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php - 查询数据库失败未选择数据库

问题描述

我收到错误没有选择数据库任何人都可以更正代码。

我正在尝试学习 php 和 mysql。所以我尝试使用
phpmyadmin 创建一个数据库并将其与我的 php 连接。这是一个简单的示例，我尝试查看数据库是否正常工作。

    <?php
    //Get values passe from form in donateform.html.
        $link;
        $Name = $_POST['Name'];
        $Mobile = $_POST['Mobile'];
        $Email = $_POST['Email'];
        $Donating =isset($_POST['Donating']);
        $Address = isset($_POST['Address']);
    //To prevent mysql injection
        $Name = stripcslashes($Name);
        $Mobile = stripcslashes($Mobile);
        $Email = stripcslashes($Email);
        $Donating = stripcslashes($Donating);
        $Address = stripcslashes($Address);

    //connect to the server and select  database.
        $link=mysqli_connect("localhost", "gooddeeds", "");
        mysqli_select_db($link,"donaters");

        $Name = mysqli_real_escape_string($link,$Name);
        $Mobile = mysqli_real_escape_string($link,$Mobile);
        $Email = mysqli_real_escape_string($link,$Email);
        $Donating = mysqli_real_escape_string($link,$Donating);
        $Address = mysqli_real_escape_string($link,$Address);

//Query the database for user
$result =mysqli_query($link,"INSERT INTO donaters (Name, Mobile, Email, Donating, Address) VALUES ('$Name', '$Mobile', '$Email', '$Donating', '$Address')")
    or die("Failed to query daabase".mysqli_error($link));

if(mysqli_query($result)){
    echo "Records inserted successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. ".mysqli_error($link);
}
?>

我明白了

Database query failed::: No database selected

这意味着这部分代码

//connect to the server and select  database.
    $link=mysqli_connect("localhost", "gooddeeds", "");
    mysqli_select_db($link,"donaters");

不起作用（我在每个 if 中放置了不同数量的“：”。任何帮助将不胜感激！谢谢！

标签： phpmysql

解决方案

这是 php 文档中的第一个示例。尝试条件检查连接。

$link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");

if (!$link) {  
    echo "Error: Unable to connect to MySQL." . PHP_EOL;  
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}

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