i got a jquery upload and crop script and i am trying to use it.
First i have a 1.html file which has a form, which requires some texts and image. After submitting the form it goes to main.php where it checks for some image properties and if successful it refreshes the page using header("location:".$_SERVER["PHP_SELF"]);
So if i place my $_POST['name'] i get the value from 1.html . Now when the image in displayed after page refresh there is one more option to select the thumbnail and upon selecting the thumbnail there is one more page refresh, to display the final images (both bigger and thumb). Now my problem is for second page refresh i am not able to get the fields which i had posted from 1.html. Any suggestion would be highly appreciated. Thanks
With that header refresh you are losing every information. Drop the refresh and do consecutive forms: you need to propagate the values you need from the first form, using hidden input fields in the forms that follow
<input type="hidden" value="<?php echo $value_from_original_post; ?>">
or you can store the value(s) of interests in session variable(s).
Alternatively, you can use an AJAX solution which requires no reload or page change, but it's a bit more work (and you might not want javascript).
You can not store states between pages unless:
Keep rolling the value(s) forward as hidden input type if it's a form submission.
Temporarily save value(s) as cookie until consumed.
Related
i got 4 pages which are company details, job requirements, job responsibilities and design. so im gonna update the data from these pages. so i retrieve the data from database to be display into form for these 4 pages. after i edit the values on first page, then i go to next page for next update. but when i go back to previous page, the data wont keep the edited value but display the old data that i retrieved from database. so how to prevent this to happen ? same goes if i refresh the pages. i know i can use the session to store edited value but still i think that form will display the old data. help me give the idea how to do this pls ! thanks
You need to follow below steps
Go to next page by submit & action attribute
In next page hold all the post value into hidden fields
Use a back button to go back.
when go button will be pressed submit the form which is holding the hidden fields
Now you are in first page and simply use $_REQUEST or $_POST to get the previous value.
Hope you under stand.
N.B: If you use browser back button browser will ask you to confirm to resubmit the form. If you confirm then no problem but if you didn't confirm then you lost your values.
Basically I have a bunch of data I get from a database and put onto my page in a table. Right now I have the user type in the name, session, etc. in the table and that is sent as post data into the next PHP page, which I then use to lookup more stuff in the DB and so on and so forth.
Obviously that's not a great user experience; it would be much easier to simply CLICK the item in the table and everything gets sent automatically into the next page.
I'm not sure how I'd go about doing this.
My tables are first and last names for now, so if you click a certain row it should go to the next page sending each cell as data.
EDIT: Some examples:
Traditionally you do this with a form
<form method="post" action="pageDataIsGoingTo.php">
to send data to the next page. However, I don't want to do this with a form; but rather when they click a URL and/or button that sends the data. I can "hide" the data from view I suppose, but I still don't know the function to actually go ahead and do that.
Would I make a javascript button/function that sets something in an invisible form?
You can use invisible/hidden form fields.
That might be your best guess.
Javascript would be a good solution if you wanted an ajax POST call, but you want to load other page.
So hidden form fields are your solution.
Parallel with table data.
You need to embed hidden fields and your visible item row within a form
(so each item row contains also a form & hidden form fields and visible submit button,
which you can style with css)
This presuming that your table contains more items which you can choose to send.
Although I would do this with backbone & jquery and do it all in ajax.
I have a page with two forms that is generated with PHP.
The first part contains text boxes, a submit button and a clear button.
The second form is just a button called "Add more text boxes" so the user can add more to his form if he needs to.
The problem is when I click the "Add more rows" which loads another page which changes a value.
This value then affects the original page when it reloads causing more text boxes to get created.
The problem is that I lose all the data that was entered.
Is there any way to preserve the data when the user clicks "Add more rows"?.
Here's a screenshot of my page.
Thanks
If you want to do it without js than you put all in one form. When you click button to add row all entered data will be available in $_POST or $_GET so you can fill form with existing data and add a row when generating new page.
Ideally you should use javascript to dynamically add new rows w/out making new requests to the server and loading new pages. But if you want to keep it javascript free. If it's all the same php script just controlled by conditions, just use the $_POST['variable'] values as the value="$_POST['variable']" in your fields. If it's handled through multiple scripts, you can use a session variable to pass the data from one page to the next.
Yes, Im having a little edit profile page, index.php?mode=profile. Lets take the username in the editprofile form as example. The username is already in the username-field. So i changed from "Peter" to "Tom" and press save.
The action is ?mode=profile&edit=true. So now when i have pressed save it has updated the column in the db from Peter to Tom. But this field keeps having the value "Peter" until if i do press refresh (or f5), then "Tom" will appear. Like it hasnt updated in the database anything, although it did but it still shows Peter until next refresh.. like it caches, but it shouldnt cache nothing?
Any help on this? Is it because its on the same "page" / file? what can i do
I think you fetch the data first, put it in the form and then update the database contents. You should update the database contents first when you submit the form. The fetch and creation of the form should be the next step.
Most likely you're rendering the page before doing any of the profile modifications. Make sure you're dealing with changes to the profile before you render it (i.e., all your database calls to update values should be before the SELECT statements to render the page).
HTTP provides two methods : POST and GET. With URL starting with ?param= you are actually using GET. "GET" should be used if and only if the form processing is idempotent query (ie the page content is NOT altered by the query).
If your form modifies data, you should use POST method.
What you can do is in each input field of your form just echo post values for respective fields.
For eg -
lets take an example of username field -
<input type="text" name="username" value='<?php echo $_POST['username']?>' />
By this way if you post any field then that value will be displayed and not from cache.
In my above example i haven't cleaned the post data, make sure that you use cleaned post data
in actual scenario.
I have a php page that has a form that asks for an e-mail. When you press the send button, it gets to another php page, which gets the form data and does its stuff. I need to then be able to go back to the old page (the one that contained the form) and give it some data so that it will be able to change itself and say "You've sent your e-mail successfully, and will not display the form.
How do I do it?
Sessions probably
http://us2.php.net/manual/en/book.session.php
You can either use sessions or cookies, to not depend on the URL cookies have always to be enabled.
Check the PHP Manual (Sessions and Cookies).
Options:
1) Set a cookie (or use a session variable, which is kind of the same thing)
2) Use a separate thank-you page. After you've processed the form, redirect to http://www.mysite.com/thankyou
3) Process the form on the same page as itself. If your form is at http://www.mysite.com/myform, then at the top of that page have a little
if ($_POST)
// process form
// display thank you
else
// display form
Good luck!
If the user is just seeing data that they've entered anyway, you can just use hidden form fields:
<input type="hidden" id="lang" name="lang" value="en" />
That way you can continue to POST new forms and pass the data down the lane. That's the easiest thing to do without having to write a single extra line of PHP code.
You could also store each section in a database and save each section as-added. That would give you the added benefit of having partial data in the case of a browser crash or whatever, depending on how many parts your form is. You could then pass just an ID to the DB table row and retrieve the data for display.