使用API数据源的php搜索表单
我想使用API构建一个PHP表单来提取数据。使用API数据源的php搜索表单,php,api,search-form,Php,Api,Search Form,我想使用API构建一个PHP表单来提取数据。 我有一个API密钥和密码。我已经通过操纵URL完成了一些基本的测试查询,以确保一切正常 我不知道如何构建PHP表单并显示结果。我打算在JS中这样做,但是API密钥和机密将在URL中显示,并且有多个搜索选项 从我看到的其他人询问php和api的情况来看,连接如下: $apikey = 'xxxx'; $secret = 'xxxx' $service_url = 'https://website.com/' .'api_key=' . $apikey
我有一个API密钥和密码。我已经通过操纵URL完成了一些基本的测试查询,以确保一切正常 我不知道如何构建PHP表单并显示结果。我打算在JS中这样做,但是API密钥和机密将在URL中显示,并且有多个搜索选项 从我看到的其他人询问php和api的情况来看,连接如下:
$apikey = 'xxxx';
$secret = 'xxxx'
$service_url = 'https://website.com/' .'api_key=' . $apikey . '&api_secret='. $secret. ‘&’ .$criteria . ‘&’ . $criteria2 . ‘&’ .$criteria3;
[{"state":"NT","date":"2016-08-17","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Darwin","file":"1234","additional_info":"1 did something stupid"},{"state":"NT","date":"2016-08-03","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Katherine"," file":"4567","additional_info":"1 did some more dumb stuff;2 did even more dumb stuff;3 really dumb stuff;4 even more dumb stuff"}]
这是我用于测试的API URL:
域名here/clients/api/search?api_key=1234&api_secret=1234&state=nsw&given_name=john&name=smith
json中的结果如下所示:
$apikey = 'xxxx';
$secret = 'xxxx'
$service_url = 'https://website.com/' .'api_key=' . $apikey . '&api_secret='. $secret. ‘&’ .$criteria . ‘&’ . $criteria2 . ‘&’ .$criteria3;
[{"state":"NT","date":"2016-08-17","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Darwin","file":"1234","additional_info":"1 did something stupid"},{"state":"NT","date":"2016-08-03","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Katherine"," file":"4567","additional_info":"1 did some more dumb stuff;2 did even more dumb stuff;3 really dumb stuff;4 even more dumb stuff"}]
我对PHP搜索表单有一个基本的了解,但是这似乎让我有些不知所措。我猜这将是一个提交查询、下拉数据和格式化数据的混合体
我要寻找的是关于如何构建此表单的答案。这将是一个搜索表格,我放在一个网站上,并有结果显示在同一页上
所以看起来是这样的:
$apikey = 'xxxx';
$secret = 'xxxx'
$service_url = 'https://website.com/' .'api_key=' . $apikey . '&api_secret='. $secret. ‘&’ .$criteria . ‘&’ . $criteria2 . ‘&’ .$criteria3;
[{"state":"NT","date":"2016-08-17","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Darwin","file":"1234","additional_info":"1 did something stupid"},{"state":"NT","date":"2016-08-03","listing_type":"listing info","place":"place info","moreinfo":"","name":"SMITH Dale John","location":"Katherine"," file":"4567","additional_info":"1 did some more dumb stuff;2 did even more dumb stuff;3 really dumb stuff;4 even more dumb stuff"}]
搜索表格:
名称:“输入名称”
姓氏:“输入姓氏”
状态:“进入状态”
结果
状态:QLD日期:2019年1月1日
挂牌类型:挂牌信息
地点:地点信息
更多信息:
姓名:史密斯·戴尔·约翰
地点:达尔文
档案:1234
附加信息:1做了一些愚蠢的事
状态:QLD
日期:2019年1月1日
挂牌类型:挂牌信息
地点:地点信息
更多信息:
姓名:史密斯·戴尔·约翰
地点:达尔文
档案:1234
附加信息:1做了一些愚蠢的事
2做了更愚蠢的事
3个非常愚蠢的东西
4更愚蠢的东西
我知道搜索表单本身应该相对简单,因为它只是将表单字段插入url。
但是如何从表单提交生成的url中提取数据并显示它们呢 要隐藏api密钥和api机密,需要添加一个新层 因此,您将有:
Apikey和secret将仅从php中使用,因此它们将是安全的url将对传递api信息的客户端可见,除非您在服务器端传递某些信息,因此我不确定您要隐藏什么。如果担心其他人知道API信息,则应使用https。至于json,我假设您使用json_encode对其进行编码,但也可以使用json_decode'()轻松显示。根据您希望如何向最终用户交付,您将决定如何向他们展示它(我知道这是一个显而易见的说法,但…)。我我会吐出json,让他们在他们认为合适的时候解析它。我只担心如果我使用Javascript,url是否可见。我会在我的问题中添加更多的信息来解释更多。感谢您的反馈。当然,最后一部分是PHP从API接收JSON数据,a)直接将该数据传递回浏览器,让JavaScript代码确定如何在页面中显示该数据,或者b)使用该数据构造一些HTML,然后将其传递回浏览器,JavaScript代码所要做的就是将其附加到页面的某个位置