I have a form inside a form and this makes the top form unresponsive. When I take off the second form (which is inside the first form), the first form works. This is the second form I have:
<form action="imgupload.php" method="post" enctype="multipart/form-data">
<h3>Upload a new image:</h3>
<input type="file" name="fileToUpload" id="fileToUpload">
<br>
<input type="hidden" value="<?php echo $row['Gallery_Id']; ?>" name="gid">
<input type="hidden" value="User" name="user">
<input type="submit" value="Upload Image" name="imgup">
</form>
Since this makes the first form not work, I was wondering if I can take off the form fields and then the submit button can send the form data to the imgupload.php like this.
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="hidden" value="<?php echo $row['Gallery_Id']; ?>" name="gid">
<input type="hidden" value="User" name="user">
<input type="submit" value="Upload Image" name="imgup" action="imgupload.php" method="post" enctype="multipart/form-data">
This does not work now. Is there a way I can get this working? If not, what's an alternative way to send this data to the other php?
Since you are uploading files, have a look at Ravi Kusuma's Hayageek jQuery File Upload plugin. It's simple, it's a Swiss Army Knife, and it works.
Study the examples.
http://hayageek.com/docs/jquery-upload-file.php
Ravi breaks down the process into three simple steps, that basically look like this:
<head>
<link href="http://hayageek.github.io/jQuery-Upload-File/uploadfile.min.css" rel="stylesheet"> // (1)
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="http://hayageek.github.io/jQuery-Upload-File/jquery.uploadfile.min.js"></script> // (1)
</head>
<body>
<div id="fileuploader">Upload</div> // (2)
<script>
$(document).ready(function(){
$("#fileuploader").uploadFile({ // (3)
url:"my_php_processor.php",
fileName:"myfile"
});
});
</script>
</body>
The final step is to have the PHP file specified in the jQuery code (in this case my_php_processor.php) to receive and process the file:
my_php_processor.php:
<?php
$output_dir = "uploads/";
$theFile = $_FILES["myfile"]["name"];
move_uploaded_file($_FILES["myfile"]["tmp_name"],$output_dir.$fileName);
Note the relationship between myfile in the PHP ($_FILES["myfile"]), and the filename specified in the jQuery code block.
Don't forget to check out the server-side code from the Server Side tab -- you need both parts (js and php).
Looking at your question again, you will probably want to use this functionality as well:
dynamicFormData: function()
{
var data ={ location:"INDIA"}
return data;
}
or
dynamicFormData: function(){
return {
newID: $("#newNID").val(),
newSubj: $("#newSubj").val(),
newBody: $("#newBody").val(),
formRole: $('#formRole').val()
};
These will appear on the PHP side, thus:
$newID = $_POST['newID'];
$subj = $_POST['newSubj'];
etc
As with any plugin, resist the temptation to just plop it into your code. Do a couple of quick-and-dirty tests with it first. Kick its tires. Fifteen minutes will save you two hours.
And don't forget to verify what was uploaded. You never know when a developing country black hat might be trying to get a new account.