PHP:使用fetchall()显示表数据

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

&NBSP;&NBSP;&NBSP;&NBSP; 1&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP; 1&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&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.

答

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);

PHP:使用fetchall()显示表数据

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