printf()

[niche]

I have this code that displays properly:

printf("<p style=\"font-size:9px;margin:0px;\"> <span style=\"background-color:yellow\">Expires %s. </span></p>",$expdat); 

Now, I need to turn the code between in <span> in to a variable with a variable.How do I code for the second variable in printf() when?:$dclaim = <span style=\"background-color:yellow\">Expires' . $expdat . '.</span>

[justsomeguy]

I don't understand your question.

[niche]

at JSG's suggestion, I'm working with printf(), as well as, eval() to try to solve my problem. This topic's about printf().When I manually code, I get the proper display based on: printf("<p style=\"font-size:9px;margin:0px;\"> <span style=\"background-color:yellow\">Expires %s. </span></p>",$expdat);Now i need to get the values from the mysql table. So based on the above success, if the <span style=\"background-color:yellow\">Expires %s. </span> value comes from the mysql table, how do I need to change my printf() script and what does the table value need to be (i assume the table value will need to be changed)?

[justsomeguy]

First, the value in the database should be the exact string, you don't need to escape quotes or anything. You should be getting this value from the database:

$db_val = '<span style="background-color:yellow">Expires %s. </span>';

Second, it doesn't matter where the string comes from, if it's hard-coded or in a database or a file or whatever else. It's just a string. Use it like you would any other string.

$replaced = sprintf($db_val, $expdat);echo $replaced;

[niche]

I understand. Thanks justsomeguy.