PHP and MYSQLI with LIKE array: does not echo all content correctly -

i have come far , seem stuck on trivial output issue, apologies in advance.

the $getexhibitions array , works when aspect of code running. have added $results based on keyword link intended array within array.

my problem output supposed show $getexhibitions->title list of each $results->sponsor underneath. java creates drop down specific sponsor information. repeat preceeding 8 years so.

as can see code getting stuck , feel though may div or { problem or may more array code. in advance.

http://tenmoregirls.com/tenmoregirls/sponsors.php

    <?php $current_url = base64_encode($url="http://".$_server['http_host'].$_server['request_uri']);   $query_getexhibitions = $mysqli->query("select * exhibitions order year desc"); if ($query_getexhibitions) {      while($getexhibitions = $query_getexhibitions->fetch_object())        {              echo "<a>$getexhibitions->year | $getexhibitions->title</a>", "<br />";  $sponsorlink = $getexhibitions->year; $result = $mysqli->query("select * sponsors (`year` '%".$sponsorlink."%')");   if ($result) {          while($results = $result->fetch_object()) {                     echo "<a href = \"javascript:void(0)\" class=\"subheading\" onclick = \"document.getelementbyid($results->id).style.display='block'\" >$results->sponsor</a>", "<br />";             echo "<div id=\"$results->id\" style=\"display: none;\">            <a href = \"javascript:void(0)\" onclick = \"document.getelementbyid($results->id).style.display='none'\"><img src=\"exhibitstyles/links/close.png\" alt=\"close\" width=\"20\"/></a><br />";                 echo '<img src="sponsorimages/'.$results->logo.'">';         echo $results->summary, "<br />";                     echo "<a href=\"http://$results->website\" target=\"_blank\">$results->website</a>", "<br /><br /><br />";  } "</div>"; } "</div>"; }}   exit(); ?> 

it's not problem php , mysqli, wrong closing div tag, outside loop. there 1 additional closing tag don't seem refer anything. try move </div> tag inside loop , delete other one.

edit: missing echo command before "</div>"; tag, because used ; on end of previous line. you can't use , between multiple echo parameters. . concatenation operator strings in php. whereas . concatenation operator strings, echo construction allows usage of , between arguments. bad.

try this:

while($getexhibitions = $query_getexhibitions->fetch_object())        {      echo "<a>$getexhibitions->year | $getexhibitions->title</a><br />";      $sponsorlink = $getexhibitions->year;     $result = $mysqli->query("select * sponsors (`year` '%".$sponsorlink."%')");       if ($result) {            while($results = $result->fetch_object()) {             echo "<a href = \"javascript:void(0)\" class=\"subheading\" onclick = \"document.getelementbyid($results->id).style.display='block'\" >$results->sponsor</a><br />";              echo "<div id=\"$results->id\" style=\"display: none;\">                 <a href = \"javascript:void(0)\" onclick = \"document.getelementbyid($results->id).style.display='none'\"><img src=\"exhibitstyles/links/close.png\" alt=\"close\" width=\"20\"/></a><br />";                     echo '<img src="sponsorimages/'.$results->logo.'">';             echo $results->summary . "<br />";             echo "<a href=\"http://$results->website\" target=\"_blank\">$results->website</a><br /><br /><br />";             echo "</div>";         }      } }