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GetAWeapon

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Posts posted by GetAWeapon

    Getting pictures out the database

    in PHP Coding Help

    Posted

    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 :)

    Getting pictures out the database

    in PHP Coding Help

    Posted · Edited by GetAWeapon

    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.

    Getting pictures out the database

    in PHP Coding Help

    Posted

    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?

    Getting pictures out the database

    in PHP Coding Help

    Posted · Edited by GetAWeapon

    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.

    Getting pictures out the database

    in PHP Coding Help

    Posted

    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 :D

     

    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> "; 
};
?>

    Split database: 20 rows a page

    in PHP Coding Help

    Posted

    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?

    Split database: 20 rows a page

    in PHP Coding Help

    Posted · Edited by GetAWeapon

    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.

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