从显示的数据库中检索id并传递到其他页面 [英] Retrieving id from database displayed and passing onto other page

问题描述

我在特定页面上显示数据库的内容（配方名称）。
我希望每个配方都可以打开到另一个页面。我知道我必须获得该特定配方的配方ID，将该ID传递到另一个页面并将配方内容打印到该页面上。我想知道如何检索特定食谱的独特配方ID ON BEING CLICKED （主要是）。

解决方案

如果您在特定页面上显示数据库的内容，那么您肯定使用了像 ASP 这样的服务器端脚本，或者 PHP 或者一些 JavaScript + AJAX 。我对此很不清楚。但是对于你的问题，我可以告诉你一件事，假如你使用PHP，那么你可能会显示链接为;

  echo'< a href =yourpage.php？recipeid ='。$ recepeid。'>'。$ recipiename。'< / a>'; 
  

其中 $ recipeid ID和 $ recipiename 是你的食谱的名称。在您的PHP文件（ yourpage.php ）中，您可以捕获这些数据，如 $ _ GET [recipeid]

您可以尝试Google搜索教程。

使用;

  $ strLink ='< a href =recipedisp.php？id ='。$ row ['id']。'>'。$ strName。' < / A>'; 
 echo< li>。$ strLink。< / li>; 
  

不要使用空格。

I'm displaying the contents of a database on a particular page(recipe names). I'd like each recipe to be a link which opens to another page. I understand I would have to obtain the recipe id for that particular recipe, pass the id to another page and print the recipe contents onto that page. I would like to know how to retrieve the unique recipe id of a particular recipe ON BEING CLICKED upon mainly.

解决方案

If you are displaying the contents of a database on a particular page, then definitely you are using a server side scripting like like ASP or PHP or some JavaScript + AJAX. I am quite unclear about that. But for your question, I can tell you one thing that, if, suppose you are using PHP, then you may display the link as;

echo '<a href="yourpage.php?recipeid='.$recepeid.'">'.$recipiename.'</a>';

where $recipeid is the variable carrying recipe id and $recipiename is the name of your recipe. In your PHP file (yourpage.php), you may catch this data like $_GET["recipeid"]

You may try a Google search for tutorials.

Use;

$strLink = '<a href="recipedisp.php?id='.$row['id'].'">'.$strName.'</a>';
echo "<li>".$strLink."</li>";

Don't use spaces.

这篇关于从显示的数据库中检索id并传递到其他页面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！