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I have list of checkboxes on web page. What I want is , after deselcting checkbox particular record from database should get deleted accordingly in php.
You'll need to use some javascript to "listen" for the uncheck action on each checkbox. You can then use AJAX to delete the record from the database on the fly, and use jQuery make the record disappear visually from the webpage, all without reloading.
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I'm trying to create a 3-question survey written in PHP/AJAX/JQuery that will basically present 6 pictures to be ordered/ranked from favorite to least favorite, then submitted & recorded to a mysql db, that will in turn drive the next 6 pictures to be ordered and submitted.
I've tried using a multi-step form, which seems ideal, but all the examples write the recorded choices at the end of the form submission, whereas I need each question in the survey to drive the next set of choices. I can't find anything like that to use as an example.
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when I sumbit data from the web site to check if it works properly web site shows that data entered sucussefully to the data abse. this is how it shows But on data base it only shows number 1
I need a solution to make this work
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How i show data postet in another website to mysql database, in real time on main page of website using PHP and that any new data is displayed in a new line.
Datas that i need show are: name, and finish time (from race).
Use .setTimeout() JS method on main page of website and do jQuery.ajax() requests to your server to update data.
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I wish to know that whether is there any possible that can allow me to delete or hide the data on my php website but when I go inside my database, the record that is being deleted will still store inside the original database?
You have to add a deleted flag to your table
For deleting you set the flag:
UPDATE table SET deleted=1 WHERE id=xy
For geting the data from the database you have to check the flag:
SELECT * FROM table WHERE deleted=0
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I have a div which value is changed in tyiping... on typing in textarea and when stopped typing its value again changed to no typing or maybe its empty.
how can I put its changing value into mysql table? or update?
any help would b appreciated.
thanks
Learn ajax requests...
By ajax I mean not only ajax. To start learning ajax you must know some jQuery.
Here are some tutorials:
Bucky's Room
Basicly get the value of the div you change while typing and send it to the database by ajax.