我正在尝试在数据库中插入数据，但它根本不起作用！我按照教程进行了操作，但它不起作用。我尝试了很多但直到现在都没有成功。
这是我的html文件
<!DOCTYPE html>
<html>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"> </script>
<body>
<div ng-app="myApp" ng-controller="myCtrl">
<form>
Name:-<input type="text" ng-model="bname" />
Phone:-<input type="text" ng-model="bphone" />
<input type="button" value="Submit" ng-click="insertData()" />
</form>
</div>
<script>
var app = angular.module('myApp',[]);
app.controller('myCtrl',function($scope,$http){
$scope.insertData=function(){
$http.post("insert.php", {
'bname':$scope.bname,
'bphone':$scope.bphone})
.success(function(data,status,headers,config){
console.log("Data Inserted Successfully");
});
}
});
</script>
</body>
</html>
我在数据库中插入数据的insert.php
<?php
$data = json_decode(file_get_contents("php://input"));
$bname = mysql_real_escape_string($data->bname);
$bauthor = mysql_real_escape_string($data->bphone);
mysql_connect("localhost", "root", "");
mysql_select_db("angular");
mysql_query("INSERT INTO ang('name', 'phone') VALUES('".$bname."','".$bauthor."')");
?>
更新
<?php
$data = json_decode(file_get_contents("php://input"));
$con = new mysqli('localhost', 'root', '', 'angularjs');
$bname = mysqli_real_escape_string($con, $data->bname);
$bphone = mysqli_real_escape_string($con, $data->bphone);
if($con->connect_errno > 0){
die('Unable to connect to database [' . $con->connect_error . ']');
}
mysqli_query($con,"INSERT INTO jstable (name, phone)VALUES ('".$bname."', '".$bphone."')");
mysqli_close($con);
?>
更新我的代码后，我收到了这个错误： -
注意：尝试在第4行的E：\ xampp \ htdocs \ insert.php中获取非对象的属性
注意：尝试在第5行的E：\ xampp \ htdocs \ insert.php中获取非对象的属性
我搜索了这个错误，发现了很多结果，但没有一个对我的情况有帮助！
答案 0 :(得分：3)
我以下列方式修改了您的代码
HTML CODE：
<!DOCTYPE html>
<html>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"> </script>
<body>
<div ng-app="myApp" ng-controller="myCtrl">
<form>
Name:-<input type="text" ng-model="bname" />
Phone:-<input type="text" ng-model="bphone" />
<input type="button" value="Submit" ng-click="insertData()" />
</form>
</div>
<script>
var app = angular.module('myApp',[]);
app.controller('myCtrl',function($scope,$http){
$scope.insertData=function(){
$http.post("test.php", {
'bname':$scope.bname,
'bphone':$scope.bphone
}).then(function(response){
console.log("Data Inserted Successfully");
},function(error){
alert("Sorry! Data Couldn't be inserted!");
console.error(error);
});
}
});
</script>
</body>
</html>
和test.php
<?php
$data = json_decode(file_get_contents("php://input"));
// $bname = mysql_real_escape_string($data->bname);
// $bauthor = mysql_real_escape_string($data->bphone);
$bname = $data->bname;
$bphone = $data->bphone;
mysql_connect("localhost", "root", "password");
mysql_select_db("test");
mysql_query("INSERT INTO `ang`(`name`,`phone`)
VALUES ('".$bname."','".$bphone."') ") or die(mysql_error());
echo $bname." ".$bphone;
?>
效果很好..
