Question

我正在为一个项目编写设计代码，过去几天都试图让它工作，但无法解决这个问题。

我正在尝试从数据库中获取前4个结果，并将它们打印到标题中。对于每个结果，导航列表编号需要增加1。例如：

<li class="nav_list" id="nav_list1">RESULT 1
<li class="nav_list" id="nav_list2">RESULT 2

这是我到目前为止所拥有的，但不知道如何让它正确显示。任何帮助表示赞赏！

$query = "SELECT id, title, description, groupid FROM category LIMIT 0, 4";
            $result = mysqli_query($dbc, $query)
            or die('Error querying database');
            while ($row = $result->fetch_row()){
                $catID = $row[0];
                $catTITLE = $row[1];

            }
            ?> 
            <ul id="nav_line1">
            <?php
            for ($i = 1; $i <= 4; $i++) {
                print('<li class="nav_list" id="nav_list'.$i.'"><a href="catIDhere">catTITLEhere</a>|</li>');
            }

Answer 1

$i = 1;
while ($row = $result->fetch_row()) {
   echo <<<EOL
<li id="nav_list{$i}"><a href="{$row[0]}"> etc...

EOL;
    $i++;
}

Answer 2

你能试试吗，

    $query = "SELECT id, title, description, groupid FROM category ORDER BY id DESC LIMIT 0, 4";
    $result = mysqli_query($dbc, $query)
    or die('Error querying database');

      echo '<ul id="nav_line1">';
      while ($row = $result->fetch_row()){
        $catID = $row[0];
        $catTITLE = $row[1];           
            print('<li class="nav_list" id="nav_list'.$catID.'"><a href="'.$catID.'">'.$catTITLE.'</a>|</li>');
      }

     echo "</ul>";

Answer 3

尝试这样的事情：

query = "SELECT id, title, description, groupid FROM category LIMIT 0, 4";
        $result = mysqli_query($dbc, $query)
        or die('Error querying database');
        $i = 1
        while ($row = $result->fetch_row()){
            $catID = $row[0];
            $catTITLE = $row[1];
            ?><ul id="nav_line1"><?php
            print('<li class="nav_list" id="nav_list'.$i.'"><a href="'.$catID.'">'.$catTITLE.'</a>|</li>');
            $i++
    }

MySQL查询结果

3 个答案: