Well i do have a mysql query in one php page(php_1) & I want to submit the variables to the query in different php page(php_2) via form action but how am I supposed to do it without redirecting to php_1..All I need is to post the data to the first php page query so that It performs some action over there & thats all... So hoping sumone who knows ajax might helpme out with a sample code or point me in the right direction.
php_1 page :- $query = ("SELECT * FROM state4 WHERE (LONG_HI<'".$_POST['Ymax']."')"); php_2 page:-
I am facing some problem with this when i am trying to sent the data in krnlAddComment1.asp page and there is only an insert query exists. after run the query successfully, response.write "...." revert back to me
I'm trying using this good example [URL] but I need to use it with more than 1 parameter. What I'm trying to do is a form with two fields (i.e. Name and city) who feed a query. The results of the query should be shown below the form.
i.e.: Enter name and city in the follow form...
SELECT * FROM partecipants WHERE name=name AD city=city
I have many tags. I want click each tag, then post/get the tag's value to another page. In another page, received the values and make a mysql query. Then return the resalt data to the first page(do not make an iframe).
All I need is a MySQL update by pressing the submit button. The user sould't get on the php site by pressing the Submitbutton and when it's possible there should't be a page refresh by pressing the submit button.
The jquery.form.js plugin also didn't work that way.
I want to be able to open a window with a variable. For example, I will use domain.com?reference=1234- This link will open a pop up window as follows:
When I open this window, it is always showing data put in last, it isnt refreshed. I wanted to open this pop up window, enter data and then click submit and make the data enter the database and then refresh the parent window. Every time I have done this, either the parent or the pop (or both) is showing old data.
I have a simple chat on a web page, but It wont refresh itself unless the user refreshes the whole page and I cannot force the page to totally refresh itself cause that would mess up the other things the user can do on the site. So I would need to make the DIV to "load itself again in for example 3sec". I believe this is quite common thing to construct, how to construct this?
this code refereshes the whole page and focus goes at the top of the page but I want the focus of the page at the same place where I perform refresh. Actually I am using AJAX to submit a form, a form opens through ajax request but I need to refresh the page to take the effect of the dynamic data, which is being saved,updated or deleted on the page. So is there any way where focus of the page remains there when I refreshed the page
when i load page it does nothing only blank page, may have code in wrong place but not sure so can someone show me a working example so i can find out what i'm doing wrong also i would like the part of page to refresh evey 120 seconds is this possibe with jquery?
I have been using asynchronous requests for a long time. so the response was processed in a callback function. I thought of not using async so i made synchronous requests. The reason is that i dont have to have two more lines for checking the status and the onreadystatechange.... my synchronous requests would be like this...
Do we have to check whether the page is navigating away while in synchronous operation and abort the request? or what could be the reason for the error. This will not happen in async requests because that is also the reason for async...
I want to have a form to select between English and Spanish on a site. Upon selecting a radio buton (I prefer to do this without a submit button - just select the radio button and move on), I want a cookie to be set for which selection has occured and then be taken to that page. On the return visit I'd like to check for this cookie and then be sent to the correct page if it exists. Anyone have any ideas? I've found a few cookie tutorials and things, but they invole a submit button or a manual page refresh after the selection.
I have a php webpage that has a single input box. When ever an item is entered and the Submit button is pressed, the entry is added to a database. Additionally, the database table contents are read and printed directly below the input box and the submit button on the same page, using AJAX. Now, I have decided to add additional functionality to the page. I want a delete button for each entry displayed in the table below. For any button that is pressed, I want to delete that entry from the database.
Here's the problem: I want the database table contents that has been printed below to reflect this change. I can achieve the delete, but in order to show the change in content caused by the delete, the table below must either be replaced or updated via ajax. How can I achieve this? I have been trying to follow these examples using jQuery and Mootools respectively, but I've been having no luck. [URL] I don't want all the fancy looks, I just want the core functionality.
I am showing recent visitors on profile page.The most recent visitor would be "1 min ago".If suddenly new user comes so i just want to append that user with the message "about a min ago" without refreshing the page.
ex- suppose my recent visitors list ll be like this-
And now smita comes as the recent most visitor i want to append her name in to this list without refreshing the the entire page like this
I know i can do this with AJAX. I have searching around the net for the any helpful resource but didn't get anything yet.
I am not sure, if this works. I make a function call with the ajax option of jquery.If I try this, the whole response data is shown in the specific #div.Is there any possibility to load only the content of "#div" from some.php?
I'd want that: at the first user's click on the #register div, the div #registerForm become visible, it should remains visible also when the user click the submit button within and it should disappear if the user click a second time on the #register.I partially solved my problems with this:
[URL]. I looked at this script and I tried playing around with it but could not figure out how to: Instead of having a check box field and then a radio box field, have 2 radio box fields, so the user picks from one group of radio boxes and then another. Then change the math behind it so the values in the first field are multiplied by 100 and then the values in the second field tack on a percentage (ex value in first field is 1.5 so 1.5x100 = 150 and value in second field is .5 so total is 225).