How to go back without asking to confirm resubmission of the form-CodePudding

I have a web page that loads all the data from a mysql database called datalist.php

From this page I can edit record by record with a button that redirects you to an editdata.php page adapted to the fid of the record.

Once edited as they want to see the changes, I don't redirect them to the main one letting them see the changes and simply clicking back or with a button they return to the datalist.php without any problem.

The button is this

echo "<p id='parrafo'><a style='padding:1px 20px'class='button rounded-0 primary-bg text-white w-0 btn_1 boxed-btn' href='javascript:history.back()  '><--</a></p>";

PROBLEM

I added a search engine where the displayed data can be filtered.

When they use the search engine from datalist.php, I direct them to a page called search engine.php where, through a post method, I store what they are looking for in a variable and the data that users want appears.

But when they edit a filtered record, it is edited without problems, but when they go back, they return to the search engine.php and the message appears:

"Confirm form resubmission In order to display correctly, this web page needs the data you entered earlier. You can submit that data again, but that will cause the page to repeat all previous actions. Press Reload to submit the data and display the page.

Hit the page refresh button to resubmit the data needed to load the page."

Of course, if they update, they come back when the filtered data comes out.

Isn't there any way to store the variable used in the search so that when I go back I don't get this error or any solution??

CodePudding user response：

simple! when user will submit form for that variable instead of making post request
option1: just use get request __url__?variable=... but this will not remember the variable when you go back
option2: store the variable in the cookie and just go to next page (eg. window.location.href = '...';). and in next page access the cookie from php.

CodePudding user response：

If you are wanting to show the form to the user as a confirmation, but without the possibility of another post, then remove the form element and the button. Display all other boxes as they are (with the values populated from the POST array).

And display another message telling them that it has been successful.

You are using PHP, you can achieve this easily with it. If you are unsure, then post a short version of your code in a separate question.