Problem with file paths

Howard007

I'm slowing getting to grips with PHP, but I'm stuck on what appears to be the world's simplest problem.

I'm trying to write some code to dynamically build a page of clickable images from the contents of a DB table. I've successfully done the image upload and DB insert steps, but I'm now stuck on what I thought was going to be the easiest bit - displaying the images. I've written the following code:

$query = "SELECT fname, name, URL from banners";

$results = mysql_query($query) or die (mysql_error());

while ($row = mysql_fetch_array($results))
{
extract($row);

$filename = "../banners/" . $fname;

echo "<a href=\"" . $URL . "\"><img src=\"" . $filename . "\"/></a>";
}

The problem I'm getting is that PHP doesn't seem to be able to handle the "../banners/" part of the string I'm trying to build up - the problem seems to be the last forward slash (I'm guessing that PHP thinks it's a regex delimiter rather than the character). The $fname statement should evaluate to something like "page1_01.jpg", so the $filename part should end up as "../banners/page1_01.jpg" and slot nicely into my "img src" tag.

If I leave the last forward slash in, then my <img src..> tag fails to evaluate and I get just the <a href> part. If I remove the final forward slash from "../banners/" and make it "../banners", then I get "../bannerspage1_01.jpg", but unfortunately that's not the path to my image.

I've tried various things like trying to escape the forward slash, but none of them have worked.

Any help would be greatly appreciated.

Thanks

chrislive

try $filename = "../banners/$fname";

Howard007

chrislive wrote:
try $filename = "../banners/$fname";

Ok, tried that and it does let me build the filename value correctly, but... I still can't get the use of it in the <img src..> tag to work right. If I use a $filename value = "../banners/page1_01.jpeg", then the <img src> tag is not visible in the HTML returned and neither is anything after it on the same line.

Has anyone got any more ideas on this?

Thanks

chrislive

try this.. i havent used the extract function so stick to what I know.. haha

$query = "SELECT fname, name, URL from banners";
$result = @mysql_query ($query) or die (mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_NUM)){
    $fname = $row[0];
    $name = $row[1];
    $url = $row[2];

$filename = "../banners/$fname";

echo "<a href=\"$url\"><img src=\"$filename\" title=\"$name\" alt=\"$name\"></a>";
}

chrislive

another tip.. when using variables ($var) you can access them inside "these quotes" so you dont need to use ". $URL . " however if you use 'these quotes' you will need to user your method.

chrislive

another thought.. is ../banners/ accessible from the webserver... try putting in a full path.. http://www.domain.com/banners/

Kardall

Anothing thing, if the database only has a few headers, or you need to make more sense of it rather than an array, get to know your friend:

/* Example Database Structure

PersonID | ClassName | PictureLoc
-------------|----------------|------------------------------|
 0             | home1        | ../classpics/pic-001.jpg |
-------------|----------------|------------------------------|
 1             | home1        | ../classpics/pic-002.jpg |
-------------|----------------|------------------------------|
 2             | home2        | ../classpics/pic-003.jpg |
*/

while( $res = mysql_fetch_assoc($query) ) {
  echo "{$res['PersonID']} was in {$res['ClassName']} and looks like <img src='{$res['PictureLoc']}'>";
}

something like that anyway 🙂

Howard007

Ok, I tried what you said i.e. doing \"$fname\", but it still doesn't work.

It seems to be because of the last forward slash in the value of fname (../banners/page1_01.jpeg). If I make this ../bannerspage1_01.jpeg, then I get the img src appearing in the HTML. With the last forward slash in I don't get the img src part at all.

Any more ideas?

Thanks

Weedpacket

Okay; here's a question. What should the src value look like (i.e., what would it say if the filename was, say, wibble.jpeg)? Put something together manually that works, then replace the bit that should come out of the database (whatever you used for "wibble.jpeg") with the appropriate variable. And take Kardall's suggestion as well and refer to the fields by name - much easier to know what $row['fname'] is than $row[0].

Howard007

Weedpacket wrote:
Okay; here's a question. What should the src value look like (i.e., what would it say if the filename was, say, wibble.jpeg)? Put something together manually that works, then replace the bit that should come out of the database (whatever you used for "wibble.jpeg") with the appropriate variable. And take Kardall's suggestion as well and refer to the fields by name - much easier to know what $row['fname'] is than $row[0].

I've done exactly that already, which is precisely how I ended up with the code I have. If I replace the $fname with "../banners/page1_01.jpeg", it doesn't work either i.e. PHP appears to be having a problem with the last forward slash in the string I've given.

In my original example I am referring to the fields by name rather than position.

Any more ideas?

Thanks

chrislive

as i said before try using an absolute path... eg http://hostname/banners/img.jpg

Howard007

Thanks for the suggestion, but that is not the user requirement and AFAIK it should be possible to do what the user actually wants i.e. what the example shows.

Please could someone try my example code in a page of their own and let me know if they get the same results as me?

I'm beginning to wonder if the '/p' that results from the last forward slash has some kind of special significance in PHP and hence makes it screw up.

Howard007

I tried doing the following:

$query = "SELECT fname, name, URL from banners";

$results = mysql_query($query) or die (mysql_error());

while ($row = mysql_fetch_array($results))
{

echo "<a href=\"" . $row['URL'] . "\"><img src=\"" . "http://www.mysite.com/banners/test.jpeg" . "\"/></a>";
}

And the result was.... the img src part is NOT displayed.

If I replace "http://www.mysite.com/banners/test.jpeg" with http://www.mysite.com/test.jpeg", then the img src part is displayed. However, I want to store the banner jpegs in a folder below the root of my site.

Now has anyone got any ideas?

chrislive

maybe you need to give more details because you are obviously not telling us everything.. your images MUST be in a place where the webserver is able to show them.. eg you cant have a website in the webroot say /var/www that is calling an image that is in /var/

you MAY be able to open the binary data of the image and print it to the screen but im not sure...

This example WILL work unless you have something else in your code that is stuffing it up.. copy the whole code and try it.

$query = "SELECT fname, name, URL from banners";
$result = @mysql_query ($query) or die (mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_NUM)){
    $fname = $row[0];
    $name = $row[1];
    $url = $row[2];

$filename = "http://www.mysite.com/banners/$fname";

echo "<a href=\"$url\"><img src=\"$filename\" title=\"$name\" alt=\"$name\"></a>";
}

once you have done that copy the page source (from your web browser) here so we can see the output.

Howard007

Ok, here are the results of using your example code:

$url is aaaa

with $filename = "http://www.mysite.com/banners$fname" - the last forward slash removed, I get:

Again, $url is aaaa, $fname = page1_01.jpeg and $name is test and test1 (for the 2 rows of data I have in my table.

Obviously, the image is actually in http://www.mysite.com/banners/, so I do get the red X where the picture should be - this I can handle

Any ideas?

chrislive

does your $fname field contain any / or \ or spaces?? what about other image types? try getting a gif image...

Howard007

$fname is page1_01.jpeg and this displays correctly if I remove the last forward slash from the string that precedes it.

I'm not sure why you think trying a different file type will fix it. It looks to me like it's a problem evaluating the string expression when it has a forward slash before the page1_01.jpeg part rather than something to do with the server not being able to handle particular filetypes.

Looks like I'm going to have to go back to blank piece of paper and come up with a workaround for this problem. I just can't believe this problem is happening when I'm doing the simplest thing possible.

Has anyone got any other ideas?

Weedpacket

One thing to keep in mind about all of this is that relative URLs will be relative to the URL of the page displaying the URL (not, e.g., to the script's own filepath).

So if there are different pages at different depths in the document tree then that's your problem: the relative URL would be different depending on where in the site the page is. A relative URL that will always expand to "http://www.example.com/banners/" on the www.example.com host is "/banners/".

Now, you say that

Howard007 wrote:
echo "<a href=\"" . $row['URL'] . "\"><img src=\"" . "http://www.mysite.com/banners/test.jpeg" . "\"/></a>";
}

And the result was.... the img src part is NOT displayed.

If I replace "http://www.mysite.com/banners/test.jpeg" with http://www.mysite.com/test.jpeg", then the img src part is displayed. However, I want to store the banner jpegs in a folder below the root of my site.

What do you mean "the img src part is NOT displayed"? Do you mean that when you look at the HTML source of the generated page using the above-quoted code all you see is

<a href="www.example.com/whatever"><img src=""/></a>

That would be weird. Any build of PHP that gets released would have no trouble concatenating two literal strings and outputting

<a href="http://www.example.com/whatever"><img src="http://www.mysite.com/banners/test.jpeg"/></a>

If that is what it's putting out and you're still not seeing the image, then that would imply that the URL itself is wrong - it would suggest that these images aren't in the /banners/ directory at all. (Since you claim that http://www.mysite.com/test.jpeg does work (at least, the "img src part" is displayed), that would also suggest that your images aren't where you think they are.)

Forward slashes / don't have any special meaning as far as PHP is concerned. Backslashes do, but not forward ones.

Howard007

By not being displayed, I mean I get:

So it's not a problem with the path to the file. I can handle that part. It's that the string concatenation fails and results in nothing being returned for the <img src part.

Weird is right!!

I can't believe how hard it is to get string concatenation to work. I had expected this part of my work to take about 5 seconds, but so far I've spent way too long on ways around it.

Houdini

Is the file you are running in http://www.mysite.com/banners/banner.php ? or is it in
http://www.mysite.com/banner.php? Because I ran your code as is except I made border =0 so there wasn't the ugly blue border and made a foler banners and put two gifs in there and they show using either

$filename = "banners/" . $fname; //the current file and banners in root.
echo "<a href=\"" . $URL . "\"><img src=\"" . $filename . "\" border=0 /></a><br />";

$filename = "../" . $fname;//current file is in root/banners/script/ folder
echo "<a href=\"" . $URL . "\"><img src=\"" . $filename . "\" border=0 /></a><br />";

$filename = "../banners/" . $fname;//current file is in root/script/ folder
echo "<a href=\"" . $URL . "\"><img src=\"" . $filename . "\" border=0 /></a><br />";

The code runs fine for me in any of the above cases depending on where I am running the script from.