php - 为什么不创建表 Simpsons？

Question

这是充当另一个 HTML 页面的“操作”的 PHP 代码，其中只有一个表单。

它没有显示任何错误，在我单击提交后出现消息“已创建数据库”，但它没有创建表“Simpsons”。为什么会这样？

请让我知道我是否应该提供更多详细信息或评论代码。

<?php

$database = 'basedados';
$username = 'root';
$password = '';

$first = $_POST['first'];
$last = $_POST['last'];
$phone = $_POST['phone'];
$mobile = $_POST['mobile'];
$fax = $_POST['fax'];
$email = $_POST['email'];
$web = $_POST['web'];

$con = mysql_connect(localhost, $username, $password);
if (!$con) {
    die('Could not connect: ' . mysql_error());
}

if (mysql_query("CREATE DATABASE IF NOT EXISTS $database", $con)) {
    echo "Database created";
} else {
    echo "Error creating database: " . mysql_error();
}

mysql_select_db($database, $con);
$create_table = "CREATE TABLE Simpsons
                (
                first varchar(15),
                last varchar(15),
                phone varchar,
                mobile varchar,
                fax varchar,
                email varchar,
                web varchar,
                )";
$insert_values = "INSERT INTO Simpsons VALUES ($first','$last','$phone','$mobile','$fax','$email','$web')";
if (table_exists(Simpsons, $database)) {
    mysql_select_db($database) or die("Unable to select database");
    mysql_query($insert_values, $con);
} else {
    mysql_select_db($database) or die("Unable to select database");
    mysql_query($create_table, $con);
    mysql_query($insert_values, $con);
}

function table_exists($table, $db) {
    $tables = mysql_list_tables($db);
    while (list ($temp) = mysql_fetch_array($tables)) {
        if ($temp == $table) {
            return TRUE;
        }
    }
    return FALSE;
}

mysql_close($con);
?>

---

更新:

好的。谢谢你们。现在我的代码看起来更像这样:

<?php

$database = 'basedados';
$username = 'root';
$password = '';

$first = $_POST['first'];
$last = $_POST['last'];
$phone = $_POST['phone'];
$mobile = $_POST['mobile'];
$fax = $_POST['fax'];
$email = $_POST['email'];
$web = $_POST['web'];

$con = mysql_connect(localhost, $username, $password);
if (!$con) {
    die('Could not connect: ' . mysql_error());
}

if (mysql_query("CREATE DATABASE IF NOT EXISTS $database", $con)) {
    echo "Database created";
} else {
    echo "Error creating database: " . mysql_error();
}

$create_table = "CREATE TABLE IF NOT EXISTS simpsons
                (
                first varchar(15),
                last varchar(15),
                phone varchar(15),
                mobile varchar(15),
                fax varchar(15),
                email varchar(15),
                web varchar(15)
                )";
$insert_values = "INSERT INTO simpsons VALUES ('$first','$last','$phone','$mobile','$fax','$email','$web')";

mysql_select_db($database, $con) or die("Unable to select database: " . mysql_error());
mysql_query($create_table, $con) or die("Error creating table: " . mysql_error());
mysql_query($insert_values, $con) or die("Error inserting values: " . mysql_error());
mysql_close($con);
?>

我认为它看起来更干净，虽然我不确定我是否正确使用了 mysql_error() 函数。 再次感谢。

Answer 1

一个逗号太多，你需要指定varchar-fields的长度

CREATE TABLE Simpsons
            (
            first varchar(15),
            last varchar(15),
            phone varchar(xx),
            mobile varchar(xx),
            fax varchar(xx),
            email varchar(xx),
            web varchar(xx)
            )