我在PHP和MySQL方面犯了什么错误?
我正在开发一个小的浏览器游戏,但是当我连接到MySQLi数据库时,它就不起作用了 在else闭包中,它应该写$name,但它没有我在PHP和MySQL方面犯了什么错误?,php,mysql,database,conditional-statements,Php,Mysql,Database,Conditional Statements,我正在开发一个小的浏览器游戏,但是当我连接到MySQLi数据库时,它就不起作用了 在else闭包中,它应该写$name,但它没有 if ($conn->connect_error){ die("Connection failed: ".$conn->connect_error); } else{ //IF CONNECTION IS GOOD, GET DATA FROM DATABASE $query = "SELECT name, separator,
if ($conn->connect_error){
die("Connection failed: ".$conn->connect_error);
}
else{
//IF CONNECTION IS GOOD, GET DATA FROM DATABASE
$query = "SELECT name, separator, description, maintenance FROM configuration";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$name = $row['name'];
//this ↓↓↓
echo $name;
}
尝试使用backticks:
$query = "SELECT name, `separator`, description, maintenance FROM configuration";
使用保留字时,请使用回勾(``)
从
对,
separator
是一个MySQL保留字
$query = "SELECT name, separator, description, maintenance FROM configuration";
$query = "SELECT name, `separator`, description, maintenance FROM configuration";