[Resolved] New post overwrites current image

kekke

Hi!

I have a problem with my code. When I do a new post on my page and upload an image the new image overwrites the current image. Is there someway I can store the filename aswell? The script itself works like a charm!

Please help!

Here is the code:

<HTML>
    <?php
    if($submit)
    {

    ####directory from web
    $direx = "my_path_from_the_web";

    ####path to image upload area from root
    $path = "my_root_path"; 

    $jpeg = ".jpg";

// database connection

    $database = 'db';  //database name
    $sqluser = 'user'; //user
    $sqlpass ='pass'; //pass
    $tablename ='front'; //table

    $db = mysql_connect("host","$sqluser","$sqlpass") OR DIE (mysql_error()); //connect to db
    @mysql_select_db ("$database", $db) OR DIE (mysql_error());
// Do this process if user has browse the
// file and click the submit button
if ($_FILES) 
    {
    $image_types = Array ("image/jpeg","image/pjpeg");
    if (is_uploaded_file ($_FILES["userfile1"]["tmp_name"])) 
        {
        }
        ELSE
        {
             die;
        }

        copy("$userfile1", "$path$image$jpeg"); 
        $imagelocation = "$direx$image$jpeg";
    }
    $sql = "INSERT INTO front (titel, text, imagelocation) VALUES ('$titel', '$text', '$imagelocation')";
    $result = mysql_query($sql);
    echo "Thank you! Information entered.\n";
}
else
    {
?>
<body bgcolor="#95C1D5">
<form method="post" action="posta_front.php" ENCTYPE="multipart/form-data">
  <p>
  <FONT FACE="verdana" SIZE="1"> 
      <B>FRONTPAGE</B> 
    <BR>
    Titel: 
    <BR>
    <input type="Text" name="titel">
    <br>
    <B>Text:</B> 
    <textarea rows="10" cols="30" name="text"></textarea>
    <br>
    <?php
          print("<table width=\"600\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">");
          print("<tr><td align=\"center\" valign=\"middle\"><font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"2\"><b>Profile Image:</b></font><br>\n");
          print("<tr><td align=\"center\" valign=\"middle\"><font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"2\">Image #1 :</font>\n");
          print("<input type=\"file\" name=\"userfile1\" size=\"35\"><br><br><br></td></tr>\n");
          print("</table><br><br>");
    ?>
    <input type="Submit"name="submit" value=" Publicera">
  </FONT>
  </p>
</form>
<?
    }
?>
</HTML>

happyme

the value of $_FILES['userfile']['name'] is set to the original name of the file that you uploaded

copy($_FILES['userfile']['tmp_name'], $path.$_FILES['userfile']['name']);

will copy the uploaded file to $path with it's original name

is that helpful?

Julia

kekke

Thanks for the reply Happyme! 😃

Now the uploaded images keeps the filename on the server. Now just one stupid question: How do I print it out on my page?

happyme

try this:

printf("<img src='%s%s'>",$imagelocation,$titel);
printf("<p class = 'body'>%s</p>",$text";

using css to format text saves a lot of trouble

kekke

Didn´t work...feels like I have tried everything now.

kekke

Here is a link to the site i am building. Rightclick the broken images and you will se that the images that i have uploaded have been assigned a temp-name. On the server the images have their original filenames.

http://kekke.pixelstorm.se/demo/

Here is the code for front.php

<HTML>
<?php 
$db = mysql_connect("host", "username", "db_pass"); mysql_select_db("my_db_name"); 
$result = mysql_query("SELECT * FROM front ORDER BY id ASC",$db); 
echo "<IMG SRC=\"front_title.jpg\"><TABLE WIDTH=\"50%\" BORDER=\"0\" bgcolor=#ADC5DE>"; 
while($myrow = mysql_fetch_array($result)) 
{

$titel=$r["titel"];
$text=$r["text"];
$imagelocation=$r["userfile1"];
echo "<table border=\"0\" width=\"400\" cellpadding=\"2\" cellspacing=\"0\">
	<tr>
	<td width=\"100%\"><B><FONT FACE=\"arial\" SIZE=\"3\"><IMG SRC=\"".$myrow["imagelocation"]."\"</IMG><BR><B>".$myrow["titel"]."</B><BR></td>
	</tr>
	<tr>
	<td width=\"100%\"><FONT FACE=\"verdana\" SIZE=\"1\">".$myrow["text"]."<BR><BR></td>
	</tr>
	</table><BR>";
}
?>
</HTML>

Any ideas? My computer is on it´s way out the window very soon!

happyme

look at the html your code is generating

the img tag isn't closed and I'm not clear exactly what you're doing. Have you put the new filename and path into the database? Do you then call the new filename and path and use it to display the image? Try using ' quotes within " quotes to make things a bit clearer

kekke

The second imagetag, the one I am concentrating on is closed. Don´t remember if I wrote what I wanted to do, but I´ll explain again.

When I am posting a new article/newsitem/puff on my page "front.php" I want to browse for a picture to go with the article/puff. When I look at the article/puff on my site I want the picture to show aswell. So the article and the picture sort of has the same id.

In my db for front.php i have the tables: id, titel, text and imagelocation. In the table "imagelocation" the path to the uploaded image is stored and on the server the images that I upload are saved correctly.

So...my problem is who do I output the right image with the right puff.

