif.

And somehow incorperate it into it displaying the username and email and the picture.

Thanks
Lothop

**Kyrnt** · 06-28-2004, 04:42 PM

lothop,

Ok, based on your previous post, you indicated that check() was just returning "1.gif" and now you are getting an error when you call check()? Are you including the file in which check() is defined?

**lothop** · 06-28-2004, 05:35 PM

I don't know which code im suppost to be using.

Do I use

PHP Code:


			
<?php 



function check($Security_level) { 

    if( $Security_level = 1 )  return '1.gif'; 

    if( $Security_level = 2 )  return '2.gif'; 

} 



$User_level = check($Security_level); 



echo "$User_level"; 



?>

And then within that code add

PHP Code:


			
<?php 

    $userLevelImg = check($level_id); 

?> 

<img src="yourImagesPath/<?= $userLevelImg ?>" />

I'm kinda confused

hehe

Thanks
Lothop

**Dark-Skys99** · 06-28-2004, 05:49 PM

First you have...

PHP Code:


			
<?php 



function check($Security_level) { 

    if( $Security_level = 1 )  return '1.gif'; 

    if( $Security_level = 2 )  return '2.gif'; 

} 



$User_level = check($Security_level); 



echo "$User_level"; 



?>

Then add...

PHP Code:


			
<img src="yourImagesPath/<?= $userLevelImg ?>" />

into the Html where you want the image to appear

Hope this helps

-James

**lothop** · 06-29-2004, 02:23 AM

It's still not working.
It displays 1.gif, and then tries to display a picture that it cant find, eventhough the image is right there.

Example http://www.lothop.clanecko.net/testing/memberstest.php

My code is

PHP Code:


			
<?

function check($level_id) { 

    if( $level_id = 1 )  return '1.gif'; 

    if( $level_id = 10 )  return '2.gif'; 

} 



$User_level = check($level_id); 



echo "$User_level";



} 

?>

              <img src="<?= $userLevelImg ?>" />

But above that is the database connection code. So it is connected to the database.

Lothop

**Kyrnt** · 06-29-2004, 02:49 AM

At this point, I would recommend you take a close look at the resulting HTML in your browser. Check the path to your image, etc. Do you see any errors there?

**lothop** · 05:27 AM

The images path is fine. As far as I know its just php code thats stuffing up.

Just wondering, if it was working but couldnt find the image their would be an image for each user that couldn't be found.

Why would it be displaying 1.gif?
Is the php code displaying an image for each member?

Umm, one question though, do I need to store the images in the database or just have the files on the server?

Other than those possible errors I can't see any others..
Thanks,
Lothop

**Dark-Skys99** · 06-29-2004, 07:44 AM

Quote:

Originally Posted by lothop

Umm, one question though, do I need to store the images in the database or just have the files on the server?

You store the pics on your server...

So the pics should be @

http://www.lothop.clanecko.net/testing/1.gif
http://www.lothop.clanecko.net/testing/2.gif

Hope this helps

-James

**lothop** · 06-29-2004, 06:34 PM

Ok, well they are stored on the server.. I still can't get this to work >.<

Thanks for your continueing help
Lothop

**Dark-Skys99** · 07:32 PM

Got an example of the coding for the whole page??

Might be able to find whats wrong

Also one thing.. On your Site I wouldn't display the email addresses of the members.. Otherwise they will recieve loads of spam.. You could always put 'you [at] yoursite.com'

-James

**lothop** · 06-29-2004, 11:52 PM

It was a test page

I'm not going to be displaying the members emails in the final run.

