[英]Trying to get a mysql table id row into php variable to base futher mysql queries from it
[英]Use variable as ID and get table from mysql (edited)
我正在使用多项选择选项表格来获取场地表。 每个场所都有一个ID,这就是我使用的ID:
<?php
require("db_access.php");
if(isset($_POST['select3']))
{
$aVenues = $_POST['select3'];
if(!isset($aVenues))
{
echo("<p>You didn't select any venues!</p>\n");
}
else
{
$nVenues = count($aVenues);
echo("<p>You selected $nVenues venues: ");
for($i=0; $i < $nVenues; $i++)
{
echo($aVenues[$i] . " ");
}
echo("</p>");
$sql = "SELECT * FROM venues WHERE id IN (" . implode(",",$aVenues) . ")";
$comma_separated = implode(",", $aVenues);
echo $comma_separated;
}
}
?>
结果是:
但是我认为代码会在下面使用这两个数字,并用我使用的id绘制一张表:/? 我想念什么吗?
$array
用于implode(",", $array);
但未在我们可以看到的其他地方定义。 它可能旨在:
implode(",", $aVenues);
更新
对于注释,它不会绘制表,因为您从未真正查询过数据库。
您构建了SQL语句,但是需要执行它并获取结果集。
// Make sure you actually have a database connection
$conn = mysql_connect('localhost', $username, $password);
mysql_select_db($database);
$sql = "SELECT * FROM venues WHERE id IN (" . implode(",",$aVenues) . ")";
$comma_separated = implode(",", $array);
echo $comma_separated;
// Execute query and fetch result rowset
$result = mysql_query($sql);
if ($result) {
$rowset = array();
while ($row = mysql_fetch_array($result)) {
$rowset[] = $row;
}
var_dump($rowset);
}
else echo mysql_error();
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