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Sending record ID to next page?
Hey, i`m building website
I have my main tab, where I have all my records from table .. (if it might help to help me - table is called - blog_articles, and id is BAID)
All the div`s I have in the main page are links.. For now I created seperate .php file (OpenArticle.php) and the links refere to it
When pressed, I want the article which the user pressed on, to open in that next page..
I suppous I do that by sending that articles ID to next page to show all the contents from it
Can anyone help me out with this one ? And I would appreceate other tips and tricks too relating to this ;)
1 Answer
- ?Lv 66 years ago
For content to open up on another page it should be typed up, and submitted to a database table, and then the content can be retrieved, and then rendered on a web page.
Regarding your id that's inside of your table. Your id should only be called id, and nothing else. If a user submits data to a database table, then your id table row is suppose to produce a number, and the number increases every time someone submits something to your database table.
I don't know if you made a table that only has an id inside of it, but if you did, then I think your table won't be useful to you. Because your not specifying inside of your table what you want to be inserted into your table.
Unfortunately using Div's alone is not good enough.
Don't link to OpenArticle.php.
What you should do is add the php code in the OpenArticle.php web page that's used to retrieve, and render articles from your database table onto the OpenArticle.php web page.
You should make one php web page that's used to process submitted articles. You would also have to make a database table to collect the articles.
First make a button that connects to a php web page that will process your request by submitting your articles to a database table, and then your articles can be retrieved, and rendered on OpenArticle.php.
In my opinion, I think the article should be typed up before someone submits it.