<html>
<head>
<title>Madox Guild - Members</title>
<meta name="description" content="A Gunbound Guild with it all! Join the Madox Guild today!">
<meta name="keywords" content="Gunbound Madox ">
<META NAME="keyphrases" CONTENT="Best gunbound guild, Gunbound tips and tricks">
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<link href="forAll.css" rel="stylesheet" type="text/css">
</head>
<body background="images/bg.gif" link="#003399" vlink="#0044D2" alink="#FF3300" leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
<div align="center">
<table width="100%" height="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td height="83" background="images/top_bg.gif"><table width="100%" border="0" align="center" cellpadding="0" cellspacing="0">
<tr valign="top">
<td width="50%"><a href="index.php"><img src="images/logo.gif" width="188" height="83" border="0"></a></td>
<td width="50%" align="right"><img src="images/statement.gif" width="202" height="25"></td>
</tr>
</table></td>
</tr>
<tr>
<td height="22" background="images/menu_bg.gif" class="forTexts"><div align="right">
<table width="100%" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="293"> </td>

<td width="707" class="forTexts">|    <a href="about.html">About
the Guild</a>     |     <a href="http://www.lothop.clanecko.net/forum/index.php" target="_blank">Forum</a>     |  &nbsp ; 
<a href="sitemap.html">Sitemap</a>     |    
<a href="contact.html">Contact</a>     | </td>
</tr>
</table>
</div></td>
</tr>
<tr>
<td valign="top"><table class="forTexts" width="1000" border="0" cellpadding="2" cellspacing="0">
<tr>
<td width="168" valign="top"> <table width="100%" border="0" cellpadding="1" cellspacing="1">
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="index.php">Home</a></td>
</tr>
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="rules.html">Rules</a></td>
</tr>
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="members.php">Members</a></td>
</tr>
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="restricted/index.php">Login</a></td>
</tr>
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="restricted/register1.php">Join</a></td>
</tr>
<tr>
<td bgcolor="#4F7DB0" class="forTexts"> </td>
<td bgcolor="#D9E3EE" class="forTexts"> <a href="javascript:window.external.AddFavorite('http ://www.madox.co.nr',
'Madox Guild')">Bookmark</a></td>
</tr>
</table>
<br> <table width="100%" border="0" cellspacing="1" cellpadding="1">
<tr>
<td height="20" bgcolor="#80A0C6" class="forTexts"> <div align="center"><font color="#FFFFFF"> Announcement</font></div></td>
</tr>
<tr>
<td bgcolor="#E9EFF5" class="forTexts"> <table width="100%" border="0" cellspacing="1" cellpadding="3">
<tr>

<td class="forTexts">Madox Guild is up and running. Still
working out a few glitches but should be sorted out
quickly. </td>
</tr>
</table></td>
</tr>
</table>
<br>
<table width="100%" border="0" cellspacing="1" cellpadding="1">
<tr>
<td height="20" bgcolor="#80A0C6" class="forTexts"> <div align="center"><font color="#FFFFFF"> News
& Events</font></div></td>
</tr>
<tr>
<td bgcolor="#E9EFF5" class="forTexts"> <table width="100%" border="0" cellspacing="1" cellpadding="3">
<tr>

<td class="forTexts">Guild started on 25/6/04</td>
</tr>
</table></td>
</tr>
</table></td>
<td width="675" valign="top">
<?
/* declare your DB variables */
$DBname = 'lothop_';
$DBuser = '***';
$DBpass = '***';
$DBhost = 'localhost';

/* Make The Connection */
mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable toconnect to database");
@mysql_select_db("lothop_") or die("Unable to select database $DBName");

/* SQL */
$sqlquery = "SELECT email,username FROM users";
$result = mysql_query($sqlquery);
$number = mysql_num_rows($result);

/* Display Results */
if($number < 1)
print "<CENTER><P>There Were No Results for Your Search</CENTER>";
else
{
while($data = mysql_fetch_array($result))
{
$email = $data['email'];
$username = $data['username'];
print "<p><b>Username:</b> $name<br><b>Email:</b> $email</p>";
}

function check($level_id) {
if( $level_id = 1 ) return '1.gif';
if( $level_id = 10 ) return '2.gif';
}

$User_level = check($level_id);

echo "$User_level";

