On Tue, 13 Mar 2012 21:23:57 +0530, Gu®u <nagendra802000@xxxxxxxxx> sent:
No Michael, your code is also not working. What you have understood is
correct. let me explain it to others too.
If variable twitter and facebook are empty don't echo anything,
if variable twitter has a value and facebook is empty echo out only twitter,
if variable twitter has no value and facebook has a value echo out facebook
only,
and finally if both has values echo out both.
Basically I want to echo out only if there is some value in the database.
But in my case both the images are echoing out.
For probably the bazillionth time, PUT REPLIES ON THE BOTTOM
On Tue, Mar 13, 2012 at 8:54 PM, Michael Stowe <me@xxxxxxxxxxxxx> wrote:
From looking at your code, the issue is that your if statements are
checking for the same criteria as your else statements, meaning that if the
string is empty ("") the if statements will be triggered, and since the if
statements are true, the elseif statement will not be. Or if the string
isn't empty, neither the if or the elseif statements will be triggered,
causing the else statement to be activated. Either way, the images would
be printed out. *
*
*
*
*Did you mean to do this?*
<?php
if($search->plugin->ListViewValue() == "" &&
$search->facebook->ListViewValue() == "") {
// Neither one has a value
} elseif ($search->plugin->ListViewValue() != "" &&
$search->facebook->ListViewValue() != "") {
// Both have a Value
echo '<a href="' . $search->plugin->ListViewValue() . '"><img
src="images/twitter.gif" width="22" height="23"/></a></a>' . '<a
href="' . $search->facebook->ListViewValue() . '"><img
src="images/facebook.gif" width="22" height="23"/></a></a>';
} elseif ($search->plugin->ListViewValue() != "") {
// Twitter has a value
} else {
// Facebook has a value (only possible option left)
echo '<a href="' . $search->facebook->ListViewValue() . '"><img
src="images/facebook.gif" width="22" height="23"/></a></a>';
}
?>
Hope that helps,
Mike
On Tue, Mar 13, 2012 at 9:44 AM, Matijn Woudt <tijnema@xxxxxxxxx> wrote:
On Tue, Mar 13, 2012 at 3:06 PM, Gu®u <nagendra802000@xxxxxxxxx> wrote:
> The issue is both the images are echoing and no if else statement is
> working.
>
First of all, please bottom post on this (and probably any) mailing list.
You should perhaps provide what the contents of
$search->plugin->ListViewValue()=="" and
$search->facebook->ListViewValue()=="" is.
Though, if I understood you correctly, it would be as simple as:
$facebookEnabled = $search->facebook->ListViewValue()!="";
$twitterEnabled = $search->plugin->ListViewValue()!="";
if($facebookEnabled)
{
echo '<a href="'.$search->facebook->ListViewValue().'"><img
src="images/facebook.gif" width="22" height="23"/></a></a>';
}
if($twitterEnabled)
{
echo '<a href="'.$search->plugin->ListViewValue().'"><img
src="images/twitter.gif" width="22" height="23"/></a></a>';
}
- Matijn
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-----------------------
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have loved you." John 15:12
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*Best,
*
*Gu®u*
Recapitulating your pseudocode above, you don't need anything so complex:
if twitter is set:
emit twitter image
if facebook is set:
emit facebook image
The two seem completely independent of each other. Thus:
<?php
$twitter = $search->plugin->ListViewValue();
if (!empty($twitter)) {
echo '<a href="'.$twitter.'"><img src="images/twitter.gif"
width="22" height="23"/></a>';
}
$fb = $search->facebook->ListViewValue();
if (!empty($fb)) {
echo '<a href="'.$fb.'"><img src="images/facebook.gif" width="22"
height="23"/></a>';
}
?>
The above will give a twitter image+link if the twitter link is set,
and a facebook image+link if facebook is set. You don't need to do
anything special if both are set, or if neither is set, as they are
completely independent of each other.
--
Tamara Temple
aka tamouse__
May you never see a stranger's face in the mirror
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