没有选择数据库错误信息 [英] No database selected error message

问题描述

我正在将所有使用 PHP MySQL 的查询更改为 MySQLi.

I am changing all my queries that are using PHP MySQL to MySQLi.

我用连接设置创建了一个名为 db.php 的文件.

I have made a file called db.php with the connection settings.

文件包含

<?php
$db = new mysqli('localhost','mysqlusername','mysqlpassword');
echo "<h1>Success database connection</h1>";
if($db->connect_errno > 0)
{
die('No connection [' . $db->connect_error . ']');
}
?>

我将文件包含在:

require_once "/location/db.php";    

之后我使用:

 if($db->connect_error)
 {
   echo "Not connected, error: ".$db->connect_error;
 }  
 else
 {
   echo "Connected.";
 }

它回显已连接，所以我认为我的连接良好.

It echo's Connected so I assume my connection is good.

我有 3 个 PHP 变量要插入到我的数据库表代码中

I have 3 PHP variables which I want to insert in my database table Code

我首先回显变量，所以我确定它们有内容.

I first echo the variables so I am sure they have content.

在我验证我的连接没问题后(返回 Connected)并回显我想要查询的变量的内容:

After I validated my connection is alright (returned Connected) and echoing the content of the variables I want to do the query with:

$sql = "INSERT INTO 'Code' (`Name`, `Code`, `Admin`)
VALUES ('$name', '$code', '$admin')"; 
echo $sql;//show query
// Performs the $sql query on the server to insert the values
if ($db->query($sql) === TRUE)
{
    echo 'User Created.';
}
else 
{
    echo 'Errorcreating : '. $db->error;
}

我收到消息 Errorcreating : No database selected

I get the message Errorcreating : No database selected

我有 echo $sql 来显示查询.

I have the echo $sql to show me the query.

如果我直接在 SQL 中复制查询，它会像它应该的那样工作.

If I copy the query directly in SQL it works like it should.

这是我第一次使用 MySQLi，所以我可能犯了一个非常愚蠢的错误，但我找不到.

This is my first time on MySQLi so it's possible I made a very dumb mistake but I can't find it.

推荐答案

打开连接时可以传递数据库名称作为第四个参数:

When opening the connection you can pass the database name as a 4th parameter:

$db = new mysqli('localhost','mysqlusername','mysqlpassword','database');

另外，你的转义字符是错误的.不要在表名周围使用单引号，而是使用反引号运算符

Also, your escape character is wrong. Don't use single quotes around table names, use backtick operator instead

$sql = "INSERT INTO `Code` (`Name`, `Code`, `Admin`)VALUES ('$name', '$code', '$admin')"; 

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