Question

我有这个php函数，它检查数据库中的表并检查用户是否在字段中有特定的数字。此用户可能有几行，其中该字段具有该数字。

这个问题是它会多次运行if语句，触发生成的函数多次运行。我怎么能阻止这个，只是让函数说出＆＃34;好的找到了al1和运行函数的匹配，转移到al2＆＃34;

function countNewBadges() {
    require "connect.php"; 

    $count = mysqli_query($connection,"SELECT users.studentid, al1, al2, al3 FROM userbadges ub INNER JOIN users ON users.id = ub.user_id WHERE studentid = '".$_SESSION["studentid"]."'")  or die(mysqli_error($connection));

    while ($data = mysqli_fetch_array($count)) { 

    if ($data['al1'] == 1)
    { 
        unlockedBadges();
    }
    else if ($data['al2'] == 1)
    {
        echo "No New Badges";
    }
    else if ($data['al3'] == 1)
    {
        echo "No New Badges";
    }
}
}

Answer 1

我怎么能让这个函数说“好的找到了al1和运行函数的匹配，转移到al2”

只需使用标志变量：

$flag = false;
while ($data = mysqli_fetch_array($count)) { 

    if ($data['al1'] == 1 && $flag === false)
    { 
        unlockedBadges();
        $flag = true;
    }
    else if ($data['al2'] == 1)
    {
        echo "No New Badges";
    }
    else if ($data['al3'] == 1)
    {
        echo "No New Badges";
    }
}

Answer 2

虽然user4035的回答会阻止多个数据条目真正只在你需要搜索后运行一个函数，然后按照标准使用（按照说法）行动。

function countNewBadges() {
    require "connect.php"; 

    $count = mysqli_query($connection,"SELECT users.studentid, al1, al2, al3 FROM userbadges ub INNER JOIN users ON users.id = ub.user_id WHERE studentid = '".$_SESSION["studentid"]."'")  or die(mysqli_error($connection));

    $foundit=false;

    while ($data = mysqli_fetch_array($count)) { 

        if ($data['al1'] == 1)
        { 
            $foundit=true;
        }
        else if ($data['al2'] == 1)
        {
            echo "No New Badges";
        }
        else if ($data['al3'] == 1)
        {
            echo "No New Badges";
        }
    }
    if($foundit){
        unlockedBadges();
    }
}

只运行一次php函数

2 个答案: