我有关于PHP的问题 - AngularJs,所以我有简单的PHP脚本
<?php
require_once '../../dbConnect.php';
$driverId = $_POST['driverId'];
if (isset($_POST['driverId'])) {
$sql = "delete from drivers where driver_id='$driverId'";
if ($mysqli->query($sql) === TRUE) {
echo mysqli_insert_id($mysqli);
} else {
echo "Error updating record: " . $mysqli->error;
}
$mysqli->close();
}
?>
目前我将数据传递给这样的脚本
return $http({
method: 'POST',
url: '/api/drivers/deleteDriver.php',
data: $.param(driverObject),
headers: {
'Content-Type': 'application/x-www-form-urlencoded'
}
});
我不喜欢那个代码,在其他Java项目上我用这样的angularjs $ resource服务将params发送到终点
var deleteDriverResouce = $resource('/api/drivers/deleteDriver.php');
function deleteDriver() {
deleteDriverResouce.save(driverObject);
}
正如您所看到的那样代码更清晰,更易于使用,我想知道我可以使用$ resource服务将对象传递给php脚本吗?
答案 0 :(得分:0)
所以我找到了解决方案,我将在这里分享,所以也许有人需要它。为了使用AngularJs $ resource服务,您只需要在PHP脚本中进行小的更改,只需添加$object = json_decode(file_get_contents("php://input"), true);
就可以访问通过$ resource发送的对象。这是一个有效的PHP脚本示例。
<?php
require_once '../dbConnect.php';
session_start();
$object = json_decode(file_get_contents("php://input"), true);
if (isset($object['email']) && isset($object['password'])) {
$email = $object['email'];
$password = $object['password'];
$query="select * from members where email='$email'";
$result = $mysqli->query($query) or die($mysqli->error.__LINE__);
$row = mysqli_fetch_assoc($result);
if($row) {
if (password_verify($object['password'], $row['password'])) {
$_SESSION["id"] = $row['id'];
echo 'Login Success!';
} else {
session_destroy();
var_dump(http_response_code(400));
}
} else {
session_destroy();
var_dump(http_response_code(406));
}
$mysqli->close();
} else {
session_destroy();
var_dump(http_response_code(400));
}
?>
在UI上我有这个简单而简单的代码:
var userLoginObject = {
email: 'login@email.com',
password: 'password123'
};
var authenticationResource = $resource('/api/authentication/authenticate.php');
function logIn() {
authenticationResource.save(userLoginObject );
}
这比使用丑陋
更好更清洁return $http({
method: 'POST',
url: '/api/drivers/authenticate.php',
data: $.param(userLoginObject),
headers: {
'Content-Type': 'application/x-www-form-urlencoded'
}
});