使用 PHP 创建动态表
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Creating a dynamic table with PHP
提问by maxxon15
I'm trying to make a dynamic table with PHP. I have a page which displays all the pictures from a database. I need the table to be of 5 columns only. If more than 5 pictures are returned, it should create a new row and the displaying of the rest of the pics would continue.
我正在尝试用 PHP 制作一个动态表。我有一个页面显示数据库中的所有图片。我只需要该表为 5 列。如果返回的图片超过 5 张，则应创建一个新行，其余图片将继续显示。
Can anyone please help?
有人可以帮忙吗？
Codes go here: Code in the main page:-
代码在此处：主页中的代码：-
<table>
<?php
$all_pics_rs=get_all_pics();
while($pic_info=mysql_fetch_array($all_pics_rs)){
echo "<td><img src='".$pic_info['picture']."' height='300px' width='400px' /></td>";
}
?>
</table>
The get_all_pics() function:
get_all_pics() 函数：
$all_pics_q="SELECT * FROM pics";
$all_pics_rs=mysql_query($all_pics_q,$connection1);
if(!$all_pics_rs){
die("Database query failed: ".mysql_error());
}
return $all_pics_rs;
This code is creating a single row. I can't think of how I can get multiple rows ... !!
此代码正在创建单行。我想不出如何获得多行......！
回答by Jules
$maxcols = 5;
$i = 0;
//Open the table and its first row
echo "<table>";
echo "<tr>";
while ($image = mysql_fetch_assoc($images_rs)) {
if ($i == $maxcols) {
$i = 0;
echo "</tr><tr>";
}
echo "<td><img src=\"" . $image['src'] . "\" /></td>";
$i++;
}
//Add empty <td>'s to even up the amount of cells in a row:
while ($i <= $maxcols) {
echo "<td> </td>";
$i++;
}
//Close the table row and the table
echo "</tr>";
echo "</table>";
I haven't tested it yet but my wild guess is something like that. Just cycle through your dataset with the images and as long as you didn't make 5 <td>'s yet, add one. Once you reach 5, close the row and create a new row.
我还没有测试过，但我的疯狂猜测是这样的。只需使用图像循环浏览数据集，只要您还没有制作 5 <td>，就添加一个。达到 5 后，关闭该行并创建一个新行。
This script is supposed to give you something like the following. It obviously depends on how many images you have and I assumed that 5 (defined it in $maxcols) was the maximum number of images you want to display in a row.
该脚本应该为您提供如下内容。这显然取决于您拥有的图像数量，我假设 5（在$maxcols 中定义）是您想要连续显示的最大图像数量。
<table>
<tr>
<td><img src="image1.jpg" /></td>
<td><img src="image1.jpg" /></td>
<td><img src="image1.jpg" /></td>
<td><img src="image1.jpg" /></td>
<td><img src="image1.jpg" /></td>
</tr>
<tr>
<td><img src="image1.jpg" /></td>
<td><img src="image1.jpg" /></td>
<td> </td>
<td> </td>
<td> <td>
</tr>
</table>
回答by Tom
$max_per_row = 5;
$item_count = 0;
echo "<table>";
echo "<tr>";
foreach ($images as $image)
{
if ($item_count == $max_per_row)
{
echo "</tr><tr>";
$item_count = 0;
}
echo "<td><img src='" . $image . "' /></td>";
$item_count++;
}
echo "</tr>";
echo "</table>";
