小编use*_*737的帖子

我正在尝试创建一个表,其名称是变量中存储的值的值$name.我尝试了很多不同的方法,但似乎没有一个适合我.这是我目前使用的代码:

 mysql_connect("localhost", "peltdyou_admin", "123456") or die(mysql_error()); 
 mysql_select_db("peltdyou_orders") or die(mysql_error()); 
 mysql_query("CREATE TABLE '" .$_POST['name']. "' ( name VARCHAR(30), age INT, car VARCHAR(30))");

我知道这与某些事情有关,'" .$_POST['name']. "'但我无法解决问题.我已经尝试'$name'了它从代码中进一步获得它的价值.

任何帮助都会很棒!

我从教程中选择了这段代码并进行了编辑,使其适合我的需求.

<?php
//if we got something through $_POST
if (isset($_POST['search'])) {
    // here you would normally include some database connection
    include('config2.php');
    $db = new db();
    // never trust what user wrote! We must ALWAYS sanitize user input
    $word = mysql_real_escape_string($_POST['search']);
    // build your search query to the database
    $sql = "SELECT name FROM $tbl_name WHERE name LIKE '%" . $word . "%'";
    // get results
    $row = $db->select_list($sql);
    if(count($row)) {
        $end_result = '';
        foreach($row as $r) {
            $result         = $r['title']; …