[SOLVED] Parse error: syntax error, unexpected ')'

[RyanSF07]

Hi All,

 

I'm getting a "Parse error: syntax error, unexpected ')'  on line 28" error that I need help unwinding. I've tried several combinations of quotes and single quotes without any luck. Needs fresh eyes. Could you please have a look and offer suggestions? I've left it plain here...

 

line 28 is this:

$resized= getimagesize(uploaded_files/$info['photo'].); 

 

And the full script is:

 

<?php

include_once ("connect.php");

$data = mysql_query("SELECT * FROM cars") or die(mysql_error());

function imageResize($width, $height, $target) {

if ($width > $height) {
$percentage = ($target / $width);
} else {
$percentage = ($target / $height);
}

$width = round($width * $percentage);
$height = round($height * $percentage);

return "width=\"$width\" height=\"$height\"";

}

$resized= getimagesize(uploaded_files/$info['photo'].); 

while($info = mysql_fetch_array( $data ))
{

Echo "<img src=http://www.website/uploaded_files/".$info['photo']." imageResize($resized[0],  
$resized[1], 150);> <br>";
Echo "<b>Make:</b> ".$info['make'] . "<br> ";
Echo "<b>Model:</b> ".$info['model'] . " <br>";
Echo "<b>Year:</b> ".$info['year'] . " <hr>";
}
?> 

 

 

Thank you in advanced for your help!

 

[Prismatic]

<?php
$resized= getimagesize("uploaded_files/". $info['photo']); 
?>

[RyanSF07]

Thank you for the quick reply Prismatic! That did fix the error.

 

The script is finding all of the images in the folder, however, images that are larger than 150px are not being displayed.

 

I think now the problem is here:

Echo "<img src=http://www.website/uploaded_files/".$info['photo']." imageResize($resized[0],  
$resized[1], 150);> <br>";

 

The imageResize function isn't running. I've tried adding php tags on either side and get a syntax error.

 

I know that this isn't right (snip below) --as it returns and extra >  --- could you please proof this and offer suggestions?:

Echo "<img src=http://www.website/uploaded_files/".$info['photo']." <? imageResize($resized[0],  
$resized[1], 150);?>> <br>";

 

Thanks!!

[Prismatic]

Try this

 

<?php
echo "<img src='http://www.website/uploaded_files/". $info['photo'] ."'". imageResize($resized[0], $resized[1], 150) ."> <br>";
?>

[RyanSF07]

Thank you again, Prismatic. That worked.

 

There is something wrong though that I have to figure out. If you have any ideas, please let me know -- though -- thank you very much for the help you've already provided!

 

I'm getting this error:

Warning: Division by zero in view.php on line 16

 

Line 16 is:

$percentage = ($target / $height);

 

 

[Ken2k7]

$percentage = $height == 0? 0 : $target / $height;

[RyanSF07]

Thanks Ken --

 

That fixed the error -- but now no images at all display 

 

Jeeze what a script!

 

Thanks for your help! -- almost there

[Ken2k7]

Should there be a space after the single quote that closed out the src attr of img?

[RyanSF07]

I added one, and got the same results.

 

The images are in the uploaded_files folder, but the name of the images are stored in the database.

 

I think the images are in fact being selected, but the resize function is breaking thus the images don't display.

 

trying a couple different things --- all suggestions welcome.

 

Thank you guys!

[Ken2k7]

What do you get if you echo $resized[0] and $resized[1]?

[RyanSF07]

nothing shows up

 

I don't know -- maybe not connecting to that folder

[Ken2k7]

Oh, $info isn't defined at that point. It's defined in the while loop after it. that's why.

[RyanSF07]

Thank you for your help, Ken.

 

Now only one item is selected from the database instead of the resize function being applied to all images -- and frustratingly so -- the actual image is not displayed, just the "image-broken" box.

 

echoing resized[0] and resized[1] now produces: 1

 

what do you think? scrap this and start over? If you have other ideas please let me know as I'm running out 

 

I have this now:

<?php

include_once ("connect.php");

$data = mysql_query("SELECT * FROM cars") or die(mysql_error());

function imageResize($width, $height, $target) {

if ($width > $height) {
$percentage = ($target / $width);
} else {
$percentage = $height == 0? 0 : $target / $height;
}

$width = round($width * $percentage);
$height = round($height * $percentage);

return "width=\"$width\" height=\"$height\"";

}

while($info = mysql_fetch_array( $data ))
$resized= getimagesize("uploaded_files/". $info['photo']);
{

echo "<img src='http://www.website/uploaded_files/". $info['photo'] ."'" . imageResize($resized[0], $resized[1], 150) ."> <br>";
Echo "<b>Make:</b> ".$info['make'] . "<br> ";
Echo "<b>Model:</b> ".$info['model'] . " <br>";
Echo "<b>Year:</b> ".$info['year'] . " <hr>";
}
echo $resized[0] and $resized[1];
?>

 

[Ken2k7]

Can you post the code you have thus far? It's confusing with so many changes done to it to keep track of what your script looks like now.

[RyanSF07]

yes -- I modified the previous post to include it. Thank you!

[Ken2k7]

Syntax much?

 

<?php

include_once ("connect.php");

$data = mysql_query("SELECT * FROM cars") or die(mysql_error());

function imageResize($width, $height, $target) {

if ($width > $height) {
$percentage = ($target / $width);
} else {
$percentage = $height == 0? 0 : $target / $height;
}

$width = round($width * $percentage);
$height = round($height * $percentage);
return "width=\"$width\" height=\"$height\"";
}

while($info = mysql_fetch_array( $data )) {
$resized= getimagesize("uploaded_files/". $info['photo']);
echo "<img src='http://www.website/uploaded_files/". $info['photo'] ."' " . imageResize($resized[0], $resized[1], 150) ."> <br>";
echo "<b>Make:</b> ".$info['make'] . "<br> ";
echo "<b>Model:</b> ".$info['model'] . " <br>";
echo "<b>Year:</b> ".$info['year'] . " <hr>";
}
?>

[RyanSF07]

 

rusty -- very rusty -- and was never very good at this to begin with.

 

Thank you so much for your help. Solved.

 

Ryan