想使用的輸入值作爲數據庫的名稱,但它不幹活菜鳥:嘗試使用輸入值設置PHP數據庫名稱
<html>
<form action="config.php" method="Post">
<table>
<tr><td>DB Name: <input type="text" name="dbname"></td></tr>
<tr><td><button type="submit" name="submit_two"> Submit</button></td></tr>
</table>
</form>
</html>
的config.php
<?php
$dbname = "";
if (isset($_POST['submit_two'])){
$dbname = test_input($_POST["dbname"]);
$data = "CREATE DATABASE IF NOT EXISTS $dbname";
if (!$data) {
echo "could not connect" . mysql_error();
} elseif ($dbname == ""){
echo "no";
}else{
echo "gud to go";
}
}
function test_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
它始終顯示「gud去」,但當我檢查數據庫時,將不會創建任何內容
在這裏做什麼'test_input'? – 2014-10-09 05:16:29
$ dbname = $ _ POST [「dbname」]; $ data = mysql_query(「CREATE DATABASE IF NOT EXISTS $ dbname」); – 2014-10-09 05:16:41
是@ankurbhadania是對的 – 2014-10-09 05:17:27