}
?>
<img src="<?= $userLevelImg ?>" />
<td width="150" valign="top">
<table width="100%" border="0" cellspacing="1" cellpadding="1">
<tr>
<td height="20" bgcolor="#80A0C6" class="forTexts"> <div align="center"><font color="#FFFFFF"> Poll</font></div></td>
</tr>
<tr>
<td bgcolor="#E9EFF5" class="forTexts"> <table width="100%" border="0" cellspacing="1" cellpadding="3">
<tr>

<td class="forTexts"><div align="center">No poll at the
moment.</div></td>
</tr>
</table></td>
</tr>
</table>
</tr>
</table>
</td>
</tr>
</table></td>
</tr>
<tr>

<td height="40" background="images/bottom_bg.gif"> <span class="forTexts">All
Rights Reserved. 2004. www.madox.co.nr Webmaster <a href="mailto:Lothop@xtra.co.nz">Lothop</a></span></div>
</div></td>
</tr></table>
</body>
</html>

The database field is called "id_level"
Image names are "1.gif" & "2.gif"
Images location on the server is in the same folder as this page.

Thanks,
Lothop

**Dark-Skys99** · 06-30-2004, 05:36 AM

Its the PHP bit thats the problem...

The code needs to be changed to...

PHP Code:


			
<? 

/* declare your DB variables */ 

$DBname = 'lothop_'; 

$DBuser = '***'; 

$DBpass = '***'; 

$DBhost = 'localhost'; 



/* Make The Connection */ 

mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable toconnect to database"); 

@mysql_select_db("lothop_") or die("Unable to select database $DBName"); 



/* SQL */ 

$sqlquery = "SELECT email,username FROM users"; 

$result = mysql_query($sqlquery); 

$number = mysql_num_rows($result); 



/* Display Results */ 

if($number < 1) 

print "<CENTER><P>There Were No Results for Your Search</CENTER>"; 

else 

{ 

while($data = mysql_fetch_array($result)) 

{ 

$email = $data['email']; 

$username = $data['username']; 

print "<p><b>Username:</b> $name<br><b>Email:</b> $email</p>";

} 



function check($level_id) { 

if( $level_id = 1 ) return '1.gif'; 

if( $level_id = 10 ) return '2.gif'; 

} 



$User_level = check($level_id); 

} 

?>

<img src="<?= $User_level ?>" />

The reason you were getting the 1.gif text was because you had echoed it 'echo "$User_level";'

Also you needed to change the <img src="<?= $userLevelImg ?>" /> to $User_level

The code should work now

Hope this helps

-James

**lothop** · 06-30-2004, 07:02 AM

Yay, it's now displaying a picture, but it is only displaying one picture. And it is the picture that is linked to a member whom has a level_id of 1.

How do I make it so it displays a picture for each member?

I have tried to put the code that shows the image into the loop in the php where it displays the members information, but everything I've tried is failed.

Thanks for your help,
Lothop

**lothop** · 07-02-2004, 01:56 AM

I've been working through this.. and I have tried to make the code simpler.

PHP Code:


			
/* SQL */ 

$sqlquery = "SELECT email,username,level_id FROM users"; 

$result = mysql_query($sqlquery); 

$number = mysql_num_rows($result); 



/* Display Results */ 

if($number < 1) 

print "<CENTER><P>There Were No Results for Your Search</CENTER>"; 

else 

{ 

while($data = mysql_fetch_array($result)) 

{ 

$username = $data['username']; 

$level_id = $data['level_id']; 

print "<p><b>Username:</b> $username<br></p>"; 

if( $level_id ==1 ) echo '<img src="1.gif">'; 

if( $level_id ==10 ) echo '<img src="2.gif">';

}

Which gets the results of http://www.lothop.clanecko.net/testing/memberstest.php

It is now displaying the correct image for the level_id 1 (members with a rank of 1) but the admin gets no picture (rank of 10) Do you have any idea why? I sure don't ><

Thanks,
Lothop