用PHP创建一个动态表 [英] Creating a dynamic table with PHP

问题描述

我正在尝试用PHP创建一个动态表。我有一个页面显示数据库中的所有图片。
我需要的表只有5列。如果返回超过5张照片，应该创建一个新行，剩下的照片的显示将会继续。

任何人都可以帮忙？

代码到这里：
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> 
  

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; 
  

此代码正在创建一行。我不能想到如何获得多行... !!

解决方案

  $ maxcols = 5; 
 $ i = 0; 
 
 //打开表和第一行
 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 ++; 
 
} 
 
 //添加空的< td>是连续的单元格数量：
 while（$ i <= $ maxcols）{
 echo< td>& nbsp;< / td>; 
 $ i ++; 
} 
 
 //关闭表行和表
 echo< / tr>; 
 echo< / table>; 
  

我还没有测试过，但我的狂野猜测是这样的。只要循环浏览您的数据集与图像，只要你没有制作5 < td> ，添加一个。一旦达到5，关闭该行并创建一个新行。

此脚本应该给你如下所示。这显然取决于你有多少图像，我假设5（在 $ maxcols 中定义）是要显示在一行中的最大图像数。

 < 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>& nbsp;< / td> 
< td>& nbsp;< / td> 
< td>& nbsp;< td> 
< / tr> 
< / table> 
  

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.

Can anyone please help?

Codes go here: Code in the main page:-

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

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;

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

解决方案

$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=\"" . $image['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>";

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.

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="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>&nbsp;</td>
        <td>&nbsp;</td>
        <td>&nbsp;<td>
    </tr>
</table>

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