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| Subject:
Javascript / PHP - Image
Category: Computers Asked by: billkellaway-ga List Price: $5.00 |
Posted:
14 Apr 2005 11:54 PDT
Expires: 14 May 2005 11:54 PDT Question ID: 509266 |
Javascript / PHP - I am trying to use Javascript and PHP to rotate
images via a link. I have a function that I have modified so that I
can pass the image height and width. Here's my function -
function imgRotateHW(imgName,height,width)
{
document.images.one.src = 'img/'+imgName;
document.images.one.height = height;
document.images.one.width = width;
}
Here's the PHP that I am passing it to ..
<?php //for ($i = 0 ; $i < count($resPPH); $i++) {
if (count($resPPH) > 1) {
for($t = 0; $t < count($resPPH); $t++) {
/* Get the width and height of the image ... */
list($width, $height) =
getimagesize($_SERVER['DOCUMENT_ROOT'].'/img/'.$resPPH[$t][photo_name]);
/* Compute the aspect ratio ... */
$imageHWAspecRatio = ($height/$width);
print ("<a href=\"#\"
onClick=\"imgRotateHW('t".$resPPH[$t][photo_name].",".$height.",".$width."')\">");
print (($t+1)."</a> ");
}
}
?>
I want to know if I can assign the image height and width with
Javascript. I know what these values are .. I can successfully
manipulate these values with PHP but want to know how I can pass them
to Javascript in a useful way.
Thanks,
Bill Kellaway |
|
| There is no answer at this time. |
|
| Subject:
Re: Javascript / PHP - Image
From: jemtallon-ga on 14 Apr 2005 15:12 PDT |
Actually, you very nearly had it already, if I understand your
question, asside from a few errors in your PHP.
You may want to rewrite your PHP to something like this and try it again:
if (count($resPPH) > 1) {
for($t = 0; $t < count($resPPH); $t++) {
/* Get the width and height of the image ... */
list($width, $height) =
getimagesize($_SERVER['DOCUMENT_ROOT'].'/img/'.$resPPH[$t][photo_name]);
/* Compute the aspect ratio (only is we aren't dividing by zero) ... */
if ($width!=0)
$imageHWAspecRatio = ($height/$width);
else
$imageHWAspecRatio = 0;
/* Uncomment this to give it a test!
width=500;
height=500;
*/
/* Note that I moved the last single-quote character to before the
first comma. Before you were sending the image name, height, and width
as one long string and javascript was treating it as an image name. */
print ("<a href=\"#\"
onClick=\"imgRotateHW('t".$resPPH[$t][photo_name]."',".$height.",".$width.")\">");
print (($t+1)."</a> ");
}
}
Hope this helps you!
Sincerely,
Jem Tallon |
| Subject:
Re: Javascript / PHP - Image
From: jemtallon-ga on 14 Apr 2005 15:16 PDT |
My appologies. I forgot to put $ signs before the height and width in
the examples and remove the t from the beginning of the line:
onClick=\"imgRotateHW('t".$resPPH[$t][photo_name]."',".$height.",".$width.")\">");
print (($t+1)."</a> ");
The real fix should be the following:
f (count($resPPH) > 1) {
for($t = 0; $t < count($resPPH); $t++) {
/* Get the width and height of the image ... */
list($width, $height) =
getimagesize($_SERVER['DOCUMENT_ROOT'].'/img/'.$resPPH[$t][photo_name]);
/* Compute the aspect ratio (only is we aren't dividing by zero) ... */
if ($width!=0)
$imageHWAspecRatio = ($height/$width);
else
$imageHWAspecRatio = 0;
/* Uncomment this to give it a test!
$width=500;
$height=500;
*/
/* Note that I moved the last single-quote character to before the
first comma. Before you were sending the image name, height, and width
as one long string and javascript was treating it as an image name. */
print ("<a href=\"#\"
onClick=\"imgRotateHW('".$resPPH[$t][photo_name]."',".$height.",".$width.")\">");
print (($t+1)."</a> ");
}
} |
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