从另一个php文件调用mysqli函数

Question

我正在导入事务数据并将其保存到数据库中。有些事务数据缺少字段，因此我使用单独文件(file2.php)中的以下函数手动更新它们：

function categoryRules() {
    //rules for transactions
    $sql = "UPDATE Transactions 
                SET cat1 = 'Shops', cat2 = 'Supermarkets and Groceries'
            WHERE name LIKE '%shipt%'";
    if (mysqli_query($conn, $sql)) {
        echo "SHIPT UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }   
    $sql = "UPDATE Transactions 
                SET cat1 = 'Shops', cat2 = 'Food and Beverage Store', cat3 = 'Beer, Wine and Spirits'
            WHERE name LIKE '%liquor%'";
    if (mysqli_query($conn, $sql)) {
        echo "LIQUOR UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
    $sql = "UPDATE Transactions 
                SET cat1 = 'Service', cat2 = 'Telecommunication Services'
            WHERE name LIKE '%vzwrls%'";
    if (mysqli_query($conn, $sql)) {
        echo "VERIZON UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }

    $sql = "UPDATE Transactions 
                SET cat1 = 'Recreation', cat2 = 'Gyms and Fitness Centers'
            WHERE name LIKE '%classpass%'";
    if (mysqli_query($conn, $sql)) {
        echo "VERIZON UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

为了清理file1.php中的代码，我使用以下方法调用文件1.php上的file2函数：

$conn = mysqli_connect($host, $user, $pass, $database);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
include 'categoryRules.php';
categoryRules();
mysqli_close($conn);

当我这样做时，我会得到以下的回应：

Error：更新事务集cat1 =‘cat1’，cat2 =‘超市和杂货’，其中的名称为'%shipt%‘ 错误:更新交易设置为cat1 =‘商店’，cat2 =‘超市和杂货’，其中的名称为'%shipt%‘ 错误:更新事务设置为cat1 =‘营业’，cat2 =‘食品饮料商店’，cat3 =‘啤酒，葡萄酒和烈酒’，其中的名称如‘%白酒%’ 错误:更新事务设置为cat1 =‘服务’，cat2 =‘电信服务’，其中名为'%vzwrls%‘ 错误:更新事务设置cat1 = 'Recreation'，cat2 =‘健身房和健身中心’，其中名为'%classpass%‘

函数似乎在文件2.php中运行，然后才进入file1.php，这是我打开、访问和关闭数据库链接的地方。有办法将它们导入文件1吗？除了创建文件2.php之外，还有更好的方法来清理文件1.php的这部分代码吗？

Answer 1

啊，当然，我应该早点想到的。

在函数之外建立一个连接，因此$conn在函数中不可见，只需将$conn作为参数传递给函数，所有这些都会神奇地工作。这是一个简单的范围问题！

categoryRules.php

function categoryRules($conn) {    // <-- add parameter
    //rules for transactions
    $sql = "UPDATE Transactions 
                SET cat1 = 'Shops', cat2 = 'Supermarkets and Groceries'
            WHERE name LIKE '%shipt%'";
    if (mysqli_query($conn, $sql)) {
        echo "SHIPT UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }   
    $sql = "UPDATE Transactions 
                SET cat1 = 'Shops', cat2 = 'Food and Beverage Store', cat3 = 'Beer, Wine and Spirits'
            WHERE name LIKE '%liquor%'";
    if (mysqli_query($conn, $sql)) {
        echo "LIQUOR UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
    $sql = "UPDATE Transactions 
                SET cat1 = 'Service', cat2 = 'Telecommunication Services'
            WHERE name LIKE '%vzwrls%'";
    if (mysqli_query($conn, $sql)) {
        echo "VERIZON UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }

    $sql = "UPDATE Transactions 
                SET cat1 = 'Recreation', cat2 = 'Gyms and Fitness Centers'
            WHERE name LIKE '%classpass%'";
    if (mysqli_query($conn, $sql)) {
        echo "VERIZON UPDATE PERFORMED";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

现在，主要脚本应该是

$conn = mysqli_connect($host, $user, $pass, $database);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
include 'categoryRules.php';
// add $conn parameter to the call
categoryRules($conn);           
mysqli_close($conn);