答案 1 :(得分：1)
**insert.php**
<!DOCTYPE html>
<html>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"> </script>
<body>
<div ng-app="myApp" ng-controller="myCtrl">
<form>
Name:-<input type="text" ng-model="uname" />
Phone:-<input type="text" ng-model="uphone" />
<input type="button" value="Submit" ng-click="insertData()" />
</form>
</div>
<script>
var app = angular.module('myApp',[]);
app.controller('myCtrl',function($scope,$http){
$scope.insertData=function(){
$http.post("testinsert.php", {
'uname':$scope.uname,
'uphone': $scope.uphone
})
.success(function (data, status, headers, config) {
console.log("Inserted Successfully!");
});
}
});
</script>
</body>
</html>
**testinsert.php**
<?php
$data = json_decode(file_get_contents("php://input"));
$uname = $data->uname;
$uphone = $data->uphone;
$con = mysql_connect("localhost","root","");
mysql_select_db("angular");
$sql = "insert into Table Name(user_name,user_phone) values('$uname','$uphone')";
$result = mysql_query($sql);
?>
答案 2 :(得分：0)
在打开数据库连接之前，您使用false emptying your values（将它们设置为mysql_real_escape_string() ...）。您应该先打开连接。
但最好的解决方案是切换到PDO或mysqli并使用预准备语句。然后你就不必再逃避了。
答案 3 :(得分：0)
像这样更改你的代码，它可能会解决你的问题。
mysql_connect("localhost", "root", "");
mysql_select_db("angular");
$data = json_decode(file_get_contents("php://input"),true);
$bname = mysql_real_escape_string($data['bname']);
$bauthor = mysql_real_escape_string($data['bphone']);
答案 4 :(得分：0)
您可以使用图片插入数据并验证输入。
<form ng-controller="insert_Ctrl" method="post" action="" name="myForm" enctype="multipart/form-data" novalidate>
<div>
<p><input type="text" class="form-control" placeholder="Name" name="name" ng-model="name" required>
<span style="color:red" ng-show="myForm.name.$invalid&&myForm.name.$touched">Enter Name</span>
</p>
</div>
<div>
<p><input type="file" ng-model="myFile" class="form-control" onchange="angular.element(this).scope().uploadedFile(this)">
<span style="color:red" ng-show="(myForm.myFile.$error.required&&myForm.myFile.$touched)">Select Picture</span>
</p>
</div>
<div>
<input type="button" name="submit2" ng-click="uploadFile()" class="btn-primary" ng-disabled="myForm.name.$invalid || myForm.myFile.$invalid" value="Insert">
</div>
</form>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script>
<script src="insert.js"></script>
<强> insert.js
var app = angular.module('myApp',[]);
app.service('uploadFile', ['$http','$window', function ($http,$window) {
this.uploadFiletoServer = function(file,info,uploadUrl){
var fd = new FormData();
fd.append('file', file);
fd.append('data', angular.toJson(info));
$http.post(uploadUrl, fd, {
transformRequest: angular.identity,
headers: {'Content-Type': undefined}
})
.success(function(data){
alert("insert successfull");
$window.location.href = ' ';//your window location
})
.error(function(){
alert("Error");
});
}
}]);
app.controller('insert_Ctrl', ['$scope', 'uploadFile', function($scope, uploadFile){
$scope.uploadFile = function() {
$scope.myFile = $scope.files[0];
var file = $scope.myFile;
var info = {
'name':$scope.name,
};
var url = "save_data.php";
uploadFile.uploadFiletoServer(file,info,url);
};
$scope.uploadedFile = function(element) {
var reader = new FileReader();
reader.onload = function(event) {
$scope.$apply(function($scope) {
$scope.files = element.files;
$scope.src = event.target.result
});
}
reader.readAsDataURL(element.files[0]);
}
}]);
<强> save_data.php
<?php
require "dbconnection.php";
$datas=$_POST['data'];
$myArray = json_decode($datas, true);
$name=$myArray['name'];
$ext = pathinfo($_FILES['file']['name'],PATHINFO_EXTENSION);
$image = time().'.'.$ext;
move_uploaded_file($_FILES["file"]["tmp_name"],"upload/".$image);
$query="insert into test_table values ('null','$name','$image')";
mysqli_query($con,$query);
?>