用于从数据库中查找通用值的 PHP 函数

Question

我将如何创建（对于 mySQL 数据库）具有两个参数的 PHP 函数：第一个是表中的列名，第二个是用户名？该函数应返回给定用户名在该列中的表值。

本质上，我需要类似下面的代码（用于 SQLite 数据库和 Java），但在 PHP 中。

public String getInformation(String information, String username) {

        SQLiteDatabase db = getReadableDatabase();

        Cursor cur = db.rawQuery("SELECT " + information + " FROM registeruser WHERE username=" + "'" + username + "'",null);

        String result = "";

        if (cur.moveToFirst()) {

            int resultColumn = cur.getColumnIndex(information);

            result = cur.getString(resultColumn);

        }

        return result;

    }

Answer 1

我会将您的代码转换为 PHP：脚步：连接到您的数据库。使用 prepare 方法准备您的陈述。将参数绑定到准备好的语句 $stmt。执行你的语句，获取结果集，提取值，然后返回它。<?phpfunction getInformation($field, $value){&nbsp; &nbsp; $conn = new mysqli("localhost", "user", "password", "database");&nbsp; &nbsp; $query ="select $field from registeruser where username=?";&nbsp; &nbsp; $stmt = $conn->prepare($query);&nbsp; &nbsp; $stmt->bind_param("s",$value);&nbsp; &nbsp; $stmt->execute();&nbsp; &nbsp; $res = $stmt->get_result();&nbsp; &nbsp; $row = $res->fetch_assoc();&nbsp; &nbsp; &nbsp;return $row[$field];}?>

Answer 2

连接到数据库：$link = mysqli_connect("yourip", "dbusername", "dbpassword", "dbname");//check connectionif($link === false){&nbsp; &nbsp; die("ERROR: Could not connect. ".mysqli_connect_error());}创建请求：$sql = "SELECT $columnsyouwant FROM $tablename WHERE username=$username";$result = $link->query($sql);浏览每个结果并做任何你需要的事情：if($result->num_rows>0){&nbsp; &nbsp; while($row = $result->fetch_assoc()){&nbsp; &nbsp; &nbsp; &nbsp; x=$row["columnname"] //or do whatever you need with this value&nbsp; &nbsp; }}