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anyone know php?
- Thread starter coolguy23
- Start date
That is very possible.
PHP is very flexible when it comes to using the query string. In fact, you don't need any more code...once you define ID to equal a number, that will be recognized within the script.
Here:
If you link to display.php?id=492, 492.gif will then be printed to the browser with that code.
You can even setup a little feature that, if no number is specified, the script uses a default one. (Make sure you put this before the code above, still between the <?php and ?> tags, though.)
Easy
PHP is very flexible when it comes to using the query string. In fact, you don't need any more code...once you define ID to equal a number, that will be recognized within the script.
Here:
Code:
<?php
echo "<img src=$id.gif>";
?>If you link to display.php?id=492, 492.gif will then be printed to the browser with that code.
You can even setup a little feature that, if no number is specified, the script uses a default one. (Make sure you put this before the code above, still between the <?php and ?> tags, though.)
Code:
if ($id == 0) {
$id = 1;
}Easy
It gets put into any old HTML file--trick is, you have to rename that file to a .php extension.
You start PHP code with <?php and end it with ?>. Normal HTML can go anywhere but inside those two operators (well, it can, but you have to "echo" or "print" it.)
Anyway, just copy and paste this code where ever you want the image to be displayed.
Again, you might want to change the default (if nothing is specified) from 1 to whatever.
You can also use .jpg rather than .gif if you need to.
You'll probably have to use Notepad for this--just make sure you save it as display.php (it doesn't really matter, though). Upload the file in ASCII mode, and simply direct your browser to display.php?id=123 to see 123.gif displayed where you put that code!
You start PHP code with <?php and end it with ?>. Normal HTML can go anywhere but inside those two operators (well, it can, but you have to "echo" or "print" it.)
Anyway, just copy and paste this code where ever you want the image to be displayed.
Code:
<?php
// If no ID is specified, use a default.
if ($id == 0) {
$id = 1;
}
// Print the image.
echo "<img src=$id.gif>";
?>Again, you might want to change the default (if nothing is specified) from 1 to whatever.
You can also use .jpg rather than .gif if you need to.
You'll probably have to use Notepad for this--just make sure you save it as display.php (it doesn't really matter, though). Upload the file in ASCII mode, and simply direct your browser to display.php?id=123 to see 123.gif displayed where you put that code!
For the title thing, you have to do it this way:
Now, for the links to the next and previous pictures, this can be done with a little simple subtraction and addition, but it will take a little more code.
Put the following code where ever you want the "Next Image" and "Previous Image" links to be generated.
You can obviously change the format of that--right now you would see something similar to this:
Next Image | Previous Image
* Notice the ?php_self variable, that saves lots of time--that variable always parses to the location of the PHP document, that way you don't have to retype/copy & paste long URLs all the time.
That should work fine, if it doesn't, I probably just made a stupid mistake...
[Edited by Niaad on 03-18-2001 at 10:28 PM]
Code:
<?php
echo "<title>Picture No. $id</title>";
?>Now, for the links to the next and previous pictures, this can be done with a little simple subtraction and addition, but it will take a little more code.
Put the following code where ever you want the "Next Image" and "Previous Image" links to be generated.
Code:
<?php
// Determine Variables
$next_id = $id + 1;
$prev_id = $id - 1;
// Print Link(s)
echo "<a href=$php_self?id=$next_id>Next Image</a>;"
echo " | ";
echo "<a href=$php_self?id=$prev_id>Previous Image</a>;"
?>You can obviously change the format of that--right now you would see something similar to this:
Next Image | Previous Image
* Notice the ?php_self variable, that saves lots of time--that variable always parses to the location of the PHP document, that way you don't have to retype/copy & paste long URLs all the time.
That should work fine, if it doesn't, I probably just made a stupid mistake...
[Edited by Niaad on 03-18-2001 at 10:28 PM]
coolguy23
Guest
i tried it but it says there is an error in line 21...can you help
please tell me the exact code i should save in the file
here it is
cgi-bin.spaceports.com/~bsvsca/display.php
Code:
<?php
// If no ID is specified, use a default.
if ($id == 0) {
$id = 1;
}
// Print the image.
echo "<img src=C:\WINDOWS\Desktop\mysite\bspics\BSpics\$id.jpg>";
?>
<?php
// Determine Variables
$next_id = $id + 1;
$prev_id = $id - 1;
// Print Link(s)
echo "<a href=$php_self?id=$next_id>Next Image</a>;"
echo " | ";
echo "<a href=$php_self?id=$prev_id>Previous Image</a>;"
?>please tell me the exact code i should save in the file
here it is
cgi-bin.spaceports.com/~bsvsca/display.php
Woofcat
New Member
This should do the same thing more safely and efficiently. Note that you probably don't want the img src to point to your desktop...
<?$id=empty($id)?1:floor($id)?>
<img src="C:\WINDOWS\Desktop\mysite\bspics\BSpics\<?=$id?>.jpg">
<a href="<?=$PHP_SELF?>?id=<?=$id+1?>">Next Image</a>
|
<a href="<?=$PHP_SELF?>?id=<?=$id-1?>">Previous Image</a>
<?$id=empty($id)?1:floor($id)?>
<img src="C:\WINDOWS\Desktop\mysite\bspics\BSpics\<?=$id?>.jpg">
<a href="<?=$PHP_SELF?>?id=<?=$id+1?>">Next Image</a>
|
<a href="<?=$PHP_SELF?>?id=<?=$id-1?>">Previous Image</a>
php switches between types so it's treating 01 as the number '1' not the string '01' the easy way is not to have any files starting with a 0. you could make sure it was a string but then the next and previous buttons would f**k.
plus having a number means you can error trap and not try to load a picture if it's not in the range of your pics - but then you'd have to update the page when you added more pictures ...
- or you could get it to check the file exists on the server .....
[knowing when to stop is important - keep it simple]
plus having a number means you can error trap and not try to load a picture if it's not in the range of your pics - but then you'd have to update the page when you added more pictures ...
- or you could get it to check the file exists on the server .....
[knowing when to stop is important - keep it simple]
TheSpaceDude
New Member
...Just wondering (i'm a bit new to php)... what does this line do?:
$id=empty($id)?1:floor($id)
...ignore this post if its a dumb question
...Just wondering (i'm a bit new to php)... what does this line do?:
$id=empty($id)?1:floor($id)
...ignore this post if its a dumb question
funny little expression
$var = (condition) ? (value1) : (value2) ;
if condition is true it equates to value1 else to value2
same as
if (condition) {$var = value1}
else {$var= value2}
it's great for making your code look complicated
can be v useful
A variable variable is something complety different.
Lets say you had a variable named pizza which equals cheese:
$pizza = "cheese";
You could use the text in the pizza variable and make it into a variable:
$$pizza
Resulting in a variable named cheese ($cheese).
That's a variable variable. =)
Lets say you had a variable named pizza which equals cheese:
$pizza = "cheese";
You could use the text in the pizza variable and make it into a variable:
$$pizza
Resulting in a variable named cheese ($cheese).
That's a variable variable. =)
Using a variable's contents as a variable's name is a bad idea for many reasons. Read Mark-Jason Dominus's articles at http://perl.plover.com/varvarname.html for more information. They're on the topic of Perl, but PHP has the same problem AFAIK. A good quote: "The real problem is that $$fields = ... is not confined at all, and if something goes wrong, it can smash any variable in the entire program, which will have a bizarre effect, possibly on something far away. "
