SQL在表中显示数据而未指定列名 [英] Sql show data in tables without specifying the colum name
问题描述
所以我有这段代码:
<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
include "../includes/db_conn.php";
$data = $_POST['tabledb'];
$sfm = $dbm->prepare('SELECT * FROM '.$data.'');
$sfm->execute();
$all = $sfm->fetchAll();
?>
<!DOCTYPE html>
<html>
<body>
<?php
foreach ($all as $row)
{
echo "<option value=\"info\">" . $row[0] . "</option>";
}
?>
</body>
</html>
如果具有下拉菜单，则可以选择上下文，如果可以选择该表是通过ajax调用发送到此页面的，则可以从数据库中选择行.这就是为什么我有
oke a littel context if have a dropdown menu where you can select rows from the database, if you select a table that table is send to this page trough a ajax call. that is why I have:
$data = $_POST['tabledb'];
在我的代码中.现在，我想显示表的内容，但是每个表都有不同数量的列和列名称.如何在不指定列名的情况下显示表的内容?请帮忙.
in my code. now I would like to show the contents of the tables, but every table has a different amount of columus and colum names. how do I show the contents of the tables without specifying the colum name? please help.
编辑
我尝试过的方法是，尝试将其设置为for循环，但是没有用.
what I tried, I tried making it a for loop, didn't work.
编辑2
我还不够清楚，我试图获得与print_r相同的结果，但带有回显.
I wasn't clear enough, I am trying to get the same result as print_r but with a echo.
推荐答案
您可以使用implode将每一行的所有元素组合为一个字符串:
You could use implode to combine all elements for each row into a single string:
foreach ($all as $row)
{
echo ‘<option value="info">’ . implode(‘|’, $row) . ‘</option>‘;
}
或者您可以使用嵌套的foreach:
Or you could use use nested foreach:
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