GetAWeapon

Thanks for your reply, but yes, my problem with that link already got fixed with this: echo "<img src='". IMAGE_PATH . $row['I_PATH']. '/' . $row['I_ID']. '.png' ."' />"; And my problem after that also got fixed a couple of minutes ago with this: $sqlget = "SELECT * FROM artikel LEFT JOIN images ON artikel.A_ARTCODE = images.I_ARTCODE LIMIT $start_from, 20 "; Anyway thank you very much for your help making some time for my problem. But for now everthing is fixed

I think my mistake is in my selection from my tables: $sqlget = "SELECT * FROM artikel, images LIMIT $start_from, 20 "; Normally to set my 2 keys the same I need to do this: WHERE artikel.A_ARTCODE = images.I_ARTCODE But when I do this, then I only got 2 products left, I just want to see all my products even when they don't got a picture.

I changed it to this: echo "<img src='". IMAGE_PATH . $row['I_PATH']. '/' . $row['I_ID']. '.png' ."' />"; But still got the same result, really strange

Noooo, I still got a bug, in my 2 tables, I got 2 keys, one array called 'A_ARTCODE' and in my other one for the images 'I_ARTCODE'. Normally those 2 ID's need to be te same.So normally every product will have his own picture and ID. But now I got this: https://www.dropbox.com/s/mwcadcjbi4ol8cr/wrongid.jpg So the only products that should have a picture, that are the first 2, I think I made a mistake by making a fixed path like this: define('IMAGE_PATH', 'images/000000-001000/'); So I need to make a relative path in this line: echo "<img src='". IMAGE_PATH . $row['I_ID']. '.JPG' ."' />"; theoretical it should be something like I_PATH + I_FILE, the path is the name of the folder and the file is the name of my picture. How can put this in php code?

Ah, nvm, it works now, I founded it didn't knew there was another map in the images map xd stupped hidden path define('IMAGE_PATH', 'images/000000-001000/'); echo "<img src='". IMAGE_PATH . $row['I_ID']. '.JPG' ."' />";

Ok thanks, now I doesn't direct me to my localhost anymore. But I still got no results, really strange. My url is now this: http://dbfact.consult4u.be/images/NWADLIDES-1005D I really thought it should work. Is it possible, it doens't work because there is no '.jpg' behind the url? My code atm is this: echo "<img src='". IMAGE_PATH . $row['I_PATH' + 'I_ID']."' />";

Oooh nvm, my collague, fckd it up. Forget my post before, It's totally worthless. Now I did this: echo "<img src='".$row['I_PATH' + 'I_ID']."' />"; I get this as result in the url: http://localhost/DBFact/NWADLIDES-1005D I think that's almost good, but you see, it goes to the localhost, but it should go to my online host, where my database is... how can I do that?

Ya but if that worked, I wouldn't putted // in front of it xD Like you say, yes indeed, when I do that I get a icon that there is no image found. But it's not that what I need, there are images behind the word 5.jpg and 6.jpg, it's really strange. I'll show you a picture in this post with the proof of it. Maybe I explained not good enough, my collague made also a version from it and had this and normally it's the same code, I'll double check it after this post. But I'm sure it was the same. https://www.dropbox.com/s/j8q8frdufj9jtn7/strange.jpg Like you see, every name end with 5.jpg in this example, but when I scroll down, a bunch of them ends with 6.jpg. So if im right, the part in front of that makes the url or path to that picture complete. So how can I display this like an image? I thougth something like this: [frontcode]+[i_FILE]=complete url/path and that should give the pictures if I'm right. I think my idea can work, but I don't know how to put this in code. I'm only a noob at php and it's my 5 day I use php in my life.

