PHP EOF showing only one result [duplicate]

This question already has an answer here: 开发者_如何学JAVA Closed 11 years ago.

Possible Duplicate:

PHP EOF shows only one result from loop

Hello

Seems like I can not find the solution for such problem.

I am using the following code.

it should display all the results given by the mySQL loop in the EOF.

But it is showing only the first results, nothing else.

What I am doing wrong?

Please help me

function getYiBAdminBanner() {
global $site;
global $dir;
$queryYiBmenu = "SELECT * FROM `(YiB)_cPanel_Menu` WHERE Type = 'top'";
$resultYiBmenu=mysql_query($queryYiBmenu) or die("Errore select menu: ".mysql_error());
$countYiBmenu = mysql_num_rows($resultYiBmenu); 
while($rowYiBmenu = mysql_fetch_array($resultYiBmenu)) {
$menu .= "<div id=\"menu\" style=\"display:none;\"><li><a href=\"".$site['url'].$rowYiBmenu['linkHref']."\" onMouseOut=\"javascript: $('#menu').hide('9000');\"><img class=\"imgmenu\" src=\"".$site['url'].$rowYiBmenu['linkIcon']."\">".$rowYiBmenu['linkTitle']."</a></li></div>";
}
if($countYiBmenu <= 0){
$menu = "No Modules Installed";
}
$bannerCode .= <<<EOF
<div style="width:520px; background-color: #EEE; height:30px;">
{$menu}
</div>
EOF;
return $bannerCode;
}

---

Though this won't fix your issue, I see you are using mysql_fetch_array(). This is rather pointless, as you are doing associative matching afterwards ($rowYiBmenu['linkHref'], for example). It will work for you, but it's a waste of resources, as the results will be loaded in a numeric array as well (making the $rowYiBmenu array twice as large, just wasting memory).

Also, you didn't declare the $menu variable, you just 'added on'. Before the while statement, put $menu = ''; (or anything else that will declare an empty string for that variable).

Lastly, use single quotes around html strings. That way you don't have to keep escaping double quotes for when adding an attribute. For example, the $menu line in the while statement should look like this:

$menu .= '<div id="menu" style="display:none;"><li><a href="'.$site['url'].$rowYiBmenu['linkHref'].'" onMouseOut="javascript: $(\'#menu\').hide(9000);"><img class="imgmenu" src="'.$site['url'].$rowYiBmenu['linkIcon'].'">'.$rowYiBmenu['linkTitle'].'</a></li></div>';

I'm not sure if I helped you with your problem at all, but I thought this would help you clean up your code a bit (the cleaner the code, the less prone to errors and the easier it is to spot errors).

---

I have tried helping in the other duplicate question.

i feel you need to implement (learn?) some basic debugging on the data coming from the database. This should be very simple to solve.

I suggest changing your while() code to something like this to help debug whats happening:

while($rowYiBmenu = mysql_fetch_array($resultYiBmenu)) {
    print "linkTitle: " . $rowYiBmenu['linkTitle'] ."<BR>";
    print "linkHref: " . $rowYiBmenu['linkHref'] ."<BR><BR>";
}
exit;

You should see output of all results from the database (not just the first or last). After establishing that the correct data is being retrieved and looped through. you can get that $menu variable concatenation and $bannerCode + HereDoc EOF code sorted out.