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Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/test/public_html/theme/basic/_side.php on line 49
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/test/public_html/theme/basic/_side.php on line 50
카테고리 부분에 이렇게 오류가 뜨네요
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<div id="widget-area"> <div class="ig_widget"> <div class="fancy_one"> <span>CATEGORIES</span> </div> <? $sql = "select bo_category_list from g5_board where bo_table = '{$blog}'"; //카테고리 $result = mysql_query($sql); $bo = mysql_fetch_array($result); $ca_menu = explode("|",$bo[bo_category_list]); ?> <ul> <? for ($c=1, $cnt=count($ca_menu); $c<$cnt; $c++) { ?> <li class="cat-item"><a href="<?php echo G5_URL;?>/<?php echo $board;?>/<?php echo $ca_menu[$c];?>/" ><?php echo $ca_menu[$c];?></a></li> <?}?> </ul> </div> <div class="ig_widget"> <div class="fancy_one"> <span>NEWEST ALBUM </span> </div> <?php echo latest("theme/side-album", "album", 1, 55); ?> </div> </div> </div><!-- END col-md-3 -->멀 어케해줘야 될지 모르겠네요 ㅠㅠ
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답변 1개
2017-02-09 (목) 12:36:35
일단은..
mysql_fetch_array
여기를 이렇게 수정
sql_fetch_array
답변을 작성하려면 로그인이 필요합니다.