Hello, like I promised, I'm back with another problem. This one will be harder xD so squeeze your brain Now I've got a webpage, with a database shown into a table, under that I've got page numbers. So I got my 20 rows for each page. https://www.dropbox.com/s/txsinttizihiwo5/stillnotok.jpg Great/awesome so far Now I've got a new problem and it's hard for me to explain this too. In my database I got allot of tables, in my webpage I'm using 2 of them, called 'artikel' and 'images". Everything from my artikel table is ok. So we don't need to look at that. But my problem is with the table 'images'. In that table I got 2 items. Like you will see in my code, I need the I_ARTCODE and I_FILE in I_ARTCODE stands: 14 and for the other image 15 in I_FILE stands: 5.jpg and for the other image 6.jpg With the code I've got now, I display the word 5.jpg and 6.jpg. But I need to display the picture behind that, the person who gave me this task sayed, I normally can do this with the pad but it's not a real url, I'll give you guys a picture how my database looks like: https://www.dropbox.com/s/lfac7gzr1uqplxd/path.jpg Over here the pad is this: 000000-001000 It doesn't seems normal to me and I can't find a sollution for this on the internet. Can someone plz help me? Ooh ya, almost forgot, this is my code: <?php include('connect-mysql.php'); if (!empty($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 20; $sqlget = "SELECT * FROM artikel, images LIMIT $start_from, 20 "; $sqldata = mysqli_query($dbcon, $sqlget) or die('error getting'); echo "<table>"; echo "<tr><th>A_ARTCODE</th><th>A_NUMMER</th><th>A_OMSCHRN</th><th>A_REFLEV</th><th>A_WINKEL</th><th>I_ARTCODE</th><th>I_FILE</th></tr>"; while($row = mysqli_fetch_array($sqldata)){ echo "<tr><td align='right'>"; echo $row['A_ARTCODE']; echo "</td><td align='left'>"; echo $row['A_NUMMER']; echo "</td><td align='left'>"; echo $row['A_OMSCHRN']; echo "</td><td align='left'>"; echo $row['A_REFLEV']; echo "</td><td align='right'>"; echo $row['A_WINKEL']; echo "</td><td align='right'>"; echo $row['I_ARTCODE']; echo "</td><td align='right'>"; echo $row['I_FILE']; //echo "<img src='000000-001000".$row['I_ID']."' />"; echo "</td></tr>"; } echo "</table>"; $sql = "SELECT COUNT(A_ARTCODE) FROM artikel"; $rs_result = mysqli_query($dbcon, $sql) or die ("mysqli query dies"); $row = mysqli_fetch_row($rs_result) or die ("mysqli fetch row dies"); $total_records = $row[0]; $total_pages = ceil($total_records / 20); for ($i=1; $i<=$total_pages; $i++) { echo "<a href='index.php?page=".$i."'>".$i."</a> "; }; ?>

Oooooh yes, ofcourse, why didn't I think of that xD I just needed to replace this: $sqlget = "SELECT * FROM artikel, images LIMIT 0, 20 "; to $sqlget = "SELECT * FROM artikel, images LIMIT $start_from, 20 "; Thanks for the fast support, I'm really greatfull, but I also got another question, but I'll guess I better make another topic for it or not?

Hello everyone, I new to the php world and I'm only working with it for 5 days now. So I'm a noob But I've got a good start atm, I already could connect a database with my webpage and show all my coloms I need. Now I got a problem with the fact, my database is way to big to display on 1 page. To much load-time, so that's the reason why I want to split up all my products. 20 for each page, like in this tutorial: http://www.phpjabbers.com/php--mysql-select-data-and-split-on-pages-php25.html?err=1 I think I'm almost there, my problem is that when I click on the number from the next page (like nr 5), it doesn't show my next 20 products. But when I look to the url, that seems ok, because I get something like this; index.php?page=1 Can someone help me to find out, what I did wrong? And plz, remember, if you give me an answer, plz keep it short and simple. I'm a noob at this. This is my code: <?php include('connect-mysql.php'); if (!empty($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 20; $sqlget = "SELECT * FROM artikel, images LIMIT 0, 20 "; $sqldata = mysqli_query($dbcon, $sqlget) or die('error getting'); echo "<table>"; echo "<tr><th>A_ARTCODE</th><th>A_NUMMER</th><th>A_OMSCHRN</th><th>A_REFLEV</th><th>A_WINKEL</th><th>I_ARTCODE</th><th>I_FILE</th></tr>"; while($row = mysqli_fetch_array($sqldata)){ echo "<tr><td align='right'>"; echo $row['A_ARTCODE']; echo "</td><td align='left'>"; echo $row['A_NUMMER']; echo "</td><td align='left'>"; echo $row['A_OMSCHRN']; echo "</td><td align='left'>"; echo $row['A_REFLEV']; echo "</td><td align='right'>"; echo $row['A_WINKEL']; echo "</td><td align='right'>"; echo $row['I_ARTCODE']; echo "</td><td align='right'>"; echo $row['I_FILE']; //echo "<img src='000000-001000".$row['I_ID']."' />"; echo "</td></tr>"; } echo "</table>"; $sql = "SELECT COUNT(A_ARTCODE) FROM artikel"; $rs_result = mysqli_query($dbcon, $sql) or die ("mysqli query dies"); $row = mysqli_fetch_row($rs_result) or die ("mysqli fetch row dies"); $total_records = $row[0]; $total_pages = ceil($total_records / 20); for ($i=1; $i<=$total_pages; $i++) { echo "<a href='index.php?page=".$i."'>".$i."</a> "; }; ?> I hope someone can help me.