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PHP Get Hidden Variable from an AJAX Image Uploader

开发者 https://www.devze.com 2023-04-05 13:23 出处:网络
I am using this AJAX Image Uploader script and what I am trying to do is pass a hidden variable. <input type=\"hidden\" name=\"test\" value=\"test\" />

I am using this AJAX Image Uploader script and what I am trying to do is pass a hidden variable.

<input type="hidden" name="test" value="test" />

But since there is no real submit button, it is all 开发者_JAVA技巧AJAX based, I cannot just call $_POST['test'];

Part of the jQuery is:

if (formdata) {  
    $.ajax({  
        url: "upload.php",  
        type: "POST",  
        data: formdata,  
        processData: false,  
        contentType: false,  
        success: function (res) {  
            document.getElementById("response").innerHTML = res;  
        }  
    });  
}  

It should probably go somewhere in there but I'm not sure on how to approach this. If anyone has any suggestions I would really appreciate it.

Thank you!


Try this: Just give your hidden input an id so we can refer to it and get the value. I used hidden_input in this case.

if (formdata) { 
    //New line is below!
    formdata.append("test",$('#hidden_input').val());
    $.ajax({  
        url: "upload.php",  
        type: "POST",  
        data: formdata,  
        processData: false,  
        contentType: false,  
        success: function (res) {  
            document.getElementById("response").innerHTML = res;  
        }  
    });  
}

Then your upload.php should be able to access the value with $_POST['test']


But since there is no real submit button, it is all AJAX based, I cannot just call $_POST['test'];

Sure you can. AJAX can send any header you want in the type parameter. If you put type:'post' just as you have it now, the field with name="test" will be available via $_POST['test']

You can have any event trigger the ajax, including the page load.

$('somenode').bind('someevent',function(){

So, say you want the ajax to post every time a field value is changed:

$('input').change(function(){
    $.ajax({  
        url: "upload.php",  
        type: "POST",  
        data: formdata,  
        processData: false,  
        contentType: false,  
        success: function (res) {  
            $('#response').html(res);  
        }  
    });  
});
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