[SOLVED] Help a newbie with a simple If / Else

[mpausch]

I'm a total newbie to php and desperatley need help. Searched forums but couldnt find the answer.

 

I have a price field (integer) and if it's $0, then I dont want it to display at all. It should only display if its >0.

 

At this time if the price is $0 in the database, the page displays $0 and I want it to show nothing.

 

My code...

 

<?php

$result = mysql_query("SELECT breedingprice FROM animals WHERE (animals.ari=123076)");

$fresult=(mysql_result($result,0));

$newresult=(number_format($fresult,0));

printf($sign.$newresult,0);

?>

 

any help will be forever appreciated!!!

 

[JSHINER]

Replace

 

$result = mysql_query("SELECT breedingprice FROM animals WHERE (animals.ari=123076)");

 

With

 

$result = mysql_query("SELECT breedingprice FROM animals WHERE breedingprice >= 1 AND (animals.ari=123076)");

[TEENFRONT]

I know this is more or less solved, but Just to add to this and explain ....

 

it should be >1 ( as stated above ) not >0  .... ">" means greater than or equal to. So this means if the price is 0 this is equal to 0 and therefore it will show a price.

[kenrbnsn]

No, the ">" operator mean "greater than", ">=" means "greater than or equal"

 

Ken

[mpausch]

Well guys I am very impressed with the speed of your replies and thanks also.

 

I replaced the code as stated and now am getting the following error on the page..

 

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 19 in /home/lcalpaca/public_html/animals/incatest.php on line 146

 

...

[JSHINER]

Please post full code.

[per1os]

$fresult=mysql_result($result,1); // should be one

try that.

[kenrbnsn]

That means that you didn't get any rows returned. A much better way of writing this would be:

<?php
$query = "SELECT breedingprice FROM animals WHERE breedingprice >= 1 AND animals.ari=123076";
$result = mysql_query($query) or die("Problem with the query <pre>$query</pre><br>" . mysql_error());
if (mysql_num_rows > 0) {
   $rw = mysql_fetch_assoc($result);
   echo $sign . number_format($rw['breedingprice']);
}
?>

 

Ken

[mpausch]

Tried that Ken and the result is that with the field 'breedingprice' populated with an amount such as '5000' or with '0' or with NULL, nothing displays on the page (the dollar sign also is not displayed).

 

Seems like it should work. Could it have something to do with the field type in the database? This field is set to...

integer

nulls are allowed

default value is null

 

This is my opening code at the start of the page...

 

<?php -

mysql_select_db($database_LCA, $LCA);

$query_Icedancer = "SELECT * FROM animals WHERE animals.ARI=123076";

$Icedancer = mysql_query($query_Icedancer, $LCA) or die(mysql_error());

$row_Icedancer = mysql_fetch_assoc($Icedancer);

$totalRows_Icedancer = mysql_num_rows($Icedancer);

$sign="$";

?>

 

Man, this is killin' me! Thanks for your help.

 

 

 

 

[kenrbnsn]

I goofed on the "if". Try this:

<?php
if (mysql_num_rows($result) > 0) {
   $rw = mysql_fetch_assoc($result);
   echo $sign . number_format($rw['breedingprice']);
} else
   echo 'Nothing to display';
?>

 

Ken

[mpausch]

Ken,

 

Outstanding!! That works like a charm!

 

I owe you...