Show Image for 2 days

[Sparchy]

I was wondering how i could have an image show for say like 2 days.. like i play at a Virtual Navy site and i help with the web site and we have new bounties each month and i would like to have a image that says New so people know there is new bounties posted.. so how can i create and If Statment to check the date issued and check to see if 2 days has past or not

 

Thank you

Sparchy

[litebearer]

just a thought (Psuedo code)

 

image table

id image_name image_date

 

 

select image_name from image_table where image_date> current_date - 2 days

if result

  display image

else

  no display

 

[teamatomic]

get the day of the month:

if($day<3){do the image}

that would show the image the first and second of each month.

 

 

HTH

Teamatomic

[Sparchy]

well here is my code i have now

 

$result = mysql_query('SELECT * FROM bounties where status = 'Active' AND $start_date > $today - 2 days'); 
		if ($result) { 
		echo '<img src="../svn/images/new.gif">'; 
		}else{ 
		echo ' '; 
		}

 

but im getting parse error in the mysql query line im not sure what im doing wrong

 

Thank you

Sparchy

[anups]

 

following problem in your query :-

1. String is not escaped properly

2. To calculate date difference you have to use "date_diff" function.

 

for `column_name` you have to give tables field name

 

 

$result = mysql_query("SELECT * FROM bounties where status = 'Active' AND  
`column_name` <= date_diff(now(), interval 2 days)");

if ($result) { 			
echo '<img src="../svn/images/new.gif">'; 
}else{echo ' '; 			
}