在while循环中显示表

时间:2013-03-19 02:15:03

标签: php mysql while-loop html-table

我试图在while循环中显示一个表,每行有3行和3个td。但我很困惑如何归档。

到目前为止,这是我的代码:

while($row = mysqli_fetch_array($result)){

    $testimonial    = $row['testimonial'];
  $cityName         = $row['city_name'];
  $name             = $row['name'];
  $imageName        = $row['image_name'];
  $member           = $row['membership_type'];
  $testimonial  = nl2br($testimonial);

    //Create new testimonial output
    $output  = "<table>\n";
    $output .= "  <tr>\n";
    $output .= "     <td>\n";
    $output .= "      <p>\n";
    $output .= "        <img src=\"".UPLOAD_DIR.$imageName."\" />";
    $output .= "         {$testimonial}";
    $output .= "      </p>\n";
    $output .= "     <p class=\"name\">{$name}<br />\n";
    $output .= "      <span>A Teacher - From {$cityName}</span></p>\n";
    $output .= "   </td>\n";
    $output .= "  </tr>\n";
    $output .= "</table>\n";

    echo $output;
}

上面的代码给了我一个表,每行有多行和1个td。但这不是我期待的结果。

有人可以告诉我如何在While loop中显示这样的表格?

3 个答案:

答案 0 :(得分:2)

当前代码为每个条目创建一个新表。你想要的是将<table></table>部分移动到循环之外,然后你想要一个计数器来跟踪你处理的细胞数量。如果是偶数,请开始新的<tr>。否则,结束当前</tr>。请务必在最后查看是否已完成</tr>,并根据需要添加空<td></td>

答案 1 :(得分:0)

试试这个

$i = 0;
echo "<table>\n<tr>";
while($row = mysqli_fetch_array($result)){
  $testimonial    = $row['testimonial'];
  $cityName         = $row['city_name'];
  $name             = $row['name'];
  $imageName        = $row['image_name'];
  $member           = $row['membership_type'];
  $testimonial  = nl2br($testimonial);
  //Create new testimonial output
  $output .= "     <td>\n";
  $output .= "      <p>\n";
  $output .= "        <img src=\"".UPLOAD_DIR.$imageName."\" />";
  $output .= "         {$testimonial}";
  $output .= "      </p>\n";
  $output .= "     <p class=\"name\">{$name}<br />\n";
  $output .= "      <span>A Teacher - From {$cityName}</span></p>\n";
  $output .= "   </td>\n";
  echo $output;
  if( $i %2 == 1 )
    echo "</tr><tr>\n";
  $i++;
}
echo "</tr>\n</table>\n";

修改

  

通常,要在每行显示n个单元格,请将if语句更改为

if( $i % n == (n - 1) )

答案 2 :(得分:0)

这要做到这一点

<?php

echo "<table>\n";

$cells = $total = 0;
$total_cells = mysqli_num_rows($result);

while($row = mysqli_fetch_array($result))
{
    $testimonial      = nl2br($row['testimonial']);
    $cityName         = $row['city_name'];
    $name             = $row['name'];
    $imageName        = $row['image_name'];
    $member           = $row['membership_type'];

    if ($cells === 0)
        echo "<tr>\n";

    //Create new testimonial output
    $output = "     <td>\n";
    $output .= "      <p>\n";
    $output .= "        <img src=\"".UPLOAD_DIR.$imageName."\" />";
    $output .= "         {$testimonial}";
    $output .= "      </p>\n";
    $output .= "     <p class=\"name\">{$name}<br />\n";
    $output .= "      <span>A Teacher - From {$cityName}</span></p>\n";
    $output .= "   </td>\n";

    echo $output;

    $cells++;
    $total++;       

    if ($cells === 3 || $total === $total_cells)
    {
        echo "</tr>\n";
        $cells = 0;
    }
}

echo "</table>\n";

?>