dozug64282 2016-02-16 17:44
浏览 39
已采纳

如何使用PHP回显基于MySQL数据库的动态html片段?

I'm trying to create an image gallery with a for loop iterated by a counter of database rows. To make it more clear: for each row in the table, get only the id(primary index) number and the image link from the server (not all the info in the row). With that information, echo an HTML image tag with the link inside the 'src=' and the id inside the 'alt='.

Two problems here:
1- the id number of the first row isn't zero.
2- I don't have a clue on how to get the total number of rows and to fetch only those two informations (id and img source).

That way, I could subtract the total number of rows minus the id number of the first row and using it to put an end on the loop.

So how to echo this dynamic html snippet based on my databse with PHP?

My code:

<?php

$link = mysqli_connect('localhost','user','pass','db');
$result = mysqli_query($link, "SELECT * FROM `table`");
$rows = mysqli_num_rows($result);

/* free result set */
mysqli_free_result($result);

$caption = mysqli_query($link, "SELECT ");

for($i=0; $i < $rows; $i++) {
    echo "<img src='$imageURL' alt='$idNumber'>";
}

?>
  • 写回答

3条回答 默认 最新

  • dongshen9058 2016-02-16 18:48
    关注

    This is a really easy task.

    First we fetch the data from the database using mysqli_query to do the query.

    Then we use mysqli_fetch_array to get an array so then we can loop through it and echo each item.

    After that, we mysqli_num_rows to get the total number of rows returned and increment it by 1 so it is not zero.

    NOTE: Since you are going to increment the id to avoid getting a '0', don't to forget to minus '1' if you intend to use that id for some server-side purpose.

    $result = mysqli_query($link, "SELECT * FROM `table`"); //query sql
    $result_array=mysqli_fetch_array($result, MYSQLI_ASSOC);//return array from the query
    $count = mysqli_num_rows($result);                      //get numbers of rows received 
    
    foreach($result as $row){ //do a foreach loop which is really simple
        echo "<img src='". $row['img_column_name_from_db'] . "'  alt='" .$row['id_column_name_from_db'] + 1 . "'>"; //echo data from the array, + 1 to "$row['id_column_name_from_db']" so that 'alt=' doesn't start from '0'.
        } 
    echo $count;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 metadata提取的PDF元数据,如何转换为一个Excel
  • ¥15 关于arduino编程toCharArray()函数的使用
  • ¥100 vc++混合CEF采用CLR方式编译报错
  • ¥15 coze 的插件输入飞书多维表格 app_token 后一直显示错误,如何解决?
  • ¥15 vite+vue3+plyr播放本地public文件夹下视频无法加载
  • ¥15 c#逐行读取txt文本,但是每一行里面数据之间空格数量不同
  • ¥50 如何openEuler 22.03上安装配置drbd
  • ¥20 ING91680C BLE5.3 芯片怎么实现串口收发数据
  • ¥15 无线连接树莓派,无法执行update,如何解决?(相关搜索:软件下载)
  • ¥15 Windows11, backspace, enter, space键失灵