[英]How to call functions from PHP in javascript, Ajax not returning anything
我一直在尝试将数据库中的数据(在PHP文件中)转移到我的javascript程序中。 我正在尝试将查询结果传递到Javascript文件,以便可以从这些结果中生成图形。
我已经尝试过使用Ajax,但它没有响应。 我不确定我要去哪里。 我的PHP文件称为MySQLDau.php。 任何帮助深表感谢! 谢谢!
PHP代码:
<?php
header("Access-Control-Allow-Origin: *");
//Class for holding queries
class MySQLDao
{
var $dbhost = null;
var $dbuser = null;
var $dbpass = null;
var $mysqli = null;
var $dbname = null;
var $result = null;
//constructor
function __construct()
{
$this->dbhost = Conn::$dbhost;
$this->dbuser = Conn::$dbuser;
$this->dbpass = Conn::$dbpass;
$this->dbname = Conn::$dbname;
}
//Attempt a connection to the database
public function openConnection()
{
//Try and connect to the database
$this->mysqli = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
//If the connection threw an error, report it
if (mysqli_connect_errno())
{
return false;
}
else
{
return true;
}
}
//Get method for retrieving the database conection
public function getConnection()
{
return $this->mysqli;
}
//Close the connection to the database
public function closeConnection()
{
//If there is a connection to the database then close it
if ($this->mysqli != null)
$this->mysqli->close();
}
//-----------------------------------QUERY METHODS-------------------------------------
public function generateRoomID()
{
$sql = "INSERT INTO room (room_id) VALUES (null);";
$result = $this->mysqli->query($sql);
if ($result == true)
{
$toReturn["status"] = true;
$toReturn["roomID"] = $this->mysqli->insert_id;
return $toReturn;
}
else
{
$toReturn["status"] = false;
$toReturn["message"] = mysql_error($this->mysqli);
return $toReturn;
}
}
public function saveRoom($data)
{
$roomID = $data[0];
$roomDescription = $data[1];
$columns = $data[2];
$rows = $data[3];
$this->mysqli->autocommit(FALSE);
$this->mysqli->query("UPDATE room SET room_description='".$roomDescription."' WHERE room_id='".$roomID."';");
for ($i = 0; $i<count($columns); $i++)
{
for ($j = 1; $j<=$rows[$i]; $j++)
{
$currentLabel = "{$columns[$i]}{$j}";
$this->mysqli->query("INSERT INTO shelf (shelf_label) VALUES ('".$currentLabel."');");
$shelfID = $this->mysqli->insert_id;
$this->mysqli->query("INSERT INTO room_shelf (room_id, shelf_id) VALUES ('".$roomID."','".$shelfID."');");
}
}
if ($this->mysqli->commit())
{
$toReturn["status"] = true;
$toReturn["message"] = "Room Created";
return $toReturn;
}
else
{
$this->mysqli->rollback();
$toReturn["status"] = false;
$toReturn["message"] = "SQL Error";
return $toReturn;
}
}
public function updateShelf($data)
{
$shelfID = $data[0];
$itemName = $data[1];
}
public function getRoomDetails($data)
{
$roomID = $data[0];
$sql = "SELECT room.room_description, shelf.shelf_label, shelf.shelf_id FROM room INNER JOIN room_shelf ON room.room_id=room_shelf.room_id INNER JOIN shelf ON shelf.shelf_id=room_shelf.shelf_id WHERE room.room_id='".$roomID."';";
$result = $this->mysqli->query($sql);
if (mysqli_num_rows($result) > 0)
{
$toReturn["status"] = true;
$toReturn["message"] = "Room Found";
$toReturn["room_description"] = $row['room_description'];
$shelves = [];
foreach ($result as $row)
{
$currentShelf["shelf_label"] = $row['shelf_label'];
$currentShelf["shelf_id"] = $row['shelf_id'];
array_push($shelves, $currentShelf);
}
$toReturn["shelves"] = $shelves;
return $toReturn;
}
else
{
$toReturn["status"] = false;
$toReturn["title"] = "Error";
$toReturn["message"] = "Room Not Found";
return $toReturn;
}
}
echo "Hello World!";
public function getResults($data){
$sql = "SELECT * FROM room";
$result = $this->mysqli->query($sql);
if (mysql_num_rows($result) == 1) {
$obj = mysql_fetch_object($result, 'obResults');
}
echo json_encode($obj);
}
}
?>
我的Javascript代码的一部分:
function callPHP() {
$.ajax ({
type: "GET",
url: "MySQLDao.php",
data: { action : 'getResults' },
success: function(output) {
alert(output);
}
});
}
您没有任何代码可以检查GET参数,实例化该类并调用该函数。 您的AJAX请求仅创建该类,但实际上不执行任何操作。
您可以做的是,在类定义的结束}之后是这样的:
$daoObj = new MySQLDao();
$daoObj->getResults(); // this function takes a $data parameter, but doesn't use it
这将执行您的getResults函数,但当前忽略了AJAX请求的“数据:{action:'getResults'}”部分。 使基本功能正常工作(即返回有用的东西),然后对其进行改进以根据AJAX请求执行不同的功能。
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