检测到拦截器损坏:拒绝时未提供配置对象:当我调用post方法时,我会收到这个错误。
我已经给出了服务、控制器和PHP代码
服务
angular.module('sbAdminApp')
.factory('Branch', function($resource){
return $resource('api/branchdetails/:branch_id',{branch_id:'@_branch_id'},{
update: {
method: 'PUT'
}
});
})
.service('popupService',function($window){
this.showPopup=function(message){
return $window.confirm(message);
}
});
控制器
angular.module('sbAdminApp')
.controller('BranchDetailsController', function($scope,$state,$stateParams,$window,Branch){
$scope.branch = new Branch();
$scope.addBranch=function(){
$scope.branch.$save(function(){
$state.go('branchdetails');
});
}
});
PHP代码
<?php
require_once('Slim/Slim.php');
require_once('dbconnection.php');
$app = new Slim();
$app->post('/branchdetails','addBranch');
$app->run();
function addBranch() {
$request = Slim::getInstance()->request();
$branch = json_decode($request->getBody());
$sql = "INSERT INTO branch(branch_name, branch_address, branch_phno, branch_mobileno, branch_contactperson, branch_createdate, branch_modifieddate) VALUES (:branch_name,:branch_address, :branch_phno, :branch_mobileno, :branch_contactperson, :branch_createdate, :branch_modifieddate)";
try {
$db = getConnection();
$stmt = $db->prepare($sql);
$stmt->bindParam("branch_name", $branch->branch_name);
$stmt->bindParam("branch_address", $branch->branch_address);
$stmt->bindParam("branch_phno", $branch->branch_phno);
$stmt->bindParam("branch_mobileno", $branch->branch_mobileno);
$stmt->bindParam("branch_contactperson", $branch->branch_contactperson);
$stmt->bindParam("branch_createdate", $branch->branch_createdate);
$stmt->bindParam("branch_modifieddate", $branch->branch_modifieddate);
$stmt->execute();
$branch->branch_id = $db->lastInsertId();
$db = null;
echo json_encode($branch);
} catch(PDOException $e) {
echo '{"error":{"text":'. $e->getMessage() .'}}';
}
}
?>
2条答案
按热度按时间5vf7fwbs1#
在你的代码中,你有一个
interceptor
for$httpProvider
,它没有一个正确的响应错误部分,类似这样:4zcjmb1e2#
当API未返回有效的JSON时,也可能发生此类错误。
AngularJS transform Response将尝试将响应解析为JSON,如果Content-Type头是
application/json
,或者响应看起来像一个有效的JSON字符串化对象或数组。首先,始终尝试从后端接收有效的JSON对象。第二,也处理回调函数中的错误。
这是很好的练习。如果您仍然没有收到有效的JSON,那么在错误回调中处理它也可以做到这一点:https://code.angularjs.org/1.7.6/docs/API/ng/service/$http#default-transformations