earl_dc10
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Posts posted by earl_dc10
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I believe you want to change
[code]
$num=mysql_numrows($result);
[/code]
to
[code]
$num=mysql_num_rows($result);
[/code] -
while being a Guru may signify knowledge it also displays the amount of quesions one asks, which is just as important, if not more than being able to answer questions
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sorry, Im still thinkin about this one, ummm.....
just for clarification, if we are calculating the price and vat of each item in a loop, what does this do?
[code]
$price_wovat1=($product['UK_Price']/100)*17.5;
$price_wovat=round($product['UK_Price']-$price_wovat1, 2);
[/code]
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looks like you might have the quotations switched around in your INSERT query
try changing
[code]
$insert = "INSERT INTO `2006` (`date`, `event`, `player`, `place`, `entries`, `points`)
VALUES ($date, $event, $player, $place, $entries, $points)";
[/code]
to
[code]
$insert = "INSERT INTO `2006` (date, event, player, place, entries, points)
VALUES ('$date', '$event', '$player', '$place', '$entries', '$points')";
[/code] -
[!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]From this URL, a request comes in for the Sample.jpg to be displayed. But before this image be displayed, some process should be done first at the back-end, then the right size of the image should will be displayed.[/quote]
so Im assuming you want an image resize script, well here's one that works well
[code]
<?php
$img = $_REQUEST['<image name>'];
header('Content-type: image/jpeg');
$source = imagecreatefromjpeg($img);
$width = imagesx($source);
$height = imagesy($source);
$max = <maximum size> // ie 250, if max size is 250x250
if(($width > $max) || ($height > $max))
{if($width > $height)
{$new = ($width/$max);}
else
{$new = ($height/$max);}
$new_width = ($width/$new);
$new_height = ($height/$new);
$new_image = imagecreatetruecolor($new_width, $new_height);
imagecopyresized($new_image, $source, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
imagejpeg($new_image);
}
else
{imagejpeg($source);}
?>
[/code] -
remove your <html> <head> and <body> tags, only have your php script in the web page, trust me I had the same problem and this cleared it up
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can I see how you are adding these values together? it may be in there I just can't see where $amoutTotal = anything.
if this is what you have, sorry, I'll suggest it anyway since Im crashin
try this for adding the total
[code]
$amountTotal += ($Price-$vat); // since if VAT doesn't apply $vat equals 0, shouldn't be a problem
// after loop is done and you want to display the total
$total = round($amountTotal, 2);
echo $total;
[/code]
and I THINK that this line becomes irrelevent since we calculate VAT individually
[code]
$vat = $amountTotal-$price_wovat1;
[/code]
Im sorry if this is entirely irrelevent to your question, Im a bit tired, hope it helps though!
-J -
is $amountTotal calculated before this query? try changing $amountTotal to the amount of the individual item for inside the loop so that the VAT calculation only applies to the individual price, not the whole.
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I would recommend having
[code]
if ($VatInfo > 0) {
//echo $amountTotal;
$vat = (0);
// $price_wovat=round($amountTotal+$price_wovat1, 2);
//$vat = $amountTotal-$price_wovat1;
} else {
$vat = ($amountTotal/100) * 17.5;
}
[/code]
inside your for() loop but add something like
[code]
$total += <price of item>;
[/code]
so everytime it runs and calculates whether it'a VAT or not, it adds that value to $total -
does the table display where it should, only the submit button is messed up?? is the hidden value underneath the table, or is it above with the submit button?
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Im not sure if textarea's can have "values", try this as a test
[code]
<textarea rows="10" cols="20" name="S1" value="It Works"></textarea>
[/code]
just another thing I noticed, you had your "cols" value inside of the "value" tag quotes, but try that. -
try this as a backup plan if nothing else works
[code]
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST"> //reloads page when submitted
<input type="text" name="USD">
<input type="submit" name="sub" value="submit">
</form>
<?php
$sub = $_POST['sub']; // makes a variable from the Posted Data
$USD = $_POST['USD'];
$rate = 46;
if(isset($sub)) // when submit is pressed, this runs
{
$INR = $rate*$USD
echo "The converted dollar is ".$INR;
// if you really wanted to be fancy you could do this:
echo $USD." in Rupee is ".$INR;
}
[/code] -
or you could use nl2br
[code]
echo nl2br($comment);
[/code]
this converts all \n's to <br>'s -
I did incorporate elements of your code into mine, and it works great now, thank you.
and I humbly accept your correction, the above script apparently doesn't work on its own, but somehow it works incorporated in my site (seriously), but however, it will be the last time I attempt something like that, so thank you -
I thought about using the modulus, but wasn't sure how to account for the odd number that would go in the last page (the second 2 in your 22 would be lost) but Im pretty sure I got it, I'll try it in the morning. thanks for your patience!
and, just so I don't feel entirely stupid, put this in a script and run it
[code]
$page = 7/(5*$x);
echo $page; // this gives an error
if($x = 1)
{echo $page;} // this gives result
// you were correct about the loop however, Im confused why but oh well
// this yields nothing
$page = 7/(5*$x);
for($x = 1; $page == 1.4; $x++)
{echo $page;}
[/code]
because it is not called outside the loop, it doesn't get a divide by 0 error, it is simply an alias for a clumsy equation, which, when you insert $page into the loop, the server reads it as 22/(5*$x) because they are one and the same -
I apologize, when copying the code (my editor doesn't like to paste to mozilla), I left out the break statement at the end of the if(), my fault, anyways, here's what it actually is
[code]
<?php
$page = $id_main/(5*$x); // divide # of rows by 5*$x
for($x = 1; $page >= 0.2; $x++) // so it still runs if there's only 1 page left
{echo "<a href=\"".$self."?pid=".$x"\" class=\"page\">".$x." </a>"; // makes all the pages except the final
if(($page < 1) && ($page > 0))
{echo "<a href=\"".$self."?pid=".$x."\" class=\"page\">".$x."</a>"; // make last page
break; // end script
}
}
?>
[/code]
$x is stated in the loop, $page is just a shorter name than $id_main/(5*$x) and so there should be no problem there. and as for it never ending, it would end after $page became less than 1, $x is being modified which in turn changes $page (example below) and how this should output is links to pages of 5 entries each (1 link = 5 entries), should look like this: 1 2 3 4.... all being links, with 1 being the most recent 5 entries and the last having the odd number that are last.
example
[code]
for($x = 1; $page >= 0.2; $x++)
// expected results if $num_main = 7
// $num_main = 7, $x = 1; => 7/(5*1) = 1.4 which is >= 0.2, script runs
// $num_main = 7, $x = 2; => 7/(5*2) = .7 which is >= 0.2, script runs, but runs the if statement
// since it is less than 1, signifying there are less than 5 entries left for the final page and script ends with break
[/code] -
hey, I have a loop to make the links for the different pages for my site, I want them in groups of 5 (except for the last one if it doesn't divide equally)
[code]
<?php
$page = $id_main/(5*$x); // divide # of rows by 5*$x
for($x = 1; $page >= 0.2; $x++) // so it still runs if there's only 1 page left
{echo "<a href=\"".$self."?pid=".$x"\" class=\"page\">".$x." </a>"; // makes all the pages except the final
if(($page < 1) && ($page > 0))
{echo "<a href=\"".$self."?pid=".$x."\" class=\"page\">".$x."</a>";} // make last page
}
?>
[/code]
Nothing displays from the loop, even if I just put in echo $x; inside the loop, it does nothing (and I do have more than 5 pages, so it's not that). I get no errors, so if you see something, please tell me
Thanks
-J -
you could set a cookie or session at the verification page and then request it at the 3rd page
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[!--quoteo(post=356615:date=Mar 20 2006, 06:30 AM:name=lessthanthree)--][div class=\'quotetop\']QUOTE(lessthanthree @ Mar 20 2006, 06:30 AM) [snapback]356615[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Yup,
Another thing to check, although obvious, often escapes some people.
If the code that inserts data into the database appears in your code AFTER you have printed out the comments / posts then it will obviously need the second refresh as the process is like below:
Submit the form
Refresh page due to POST
Display the current posts
Insert the data.
You can clearly see that it can't display a post that has not been inserted despite the fact that you have refreshed the page and submitted the form. Obviously then, when the page is refreshed a second time, you can see the data that was inserted on the first refresh. Therefore make sure your process is:
Submit the form
Refresh due to POST
Insert the data
Display the current posts.
Hope that makes sense.
[/quote]
haha, yeah, I guess that puts me in my place ;) I guess I should've noted where in the script I was placing it! Thanks for your inputs! -
I have a comments system on my site, it displays the 10 most recent comments and you can add a comment via a form underneath it. When I go to insert a comment, the page refreshes with nothing, then if you refresh the page a second time, it appears, there is nothing WRONG with that, it's actually exactly how it should execute, Im just wondering if there's an alternative without having a referring page, here's my code
[code]
<form name="comment_form" class="comment_form" action="<?php echo $self;?>" method="POST">
Name <input type="text" name="name_comment_form" size="7"><br>
<textarea name="entry_comment_form" rows="2" cols="18" wrap="virtual">
</textarea><br>
<input type="submit" name="comment_form_sub" value="Input>
<?php
$name_comment_form = $_POST['name_comment_form'];
$entry_comment_form = $_POST['entry_comment_form'];
$insert = "INSERT INTO $table VALUES ('$name_comment_form', '$entry_comment_form', '$num_comment')";
$insert_query = mysql_query($insert, $link)
or die("Couldn't insert comment ".mysql_error() );
[/code]
I have tried this, I may be in the right direction, but not sure....
[code]
<form name="comment_form" class="comment_form" action="<?php echo $self;?>" method="POST">
Name <input type="text" name="name_comment_form" size="7"><br>
<textarea name="entry_comment_form" rows="2" cols="18" wrap="virtual">
</textarea><br>
<input type="submit" name="comment_form_sub" value="Input OnClick="<?php
$name_comment_form = "window.document.main_table.comment_form.name_comment_form.value";
$entry_comment_form = "window.document.main_table.comment_form.name_comment_form.value";
// same insert query
[/code]
whenever I push submit on that code, only the "-" is entered that goes after the name, but once I refresh the page, it is then entered, on a new post, and if I load the page without submitting anything, the "window.document......" javascript stuff is entered into the values
and a little off-topic, but can "this, " replace "window.document.main_table.comment_form"?
Thanks!
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a quick google search "how to upload on a form" yielded a number of results, the first one looked like a winner
[a href=\"http://cgi-lib.berkeley.edu/ex/fup.html\" target=\"_blank\"]here[/a] -
check any nearby loops (ie, perhaps a loop to fill the array) I normally time out on neverending loops if I forget something, but the actual line that it displays as timing out is sometimes only related to the loop, not actually in it
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I figured it out, apparently there's an html tag called valign, it conveniently does things like this
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one thing you could do is use CSS max-height in conjunction with overflow
mysql vs mysqli
in PHP Coding Help
Posted