I have a strong feeling that I have missed something (duh)!

happyme

how are you saving the image name and path to the database?

also where does $r[] come from?

$titel=$r["titel"];
$text=$r["text"];
$imagelocation=$r["userfile1"];
[/code]

did you mean

    $titel=$myrow["titel"]; 
    $text=$myrow["text"]; 
$imagelocation=$myrow["userfile1"];

kekke

I use a form to post the puffs. You can see the code in my first post.

The $r[] is how i have learned. I think your way is easier!

happyme

$imagelocation =$path.$_FILES['userfile']['name'];

will set the value of $imagelocation correctly when you upload the image, assuming you have set $path correctly


if ($myrow["imagelocation"])
	{
	printf("<img src = '%s' align = 'left'>",$myrow["imagelocation"]);
	}

will display the image

kekke

Got this error:

Parse error: parse error, unexpected T_IF, expecting ',' or ';' in /my/path/demo/front.php on line 14

I have tried with everything (i think) to get it to work...but no!

happyme

I'll write up a simple version of the script for you later this afternoon

we'll get this working, don't worry

Julia

kekke

Thanks Julia...you´re the best!

happyme

OK, got a bit of time to look at this:

your upload script is working fine, but your database entry is not, so try this:


//$titel and $text are set when the form is processed, we need to set $imagelocation

//set $path to the folder where your images are stored
$folder = "set this value to the folder where the files are uploaded to";
$path = $_SERVER['DOCUMENT_ROOT'].$folder;

$filename = $_FILES['userfile']['name'];

$imagelocation = $path.$filename


//add new record to database

$sql = "INSERT INTO front (titel, text, imagelocation) VALUES ('$titel', '$text', '$imagelocation')";

echo $sql; //take this out when it's working

$result = mysql_query($sql) or die("Technical error, ".mysql_error());

then look at the output from the database:


//connect to database
$db = mysql_connect("host", "username", "db_pass"); mysql_select_db($dbname,$db);

//select the approriate records
$sql = "select * from $tablename";

$result = mysql_query($sql)or die("Technical error, ".mysql_error());

while ($myrow = mysql_fetch_array($result))
{
printf("<img src = '%s' align = 'left'>",$myrow["imagelocation"]);
printf("<p>titel: %s</p>", $myrow["titel"]);
printf("<p>text: %s</p>", $myrow["text"]);

}

add in the formatting etc when you've got it working

I just typed this in, it's not been tested so if you get errors, can you post the error and the lines before the error too

kekke

I tried this code:

<HTML>
    <?php

$folder = "http://kekke.pixelstorm.se/demo/bilder"; 
$path = $_SERVER['/HERE_I_PUT_MY_PATH_TO_ROOT/'].$folder; 
$filename = $_FILES['userfile']['name']; 
$imagelocation = $path.$filename 

// database connection

    $db = mysql_connect("host", "username", "pass") OR DIE (mysql_error());
    @mysql_select_db ("database", $db) OR DIE (mysql_error());

$sql = "INSERT INTO front (titel, text, imagelocation) VALUES ('$titel','$text','$imagelocation')";
$result = mysql_query($sql) or die("Technical error, ".mysql_error());
}
else
    {
?>
<body bgcolor="#95C1D5">
<form method="post" action="posta_front.php" ENCTYPE="multipart/form-data">
  <p>
  <FONT FACE="verdana" SIZE="1"> 
      <B>FRONTPAGE</B> 
    <BR>
    Titel: 
    <BR>
    <input type="Text" name="titel">
    <br>
    <B>Text:</B> 
    <textarea rows="10" cols="30" name="text"></textarea>
    <br>
    <?php
          print("<table width=\"600\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">");
          print("<tr><td align=\"center\" valign=\"middle\"><font face=\"Verdana, Arial, Helvetica, sans-serif\" 

size=\"2\"><b>Profile Image:</b></font><br>\n");
          print("<tr><td align=\"center\" valign=\"middle\"><font face=\"Verdana, Arial, Helvetica, sans-serif\" 

size=\"2\">Image #1 :</font>\n");
          print("<input type=\"file\" name=\"userfile1\" size=\"35\"><br><br><br></td></tr>\n");
          print("</table><br><br>");
    ?>
    <input type="Submit" name="submit" value="Publicera">
  </FONT>
  </p>
</form>
</HTML>

...and got this errormessage:

Parse error: parse error, unexpected T_VARIABLE in /MY_PATH_TO_ROOT/posta_front.php on line 11

happyme

there's a ; missing after
$imagelocation = $path.$filename

happyme

also set $folder to "/demo/bilder";

kekke

Ok...the posting work ok now! Now let´s go on to the frontpage.

Tried this code:

<HTML>
<?php

//connect to database 
$db = mysql_connect("host", "user", "pass");
mysql_select_db(my_db,$db);

//select the approriate records 
$sql = "select * FROM front";

$result = mysql_query($sql)or die("Technical error, ".mysql_error());

while ($myrow = mysql_fetch_array($result))
{ 
printf("<img src = '%s' align = 'left'>",$myrow["imagelocation"]);
printf("<p>titel: %s</p>", $myrow["titel"]);
printf("<p>text: %s</p>", $myrow["text"]);

}

?>

</HTML>

...and got a few broken images...easier for you to understand if you see it for yourself:

http://kekke.pixelstorm.se/demo/

happyme

can you post the whole code? The problem is still with putting data into the database