[ZF 1.7] Faire passer les données POST dans un contentPane

Pour ceux que ça intéresse upload avec zend upload en ajax.
Déclaration du formulaire.

Code:

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        $this->setName('Uploadphoto');
        $this->setMethod('post');
        $this->setAttrib('enctype', 'multipart/form-data');
 $file = new Zend_Form_Element_File('filesend');
        $file->setLabel('File')
                 ->setRequired(true)
                 ->addValidator('NotEmpty')
                 ->addValidator('Size', false, array('min' => 20, 'max' => 200000))
                // ->addValidator("Extension",".jpg")
                ->addValidator('Extension', false, array('jpg', 'png','gif'))
                ->addFilter("Rename",array('target' => $path, 'overwrite' => true));
               // ->setDestination('C:\xampp\htdocs\zentest\public\files');
 
       	$image_url = new Zend_Dojo_Form_Element_ValidationTextBox('IMAGE_URL');
        $image_url->setLabel('image_url')
        ->addFilter('StringTrim')
        ->addFilter('StringToLower');
 
 
        $submitupload = new Zend_Dojo_Form_Element_Button('submitupload');
        $submitupload->dijitParams=array('label' => "Upload");
        $submitupload->setAttrib('onclick', "submitFormsendfile('Uploadphoto','utilUploadPhotoresponse','".$conf->pathweb."skillblog/compte/uploadphoto-ajax');return true");;
 
 
        $this->addElements(array( $file,$submitupload));

La vue correspondante (le petit script java pour charger l'image, précédemment uploadé,dés le chargement de page):

Code:

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<?php echo $this->formUploadPhoto ;?><div id="utilUploadPhotoresponse"></div><div id="UploadPhotoApercu"></div></div>
<script type='text/javascript'>
//<!--
dojo.byId('UploadPhotoApercu').innerHTML =  '<img src="<?php echo $this->UploadPhotoApercu ; ?>">';
//-->
</script>

puis la fonction javascript :

Code:

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<script type='text/javascript'>
//<!--
dojo.require("dojo.io.iframe");
dojo.require("dojo.parser");
function submitFormsendfile(formname,contentresponsename,urladresse) {
 
	dojo.io.iframe.send ({
                        // The page that parses the POST request
                        url: urladresse,
                        contentType: "multipart/form-data", 
                        handleAs: "html",
 
                        // Name of the Form we want to submit
                        form: dojo.byId(formname),
 
                        // Loads this function if everything went ok
                        handle: function (data,ioArgs) {
                                // Put the data into the appropriate <div>
		 						//console.debug("Uploaded");
 
                                dojo.byId(contentresponsename).innerHTML = data.body.innerHTML; 
                                var foo = dojo.fromJson(data.body.innerHTML);
 
 
            					if (foo.status == "success"){
            						dojo.byId('UploadPhotoApercu').innerHTML =  "<img src=\"" + foo.details.href + "\">";
 
                        },
                        // Call this function if an error happened
                        error: function (error) {
                                console.error ('Error: ', error);
                        }
            });
        }
//-->
 
</script>

J'espère que ça aideras quelqu'un.. :p