使用 PHP 获取表中的照片

发布于 2024-12-10 22:18:24 字数 636 浏览 0 评论 0原文

我正在尝试以这种方式获取我的 mysql 数据，每行 5 张照片，最多 2 行，这似乎不起作用，因为循环不断在我拥有的 2 行中重复照片。

这是我的代码：

    <table border="1" width="%">
<?php
while($info = mysql_fetch_array( $data )) 
                    {

for ($tr=1;$tr<=3;++$tr)
            {
                echo "<tr>";
                for ($td=1;$td<=5;++$td)
                {   


                echo "<td><img width=125 height=125 src=images/".($info['photo']). "></td>";

                }

                echo "</tr>" ;
            }

 }
            ?>

</table>

请帮我解决这个问题，结果一直处于这种状态：

原文

I am trying to fetch my mysql data on this way, 5 photos each row, 2 rows max this seems doesn't work with as the loop keeps to duplicate photos in the 2 rows I have.

Here is my code :

    <table border="1" width="%">
<?php
while($info = mysql_fetch_array( $data )) 
                    {

for ($tr=1;$tr<=3;++$tr)
            {
                echo "<tr>";
                for ($td=1;$td<=5;++$td)
                {   


                echo "<td><img width=125 height=125 src=images/".($info['photo']). "></td>";

                }

                echo "</tr>" ;
            }

 }
            ?>

</table>

Please help me to solve this, the results keep coming in this status:

results to my current code, how can I stop the duplication

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泪是无色的血 2024-12-17 22:18:24

嗯...您的 while 循环会迭代所有照片。对于每张照片，由于 for 循环，您会执行 3 * 5 次迭代。你实际上可能想做的是这样的：

$column = 0;
$row = 0;
while($info = mysql_fetch_array( $data ))  {
    $column++;
    if ($column == 5) {
        echo '</tr><tr>';
        $column = 1;
        $row++;
    }
    if ($row == 2) {
        break;
    }
    // output your image here
}

Well...your while loop iterates over all photos. For each photo you do 3 * 5 iterations due to your for loops. What you actually might want to do is something like this:

$column = 0;
$row = 0;
while($info = mysql_fetch_array( $data ))  {
    $column++;
    if ($column == 5) {
        echo '</tr><tr>';
        $column = 1;
        $row++;
    }
    if ($row == 2) {
        break;
    }
    // output your image here
}

回复收藏 0 原文

你怎么敢 2024-12-17 22:18:24

采用非常简单的方法..首先将获取的图像复制到一维数组中，然后将它们相应地放入表中

while ($ans = mysql_fetch_array($data)) {
  $image[] = $ans['photo'];
}
echo '<table>';   
for ($tr = 1; $tr<=2; $tr++) { 
  echo '<tr>';
  for ($td = 1; $td <= 5; $td++) {

    $img = 'images/' . $info[$td*$tr];
    echo '<td><img src="$img" /></td>';
  }
  echo '<tr/>';
}
echo '</table>';

following a very easy approach .. first copy the fetched images into a single dimensional array and then place them into the table accordingly

while ($ans = mysql_fetch_array($data)) {
  $image[] = $ans['photo'];
}
echo '<table>';   
for ($tr = 1; $tr<=2; $tr++) { 
  echo '<tr>';
  for ($td = 1; $td <= 5; $td++) {

    $img = 'images/' . $info[$td*$tr];
    echo '<td><img src="$img" /></td>';
  }
  echo '<tr/>';
}
echo '</table>';

回复收藏 0 原文

~没有更多了~