我想显示我使用PHP从mySQL数据库中选择的数据。数据应以角度变量显示。 单击我安装的按钮时,我可以看到JSON。但是数据不会出现在角度变量中。所以我认为问题是php和angular之间的沟通,从php回到angular。
以下是我使用的文件。我必须在哪里做出改变才能得到我想要的东西?
谢谢!
<!DOCTYPE html>
<html ng-app="MyApp">
<head>
<title>nutrimeal</title>
<link rel="stylesheet" href="css/bootstrap.min.css">
<link rel="stylesheet" href="css/main.css">
</head>
<body>
<div class="container-fluid">
<div class="col-sm-10 content">
<h1>M E A L S</h1>
<form action="./php/meal_select.php" method="post">
<input class="btn btn-default" type="submit" value="display meals"/>
</form>
<div ng-controller = "MealsController as meals">
<p>The meals should be shown underneath</p>
<div ng-repeat="meal in meals">
{{ meal.name }}
{{ meal.daytime }}
</div>
</div>
</div>
</div>
<!-- Javascript references -->
<script src="./js/vendors/bootstrap.js"></script>
<script src="./js/vendors/angular.js"></script>
<script src="./js/app.js"></script>
</body>
</html>
var app = angular.module('MyApp', ['ngRoute'])
// MealsController
app.controller('MealsController', [ '$http', function($scope, $http) {
$scope.displayData = function(){
var meals = $http.get('./backend/meal_select.php')
// .then(function (response) {$scope.accounts = response.data.records;} );
.then(function (data) {
$scope.meals = data });
};
$scope.displayData()
}]);
<?php
// connection settings
$servername = "localhost";
$username = "root";
$password = "qwertz";
$dbname = "example_nutrition";
// create connection to db
$conn = new mysqli($servername, $username, $password, $dbname);
// check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// set SQL command
$sql = "SELECT meal_id, name, daytime FROM meal";
// query db
$result = $conn->query($sql);
// show results
if ($result->num_rows > 0) {
// output data of each row
$r=array()
while($row = $result->fetch_assoc()) {
$r=[] = $row;
//simple output
// echo " Account Name: " . $row["account_name"].
// " Balance: " . $row["initial_balance"].
// " Currency: " . $row["currency"]. "<br>";
// output as JSON
//$json_array = json_encode($row);
$json_array = json_encode($r);
header('Content-type: application/json');
echo $json_array;
}
}
// if nothing was found
else {
echo "0 results";
}
// close connection to db
$conn->close();
?>
单击按钮后显示{ “meal_id”: “1”, “名称”: “干酪汉堡包”, “白天”: “午餐”} {“meal_id”:“2”,“name”:“辣椒酱”,“白天”:“午餐”}
答案 0 :(得分:0)
修改强>
PHP
$row = array();
$sql = "SELECT * FROM ...";
$result = mysqli_query($conn,$sql);
if($result){
while($r = mysqli_fetch_assoc($result)) {
$rows[] = $r;
}
print json_encode($rows);
角
$scope.showData = function(){
$http.get('data.php').then(function(response){
$scope.data = response.data;
}
$scope.showData();
});
HTML
<ul ng-repeat ="d in data">
<li> {{d.name}}</li>
<li> {{d.age}}</li>
...
</ul>
我认为上面的代码很简单,你可以理解,但我会解释一下。在角度,我们有两个ng-repeat选项:
1.Object
2.Array
如果您在object
内使用ng-repeat
,则需要使用以下内容:
<ul ng-repeat="(key, value) in data">
<li> {{ key }} : {{ value }}</li>
</ul>
如果您在array
内使用ng-repeat
,那么您可以这样做:
<ul ng-repeat="d in data">
<li>{{ d.name}}</li>
<li>{{ d.age}}</li>
</ul>
现在了解这里发生了什么,你从服务器得到一个响应,它是一个对象,并试图像一个数组一样打印它,这就是它无法工作的原因(你忘了调用函数)。这就是为什么我说你在php中添加两行代码
$row = array();
$rows[] = $r;
如果你没有两行,你的json就是对象而你不能像数组那样打印它。但是如果你有两行,你的json就是数组,你可以像数组一样打印它