PHP MySQL - Error: No Database selected

问题

I am trying to read and write to a database. Here is the code I have so far:

$mysql = mysqli_connect("example.com", "johndoe", "abc123"); // replace with actual credidentials
$sql = "CREATE DATABASE IF NOT EXISTS dbname";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating database: " . mysqli_error($mysql);
}
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($mysql);
$mysql = mysqli_connect("example.com", "johndoe", "abc123", "dbname"); // replace with actual credidentials
$sql = "CREATE TABLE IF NOT EXISTS Users(ID INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(ID), username CHAR(15), password CHAR(15), email CHAR(50))";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating table: " . mysqli_error($mysql);
}
$sql = "INSERT INTO Customers(username, password, email) VALUES(" . $username . ", " . $password . ", " . $email . ")";
if (!mysqli_query($mysql, $sql)) {
    echo "Error: " . mysqli_error($mysql);
}
mysqli_close($mysql);

However, when I try to run it, it has an error:

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' , )' at line 1

Could anybody tell me how to fix this?

回答1:

First check mysqli_select_db if it returns false then create database.

try like this:

 $mysql = mysqli_connect("example.com", "johndoe", "abc123") or die(mysqli_connect_error()); // replace with actual credidentials

if (!mysqli_select_db($mysql,'hardestgame_accounts')) {
 $sql = "CREATE DATABASE IF NOT EXISTS hardestgame_accounts";
 if (!mysqli_query($mysql, $sql)) {
    echo "Error creating database: " . mysqli_error($mysql);
 }
}
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "CREATE TABLE IF NOT EXISTS Users(ID INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(ID), username CHAR(15), password CHAR(15), email CHAR(50))";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating table: " . mysqli_error($mysql);
}
mysqli_close($mysql);

here is a good answer: Php mysql create database if not exists

回答2:

Please check syntax of mysqli, it takes 4 parameters.You also have to provide database name.

$link = mysqli_connect("myhost","myuser","mypassw","my_db");

回答3:

You're missing database in mysqli_connect() call

$link = mysqli_connect("hostname","username","password","database") or die("Error " . mysqli_error($link));

回答4:

Try using mysql_select_db in between the database connection and the table creation.

Also, mysql_ is deprecated, please use mysqli_ instead

回答5:

Obviously the answer is in your error. You didn't select any database. When using this function mysqli_connect specify the database you want to connect to.

Here is the syntax of the function: http://www.w3schools.com/Php/func_mysqli_connect.asp .

1) Create your database outside of your application

2) Specify mysqli_connect with the database you want to select.

You can also use another function called mysqli_select_db . You can find the sytanx here : http://www.w3schools.com/php/func_mysqli_select_db.asp .

As already stated in the comment, you will also have to replace : "example.com" with your ip address, if you are running locally replace it with 127.0.0.1:3306 , if you didn't change the port when you installed your mysql database / "johndoe" with your database account, you can change that to "root" / "abc123" with your root account password DEFAULT : "" .

Good luck !

回答6:

Use ` backticks for MYSQL reserved words... your table name is reserved word for MYSQL... Change your table name.

来源：https://stackoverflow.com/questions/21972146/php-mysql-error-no-database-selected

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php

mysql

create-table