About file Upload $ _ FILES []

About file Upload $ _ FILES! $ _ FILES [& quot; userfile & quot;] Here, userfile must be the front-end control name & lt; inputname = & quot; userfile & quot; type = & quot; file & quot;/& gt; the problem is. I used JS to obtain the file control about file Upload $ _ FILES!
$ _ FILES ["userfile"] Here, userfile must be the name of the front-end control.


The problem is. I use JS
Obtain the value of the file control userfile = document. getElementById ("userfile"). value;
Then use ajax to process it in the background
$. Ajax ({
Url: from. attr ('action '),
Data: "HeadImg =" + userfile + "",
Type: "post ",
DataType: "text ",
Success: function (data ){
Document. write (data );
// Alert (data)
If (data = "1 "){
Window. location. href ("home. php ");
}
}
});
Then the background
$ POST ['uploadmg '] can accept the value of userfile

However
Write in this way

Move_uploaded_file ($ _ FILES [$ POST ['uploadmg '] [temp_name], dirname (_ FILE __). $ _ FILES [$ POST ['uploadmg '] ['name']);


The file cannot be uploaded ..

$ _ FILES [$ POST ['uploadmg '] [temp_name] cannot be written in this way?



------ Solution --------------------
In this case, the client uploads files to the server through http post and stores them in the temporary directory of the server (php. ini). The file address is $ _ FILES ['updataheadimg '] ['tmp _ name'] (the file name is encrypted)
Move_uploaded_file () is used only to move files in the temporary directory to a new directory. In addition, the client can only upload files to the server, because the server cannot actively read files from the client.