用 PHP 创建动态表

Creating a dynamic table with PHP(用 PHP 创建动态表)

问题描述

我正在尝试使用 PHP 制作动态表.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.

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.

有人可以帮忙吗?

代码在这里:主页中的代码:-

Codes go here: Code in the main page:-

    <table>
    <?php
        $all_pics_rs=get_all_pics();
        while($pic_info=mysql_fetch_array($all_pics_rs)){
        echo "<td><img src="Ii4kcGljX2luZm9b"picture']."' height='300px' width='400px' /></td>";
            } 
?>
</table>

get_all_pics() 函数:

The get_all_pics() function:

$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;

此代码正在创建单行.我想不出如何获得多行...... !!

This code is creating a single row. I can't think of how I can get multiple rows ... !!

推荐答案

$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="IiAuICRpbWFnZVs="src'] . "" /></td>";

    $i++;

}

//Add empty <td>'s to even up the amount of cells in a row:
while ($i <= $maxcols) {
    echo "<td>&nbsp;</td>";
    $i++;
}

//Close the table row and the table
echo "</tr>";
echo "</table>";

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 个 <td>，就添加一个.到达 5 后，关闭该行并创建一个新行.

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.

这个脚本应该给你类似下面的东西.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.

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.

<table>
    <tr>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
    </tr>
    <tr>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td><img src="aW1hZ2UxLmpwZw==" /></td>
        <td>&nbsp;</td>
        <td>&nbsp;</td>
        <td>&nbsp;<td>
    </tr>
</table>

这篇关于用 PHP 创建动态表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！