我用 php/mysql 的工作不多，但我需要一个我认为相对简单的任务:检查一个表是否存在，如果不存在则创建它。我什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 = mysql_query($mysql_connexn, $query);
    $result = mysql_query($query, $mysql_connexn);

    if(empty($result)) {
        echo "<p>" . $table . " table does not exist</p>";
        $query = "CREATE TABLE IF NOT EXISTS WEIGHIN_DATA (
            id INT NOT NULL AUTO_INCREMENT,
            PRIMARY KEY(id),
            DATE    DATE NOT NULL,
            VALUE   SMALLINT(4) UNSIGNED NOT NULL
        )"
    }
    else {
        echo "<p>" . $table . "table exists</p>";
    } // else

?>

最佳答案

一些事情。

在 ) 中和末尾缺少一个分号 ;

if(empty($result)) {
    echo "<p>" . $table . " table does not exist</p>";
    $query = "CREATE TABLE IF NOT EXISTS WEIGHIN_DATA (
        id INT NOT NULL AUTO_INCREMENT,
        PRIMARY KEY(id),
        DATE    DATE NOT NULL,
        VALUE   SMALLINT(4) UNSIGNED NOT NULL
    )" // <--- right there

这会导致/抛出解析错误，例如:

Parse error: syntax error, unexpected '}' in...

我对您最初发布的代码的评论中显示的其他错误。

另外，您在创建表时没有使用 mysql_query。

这是一个mysqli_方法，我在其中注释掉了你的原始代码。

旁注:您正在为 $query = "SELECT ID FROM "中的列使用 ID。 $table; 然后你创建你的表和列作为小写的 id;两个字母大小写必须匹配。

<?php

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

$DB_HOST = "xxx"; // put your own data
$DB_NAME = "xxx"; // put your own data
$DB_USER = "xxx"; // put your own data
$DB_PASS = "xxx"; // put your own data


$conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($conn->connect_errno > 0) {
  die('Connection failed [' . $conn->connect_error . ']');
}



/*

    // 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 " . $table; // that should be id and not ID
    //$result = mysql_query($mysql_connexn, $query);
    $result = mysql_query($query, $mysql_connexn);


*/


    $table = "WEIGHIN_DATA";
    $query = "SELECT ID FROM " . $table; // that should be id and not ID
    //$result = mysql_query($mysql_connexn, $query); // your original code
    // however connection comes last in mysql method, unlike mysqli
    $result = mysqli_query($conn,$query);


if(empty($result)) {
    echo "<p>" . $table . " table does not exist</p>";
    $query = mysqli_query($conn,"CREATE TABLE IF NOT EXISTS WEIGHIN_DATA (
        id INT NOT NULL AUTO_INCREMENT,
        PRIMARY KEY(id),
        DATE    DATE NOT NULL,
        VALUE   SMALLINT(4) UNSIGNED NOT NULL
    )");
    }
    else {
        echo "<p>" . $table . "table exists</p>";
    } // else

?>

关于php - 如果 mysql 表不存在则创建，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25145574/