問題描述

我正在將所有使用 PHP MySQL 的查詢更改為 MySQLi.

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

我用連接設置創建了一個名為 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 . ']');
}
?>

我將文件包含在:

require_once "/location/db.php";    

之后我使用:

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

它回顯已連接，所以我認為我的連接良好.

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

我有 3 個 PHP 變量要插入到我的數據庫表代碼中

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

我首先回顯變量，所以我確定它們有內容.

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

在我驗證我的連接沒問題后(返回 Connected)并回顯我想要查詢的變量的內容:

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 來顯示查詢.

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

如果我直接在 SQL 中復制查詢，它會像它應該的那樣工作.

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

這是我第一次使用 MySQLi，所以我可能犯了一個非常愚蠢的錯誤，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.

推薦答案

打開連接時可以傳遞數據庫名稱作為第四個參數:

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

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

另外，你的轉義字符是錯誤的.不要在表名周圍使用單引號，而是使用反引號運算符

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')"; 

這篇關于沒有選擇數據庫錯誤信息的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！