Question

我在PHP中编写了一个帮助器类。它有一个构造函数，用于初始化将在方法中填充的数组。有两种方法，一种是填充数组，另一种是返回数组。问题是当我在另一个网页中创建对象并调用方法时，get方法返回错误。而且我得到的错误。请帮助

注意：未定义变量：“C /....."

中的startlatitude

<?php 

session_start();
class TrafficData
{
    function __construct()
    {
        $startlatitude = array();
        $startlongitude = array();
        $endlatitude = array();
        $endlongitude = array();
        $delay = array();
        $accidenttype = array();        
     }
    function getdatabyseverity($dayofweek, $zipcode)
    {
        $host="localhost"; // Host name
        $username="kush"; // Mysql username
        $password="password"; // Mysql password
        $db_name="trafficprediction"; // Database name
        $tbl_name="livecongestion"; // Table name

       // Create connection
       $conn = new mysqli($host, $username, $password, $db_name);
       // Check connection
       if ($conn->connect_error)
       {
           die("Connection failed: " . $conn->connect_error);
        }


       $sql = "SELECT `Street`, `CongestionType` , `StartLatitude`, `StartLongitude`, `EndLatitude`, `EndLongitude`, `Delay` FROM `" . $tbl_name . "` WHERE `ZipCode`=\"" .$zipcode . "\" and `Day`=\"" .$dayofweek."\"";
       $result = $conn->query($sql);

        if ($result->num_rows > 0)
        {
            // output data of each row
            while($row = $result->fetch_assoc())
            {
                $startlatitude[] = $row["StartLatitude"];
                $startlongitude[] = $row["StartLongitude"];
                $endlatitude[] = $row["EndLatitude"];
                $endlongitude[] = $row["EndLongitude"];
                $delay[] = $row["Delay"];
                $accidenttype[] = $row["CongestionType"];
            }
        } else
        {
                echo "0 results";
         }
    }`

    function getarray($text)
    {
        if($text == "START_LAT")
        {
            return $startlatitude;
         }
     }

现在，下面的代码就是我用来创建对象的代码。

<?php 
 include('DataClass.php');
function getdata()
{
    $zipcode=$_POST['zipcode'];
    $dayofweek=$_POST['day'];
    $severity=$_POST['severity'];

    $dataobject = new TrafficData();
    $dataobject->getdatabyseverity($dayofweek, $zipcode);

    $startlatitude = array();
    $startlatitude = $dataobject->getarray("START_LAT");
}

Answer 1

您创建object的代码很好。但是你没有返回类变量，而是函数getarray()尝试返回在函数范围内不可用的局部变量$startlatitude。

我猜你忘了$this来访问类变量了。 OOPS概念！

<?php 

session_start();
class TrafficData
{
    public var $startlatitude;

    function __construct()
    {
        $this->startlatitude = array();
        $startlongitude = array();
        $endlatitude = array();
        $endlongitude = array();
        $delay = array();
        $accidenttype = array();        
     }
    function getdatabyseverity($dayofweek, $zipcode)
    {
        $host="localhost"; // Host name
        $username="kush"; // Mysql username
        $password="password"; // Mysql password
        $db_name="trafficprediction"; // Database name
        $tbl_name="livecongestion"; // Table name

       // Create connection
       $conn = new mysqli($host, $username, $password, $db_name);
       // Check connection
       if ($conn->connect_error)
       {
           die("Connection failed: " . $conn->connect_error);
        }


       $sql = "SELECT `Street`, `CongestionType` , `StartLatitude`, `StartLongitude`, `EndLatitude`, `EndLongitude`, `Delay` FROM `" . $tbl_name . "` WHERE `ZipCode`=\"" .$zipcode . "\" and `Day`=\"" .$dayofweek."\"";
       $result = $conn->query($sql);

        if ($result->num_rows > 0)
        {
            // output data of each row
            while($row = $result->fetch_assoc())
            {
                $this->startlatitude[] = $row["StartLatitude"];
                $startlongitude[] = $row["StartLongitude"];
                $endlatitude[] = $row["EndLatitude"];
                $endlongitude[] = $row["EndLongitude"];
                $delay[] = $row["Delay"];
                $accidenttype[] = $row["CongestionType"];
            }
        } else
        {
                echo "0 results";
         }
    }`

    function getarray($text)
    {
        if($text == "START_LAT")
        {
            return $this->startlatitude;
         }
     }

在PHP中获取返回方法的“未定义变量”

1 个答案: