do this if 1

[digatle]

I'm having trouble with one piece of my php code. My problem is I have this routeen:

---
print &quot;<select name=\&quot;app\&quot; style=\&quot;font-size: 10px\&quot; onChange=\&quot;location=document.jump.app.options[document.jump.app.selectedIndex].value;\&quot; value=\&quot;GO\&quot;>&quot;;
print &quot;<option value=\&quot;application\&quot;>Applications</option>&quot;;
while ($a_row = mysql_fetch_array( $applications ) ) {
$app_url=$a_row['url'];
$app_title=$a_row['description'];
print &quot;<option value=\&quot;$app_url\&quot;>$app_title</option>&quot;;
}
print &quot;</select>&quot;;
---

What I need it to do though is to show this <select></select> if there is something there. If there isn't anything there I don't want anything to show up. But I can't write the code without saying &quot;select&quot; with nothing in it.

Help

Digatle

 

[jamesp0tter]

let me see if i got your point, if your query doesn't return any application you don't want the <select> do show up ? just use a

if (mysql_num_rows($application) != 0) {


// that code here


}

that way if there are 0 applications the print code will be avoided

jamesp0tter,
jamespotter@netcabo.pt

p.s.: sorry for my (sometimes) bad english

 

[digatle]

Yeah I want it so if the results are 0 that it does not perform the <select> option. I'll try this and let you know, thank you!!

 

[digatle]

I just tried the code and got the following error:

Notice: Undefined variable: application in /var/

http://www/virtualhosts/peas/html/includes/open/header.php

on line 41

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/

http://www/virtualhosts/peas/html/includes/open/header.php

on line 41


if (mysql_num_rows($application) != 0) {
print &quot;<select name=\&quot;menu\&quot; style=\&quot;font-size: 10px\&quot; onChange=\&quot;location=document.jump.menu.options[document.jump.menu.selectedIndex].value;\&quot; value=\&quot;GO\&quot;>&quot;;
print &quot;<option>$siteID Applications</option>&quot;;
while ($a_row = mysql_fetch_array( $menubar ) ) {
$menu_url=$a_row['url'];
$menu_title=$a_row['description'];
print &quot;<option value=\&quot;$menu_url\&quot;>$menu_title</option>&quot;;
}
print &quot;</select>&quot;;
}


The if statement is line 41.

Digatle

 

[digatle]

Figured it out. In fact the code works perfectly now. The application wasn't changed to menubar. Perfect!!

Digatle