如何使用angularjs php和mysql插入记录？
如何使用angularjs php和mysql插入记录？
提问于 2017-01-27 05:20:06
，我使用带有mysql数据库的angularjs表单来用mysql存储和检索数据，但我无法使用angularjs完成crud操作。我刚接触过天使，请帮我做同样的事。下面是我的表单代码.
<form >
<div class="control-group">
<label class="control-label" >Referal code:</label>
<div class="controls">
<input required type='text' class='span2' ng-model='tempUserData.referal_code'>
</div>
</div>
<div class="control-group">
<label class="control-label">Name:</label>
<div class="controls">
<input required type="text" ng-model="tempUserData.emp_name" class="span2">
</div>
</div>
<div class="control-group">
<label class="control-label">Date:</label>
<div class="controls input-group">
<datepicker>
<input required type="text" ng-model="tempUserData.date" class="span2">
</datepicker>
</div>
</div>
<div class="control-group">
<label class="control-label" >Date Of Joining:</label>
<div class="controls">
<datepicker>
<input required type="text" ng-model="tempUserData.dateofjoining" class="span2">
</datepicker>
</div>
</div>
<div class="control-group">
<label class="control-label" >Admin Name:</label>
<div class="controls">
<input required type="text" ng-model="tempUserData.admin" class="span2">
</div>
</div>
<div class="control-group">
<label class="control-label" >Address:</label>
<div class="controls">
<input required type="text" ng-model="tempUserData.address" class="span2">
</div>
</div>
<div class="control-group ">
<label class="control-label">Company Name:</label>
<div class="controls">
<select ng-model="tempUserData.company_name" class="span2">
<option ng-repeat="x in names" value="{{x.company_name}}">{{x.company_name}}</option>
</select>
</div>
</div>
<div class="control-group">
<label class="control-label" >Title:</label>
<div class="controls">
<input required type="text" ng-model="tempUserData.title" class="span2">
</div>
</div>
<button class="btn btn-primary" type="submit" ng-hide="tempUserData.id" ng-click="addUser()"> Add Letter</button>此表单的脚本：
<script>
var app = angular.module('myApp', ['720kb.datepicker']);
app.controller('myCtrl', function($scope, $http)
{
$http.get("http://www.adhr.adnacgroup.com/ADHRM/companyJson.php")
.then(function(response)
{$scope.names = $scope.names = response.data.service;});
$scope.addTask = function (task) {
$http.post("addTask.php?task="+task).success(function(data){
getTask();
$scope.tempUserData.referal_code= "";
$scope.tempUserData.emp_name = "";
$scope.tempUserData.date = "";
$scope.tempUserData.dateofjoining = "";
$scope.tempUserData.admin = "";
$scope.tempUserData.address = "";
$scope.tempUserData.company_name = "";
$scope.tempUserData.title = "";
});
};
});
</script>插入查询：
if(isset($_GET['task'])){
$task1 = $_GET['task'];
$task2 = $_GET['task'];
$task3 = $_GET['task'];
$task4 = $_GET['task'];
$task5 = $_GET['task'];
$task6 = $_GET['task'];
$task7 = $_GET['task'];
$task8 = $_GET['task'];
$query9=mysql_query("INSERT INTO `offer_letter`(`org_id`, `com_id`, `id`, `admin_name`, `address`, `offer_referal_code`, `com_name`, `title`, `date`, `dateofjoining`) VALUES ('".$_SESSION[org_id]."','".$task1."','".$task2."','".$task3."','".$task4."','".$task5."','".$task6."','".$task7."','".$task8."','".$task9."')");
$result = mysql_affected_rows($query);
echo $json_response = json_encode($result);
}回答 1
Stack Overflow用户
发布于 2017-01-27 05:34:06
要执行CRUD操作，您必须将数据发送为post，如
$http({
url:'addTask.php',
method: 'POST',
params:task
})在php脚本中，可以获得如下参数
$task1 = $_POST['task'];页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/41887383
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