You can use:
if($row[1]){print "Email:$row[1]";}else{print " ";}
Jason White
----- Original Message -----
From: "Matthew K. Gold" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, July 12, 2002 9:23 PM
Subject: [PHP] is_null question
> Hi Everyone,
>
> Here's my problem: I'd like to make the printing of some text dependent
on
> whether or not a variable is null. In the following example, when $row[1]
> is null, what gets printed on the page is "Email: ". I'd like the
> script to not print "Email:" if row[1] is null.
>
> It looks like I should be using is_null(), but I'm not sure how to use it,
> especially since all of this is embedded in a print statement.
>
> while ($row = mysql_fetch_row ($result))
> {
> print ("\t<tr bgcolor=\"#ff9900\">\n\t\t<th>$row[0]</th>\n</tr>\n\t<tr
> class=\"default\" bgcolor=\"#ffffff\">\n\t\t<td><p>$row[10]</p><p>Email:
<a
> href=\"mailto:$row[1]\";>$row[1]</a><br />Phone: $row[2]</p>\n\n<p>Address:
> $row[3]<br />$row[4]<br />$row[5] $row[6] $row[7]</p> <p>URL: <a
> href=\"$row[8]\" target=\"_blank\">$row[8]</a></p></td></tr>");
> }
>
>
> Instead of just printing Email: $row[1], I'd like to test whether it's
null
> first--maybe something like this?
>
> if (is_null($row[1])) print " " else print ("Email: $row[0]")
>
> but can I put that line inside of another print statement? if so, do I
have
> to escape the quotation marks? Please forgive my confusion--I'm new to
> programming in general and PHP in particular.
>
> Thanks in advance for your help.
>
> best,
>
> Matt
>
>
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