产品已成功插入但是 无法显示数据库中的产品 它没有显示任何错误...
这是我的main.html页面 首先在angularjs函数中有一些问题然后我将函数从Success()更改为then()然后值插入工作但现在无法从数据库加载数据并显示它
<!DOCTYPE html>
<html>
<head>
<title>Office Essentials</title>
</head>
<body>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.4/angular.min.js"></script>
<h2>Office Essentials</h2>
<div ng-app="myApp" ng-controller="cntrl">
<form>
Product Name:<br>
<input type="text" ng-model="name" name="name"><br>
Quantity:<br>
<input type="text" ng-model="quantity" name="quantity"><br>
Price:<br>
<input type="text" ng-model="price" name="price"><br>
Details:<br>
<textarea ng-model="details" name="details"></textarea>
<br><br>
<input type="submit" ng-click="insertData()" value="Submit">
<br><br>
<input type="submit" ng-click="displayProduct()" value="Show">
{{msg}}
</form>
这是用于显示数据库中的产品
<table border="1">
<tr>
<th>Id</th>
<th>Product Name</th>
<th>Quantity:</th>
<th>Price</th>
<th>Details</th>
<th>Status</th>
</tr>
<tr ng-repeat="product in data">
<td>{{product.id}}</td>
<td>{{product.name}}</td>
<td>{{product.quantity}}</td>
<td>{{product.price}}</td>
<td>{{product.details}}</td>
<td>{{product.status}}</td>
</tr>
</table>
</div>
这是我在main.html页面中使用的sqript
<script >
var app=angular.module('myApp',[]);
app.controller('cntrl',function($scope,$http){
$scope.insertData = function(){
$http.post("insert.php",
{
'id':$scope.id,
'name':$scope.name,
'quantity':$scope.quantity,
'price':$scope.price,
'details':$scope.details
})
.then(function onSuccess(){
$scope.msg = "data Inserted";
})
}
$scope.displayProduct=function(){
$http.get("select.php")
.then(function onSuccess(data){
$scope.data=data
})
}
});
</script>
</body>
</html>
这是我的select.php页面
<?php
include "connectdb.php";
$query = "SELECT * FROM officeessentials";
$result=$dbhandle->query($query);
while($row = $result->fetch_assoc()){
$data[]->$row;
}
print json_encode($data);
?>
答案 0 :(得分:0)
更新select.php
<?php
include "connectdb.php";
$query = "SELECT * FROM officeessentials";
$result = $dbhandle->query($query);
$data = [];
while($row = $result->fetch_assoc()){
$data[] = $row;
}
echo json_encode($data, true);
?>