PHP:使用fetchall()显示表数据 [英] PHP: Displaying table data using fetchall()

问题描述

PHP/MySQL快速问题:

Quick PHP/MySQL question:

我试图将MySQL数据库表中的数据显示为HTML表，由于某种原因，我的代码使每条数据的输出加倍.

I'm trying to display the data from a MySQL database table as an HTML table, and for some reason my code doubling the output of each piece of data.

这是我的代码:

$rowarray = $statement->fetchall();
print "<tr>\n";  
foreach ($rowarray as $row) {
    foreach ($row as $col) {
        print "\t<td>$col</td>\n";
    }
    print "</tr>\n";
}  

我的结果看起来与此类似:

My results look something similar to this:

用户标识|用户名|名|姓氏

User ID | User Name | First Name | Last Name

     1               1     &NBSP ;    用户名     名字&nspsp ;      Last Name

    1              1            User Name   User Name     First Name     First Name     Last Name     Last Name

等等.您明白了.为什么会这样呢?顺便说一句，如果我通过参考row []下标0-3手动添加列信息，则所有内容都会正确显示；只有当我使用嵌套的foreach语句时，数据才会重复.

Etc etc. You get the idea. Why this is happening? By the way, if I manually add the column information by referring to row[] subscripts 0-3, everything is displayed properly; it's only when I use the nested foreach statements that the data is duplicated.

推荐答案

您从

You are getting both a numeric and a name-indexed value back from PDO fetchall. To get just the numeric index:

$rowarray = $statement->fetchall(PDO::FETCH_NUM);

这篇关于PHP:使用fetchall()显示表数据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IT屋！