File Upload

[Steve2106Participant]

Hi There,
Where can I get the filename of the file chosen in the fileupload widget.
I can get all the other details from a form but not the fileupload details.
This is a bit of php I’m using.
` <h1>Thank You</h1>
<p>Here is the information you have submitted:</p>

1. Name: <?php echo $_POST[“Name”]?>
2. Address: <?php echo $_POST[“Address”]?>
3. Address Line 2: <?php echo $_POST[“addressLineTwo”]?>
4. Post Code: <?php echo $_POST[“postCode”]?>
5. Service: <?php echo $_POST[“dropdownValue”]?>
6. Message: <?php echo $_POST[“radiobuttonValue”]?>
7. Check 1: <?php echo $_POST[“checkboxValue1”]?>
8. Check 2: <?php echo $_POST[“checkboxValue2”]?>
9. Check 3: <?php echo $_POST[“checkboxValue3”]?>
10. FileName: <?php echo $_POST[“fileToUpload”]?>

Thanks for any help.
Best Regards,

Steve.

[adminKeymaster]

Hi Steve,

It is $(“#fileUpload”).jqxFileUpload(“fileInputName”);

Regards,
Peter

Best regards,
https://www.jqwidgets.com/

[Steve2106Participant]

Hi Peter,

That does not work in my scenario.
I’ve added the code you suggest at the bottom of the code below and I get an error of “Parse error: syntax error, unexpected ‘(‘, expecting variable (T_VARIABLE) or ‘{‘ or ‘$’ in C:\xampp\htdocs\SteveTest\Results2.php on line 21”

This is the code:
<h1>Thank You</h1>
<p>Here is the information you have submitted:</p>

1. Name: <?php echo $_POST[“Name”]?>
2. Address: <?php echo $_POST[“Address”]?>
3. Address Line 2: <?php echo $_POST[“addressLineTwo”]?>
4. Post Code: <?php echo $_POST[“postCode”]?>
5. Service: <?php echo $_POST[“dropdownValue”]?>
6. Message: <?php echo $_POST[“radiobuttonValue”]?>
7. Check 1: <?php echo $_POST[“checkboxValue1”]?>
8. Check 2: <?php echo $_POST[“checkboxValue2”]?>
9. Check 3: <?php echo $_POST[“checkboxValue3”]?>
10. FileName: <?php echo $_POST[“fileToUpload”]?>
11. FileName2: <?php echo $(“#fileUpload”).jqxFileUpload(“fileInputName”);?>

Thanks for your help.

Best Regards,

Steve.

[adminKeymaster]

Hi Steve,

I wrote you how to get the fileInput’s name in Javascript. I do not think that you can PHP echo it like that.
For example, in the docs https://www.jqwidgets.com/jquery-widgets-documentation/documentation/jqxfileupload/jquery-file-upload-getting-started.htm,
we create it like that: $(‘#jqxFileUpload’).jqxFileUpload({ width: 300, uploadUrl: ‘upload.php’, fileInputName: ‘fileToUpload’ });.
This means that when you submit a Form, the name is ‘fileToUpload’.

Regards,
Peter

Best regards,
https://www.jqwidgets.com/

[TwicsykoBlocked]

Thank you

[Author]

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