Hi,
I want to update one page according to submission of another page.
i.e
I have a page say A that asks for the no of files to be uploaded.
In page A, as soon as I enter the no of files to be uploaded, another page opens that Display that number of browse buttons, where I can upload the files.
As soon As I click on the Page with browse buttons, the page A should be updated, showing the files that have been uploaded.
How do I do it?
----
Page A
<form a>
<input type="text\" name="nofiles" onchange="window.open(a.php)">
</form>
.
.
.
Want to display the entered files here with checkboxes so that the files can be deleted
e.g
File 1 [Delete]
File 2 [Delete]
----------
Page B( to upload files)
//$COUNT id no enterd in textbox of Page A
for($i=1;$i<=$count;$i++)
{
<input type="file" name="inputfile[]" >"
}
//code for submit buttonyup,
Page B( to upload files)
<form action="a.php" method="post" enctype="multipart/form-data" >
<?php
//$COUNT id no enterd in textbox of Page A
for($i=1;$i<=$count;$i++)
{
<input type="file" name="inputfile[]" >"
}
?>
</form>
that is how you do it. but you have to POST the info to a page so you can uplod it. if this is apopup then you would post to the parent and the parent would do the uploaded and whatever you have it doing, then refresh to show the results.Thanks scoutt.
But I don't understand how to
post to the parent and the parent would do the uploaded and whatever you have it doing, then refresh to show the results.
Could you please explain this.
Thanks.after the user selects the files to upload then they hit save or what ever. the forms action will point to the parent window. easy as that.
I think it would be
<form onsubmit="opener.location=a.php" method="post">
but I would check in the client side forum to see how to submit to the parent from a popup.
I want to update one page according to submission of another page.
i.e
I have a page say A that asks for the no of files to be uploaded.
In page A, as soon as I enter the no of files to be uploaded, another page opens that Display that number of browse buttons, where I can upload the files.
As soon As I click on the Page with browse buttons, the page A should be updated, showing the files that have been uploaded.
How do I do it?
----
Page A
<form a>
<input type="text\" name="nofiles" onchange="window.open(a.php)">
</form>
.
.
.
Want to display the entered files here with checkboxes so that the files can be deleted
e.g
File 1 [Delete]
File 2 [Delete]
----------
Page B( to upload files)
//$COUNT id no enterd in textbox of Page A
for($i=1;$i<=$count;$i++)
{
<input type="file" name="inputfile[]" >"
}
//code for submit buttonyup,
Page B( to upload files)
<form action="a.php" method="post" enctype="multipart/form-data" >
<?php
//$COUNT id no enterd in textbox of Page A
for($i=1;$i<=$count;$i++)
{
<input type="file" name="inputfile[]" >"
}
?>
</form>
that is how you do it. but you have to POST the info to a page so you can uplod it. if this is apopup then you would post to the parent and the parent would do the uploaded and whatever you have it doing, then refresh to show the results.Thanks scoutt.
But I don't understand how to
post to the parent and the parent would do the uploaded and whatever you have it doing, then refresh to show the results.
Could you please explain this.
Thanks.after the user selects the files to upload then they hit save or what ever. the forms action will point to the parent window. easy as that.
I think it would be
<form onsubmit="opener.location=a.php" method="post">
but I would check in the client side forum to see how to submit to the parent from a popup.