如果不存在则创建mysql表 [英] create mysql table if it doesn't exist

问题描述

我在php/mysql上工作不多，但是我需要一个我认为相对简单的任务:检查一个表是否存在，如果不存在则创建一个表.我什至没有收到有用的错误消息，并且数据库中没有创建表.我的语法显然有问题.

I don't work much with php/mysql, but I need what I thought would be a relatively straightforward task: to check if a table exists and create it if it doesn't. I can't even get a useful error message and there's no table being created in the db. There is obviously something wrong with my syntax.

<?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

?>

推荐答案

一些事情.

)"和

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

可能会导致/引发解析错误，例如:

which would have caused/thrown a parse error, such as:

解析错误:语法错误，...中出现意外的'}'

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

在您最初发布的代码中我的评论中还显示了其他错误.

Amongst other errors as shown in my comments from your originally posted code.

此外，您在创建表时没有使用mysql_query.

Plus, you were not using mysql_query in your table creation.

这是一个mysqli_方法，在这里我注释掉了您的原始代码.

Here is a mysqli_ method, where I commented out your original codes.

旁注:您正在对$query = "SELECT ID FROM " . $table;中的列使用ID，但是您将表和列创建为小写的id；两个字母都必须匹配.

Sidenote: You're using ID for your column in $query = "SELECT ID FROM " . $table; and yet you create your table and column as id in lowercase; both lettercase must match.

<?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

?>

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