使用Angular/PHP将JSON对象插入MySQL表 [英] Inserting JSON object into MySQL table with Angular/PHP

查看:112
本文介绍了使用Angular/PHP将JSON对象插入MySQL表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

所以我有一个JSON对象,我想将其插入MySQL表的字段中.当我执行"console.log(thatjson)"时,它将在控制台中显示带有文本[Object]的JSON对象,并在单击该对象时提供所有信息.但是,当我执行"alert(thatjson)"操作时,它会在警报框中打印"object Object",而当我在phpMyAdmin中检查该表时,varchar(2000)字段仅存储了一个字符串,该字符串表示[object Object]

So I have a JSON object I want to insert into a field in a MySQL table. When I do a "console.log(thatjson)", it prints the JSON object in the console with the text [Object], with all the information available when you click on the object. However, when I do a "alert(thatjson)" it prints "object Object" in the alert box, and when I check that table in phpMyAdmin, the field which is varchar(2000) only stored a string which says [object Object].

在我的php中,我有$ property = json_decode($ request-> getBody());解码角度传递的json数据,显然没有帮助. :(

In my php, i have $property = json_decode($request->getBody()); to decode the angular passed json data, apparently that did not help. : (

有什么想法吗?

相关代码:

$scope.update = function(thatjson){                
             $http.put('api/envisiondb/0', thatjson).success(function(data) {
                $scope.properties = data;
          });
        };

PHP Slim:

function editProperty($id) {
    $request = Slim::getInstance()->request();
    $property = json_decode($request->getBody());
    $sql = "UPDATE db SET username=:username..."

JSON对象只是日常的常规JSON对象

The JSON object is just a regular everyday JSON object

推荐答案

请确保您通过JSON.stringify()函数传递请求数据.

Make sure you pass your request data through the JSON.stringify() function.

这篇关于使用Angular/PHP将JSON对象插入MySQL表的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