Question

在PHP MySQL中将来自同一个表的同一个索引的一个SQL查询组合起来，但我只是尝试索引名称新闻图像选择表而不是获取此输出视图。

这个输出视图图像到达数据库都是图像但是两个分组视图它唯一的查询如何抛出

[first image get to databese][1]    [third image get to databese][3]  [five  image get to databese][5]    five 7 img ...............
[second image get to databese][2]    [fourth image get to databese][4] [six image get to databese][6]      six 8   img...............

但我尝试此查询但没有工作

$numrow =0;
$vali=0;
$sql_q = mysqli_query($conn,"SELECT * FROM tbl_news where cat_id='$cat_id' and cat_id !=19 order by nid DESC"); 
$ny = mysqli_fetch_array($sql_q);
if mysqli_num_rows($sql_q) > 0) {
   while($newsRowi = mysqli_fetch_array($sql_q)){
            $row[]=$newsRowi;
   }
   $n=0;$val=0;
   foreach($row as $newsRow ){    
     $news_img     = $newsRow['news_image'];
     $f_id     = $newsRow['nid'];
     $news_heading = substr($newsRow['news_heading'],0,50);
     $sq = mysqli_query($conn,"SELECT * FROM tbl_news where nid='ny[0]'");
     $ya=mysqli_fetch_array($sq);
     $val = $ya['news_image'];
     $n++;
     $n++;
 }
?>

请帮帮我

如何为同一个表相同的索引组编写查询

0 个答案: