标签:create-table php mysql
我正在尝试读取和写入数据库.这是我到目前为止的代码:
$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);
但是,当我尝试运行它时,它有一个错误:
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
谁能告诉我如何解决这个问题?
解决方法:
首先检查mysqli_select_db,如果它返回false,则创建数据库.
试试这样:
$mysql = mysqli_connect("example.com", "johndoe", "abc123") or die(mysqli_connect_error()); // replace with actual credidentials
if (!mysqli_select_db($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);
这是一个很好的答案:Php mysql create database if not exists
标签:create-table,php,mysql 来源: https://codeday.me/bug/20191003/1850463.html
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