小编use*_*163的帖子

我对php/mysql的工作量不大,但我需要的是一个相对简单的任务:检查表是否存在,如果不存在则创建表.我甚至无法获得有用的错误消息,并且数据库中没有创建表.我的语法显然有问题.

<?php

    session_start();
    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    // 1. CONNECT TO THE DB SERVER, confirm connection
    mysql_connect("localhost", "root", "") or die(mysql_error());
    echo "<p>Connected to MySQL</p>";
    $mysql_connexn = mysql_connect("localhost", "root", ""); // redundant ?

    // 2. CONNECT TO THE SPECIFIED DB, confirm connection
    $db = "weighttracker";
    mysql_select_db($db) or die(mysql_error());
    echo "<p>Connected to Database '$db'</p>";
    $db_connexn = mysql_select_db($db)or die(mysql_error("can\'t connect to $db"));

    // 3. if table doesn't exist, create it
    $table = "WEIGHIN_DATA";
    $query = "SELECT ID FROM " . $table;
    //$result = …