[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)
我正在使用多項選擇選項表格來獲取場地表。 每個場所都有一個ID,這就是我使用的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;
}
}
?>
結果是:
但是我認為代碼會在下面使用這兩個數字,並用我使用的id繪制一張表:/? 我想念什么嗎?
$array
用於implode(",", $array);
但未在我們可以看到的其他地方定義。 它可能旨在:
implode(",", $aVenues);
更新
對於注釋,它不會繪制表,因為您從未真正查詢過數據庫。
您構建了SQL語句,但是需要執行它並獲取結果集。
// 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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