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help with ajax

Hi
i have a view with a table of users. with each row there are three links to view details, update and delete. each open a jquery ui dialog and when it is pressed the user is deleted by ajax call and so on.

i have a link to add a user. which open a dialog too with a form.
after the form is filled and the add button pressed it is added to the database via ajax.

all works but 2 things:

1- if i get success call i prepend a row to the user table to display the added user. it works but the three links (view update delete) does not open a dialog when they are pressed. instead i am redirected to the corospondant page as if javascript is disabled. after i refresh the page that same record does open the dialogs.
what am i missing here?

2- i have client side validation and a server side validation. but i cannot display the server side validation .

i attach my ajax call and the php code for the server validation.

best regards

this is the ajax call:
     $.ajax({
          type: 'POST',
          url: 'addforajax',
          data: {
			  username: username,
			  password: password,
			  firstname: firstname,
			  lastname: lastname,
			  email: email,
			  phone: phone,
			  mobile: mobile,
			  short_desc: short_desc,
			  description: description,
			  imgpaths: paths,
			  role: role
			  },
         // contentType: "application/json; charset=utf-8",
          datatype: 'json',
          success:function(data){
            //console.log(data);
         var userdata = JSON.parse(data);
		 //console.log(userdata[0].first_name); 
          //setTimeout(function() { location.reload() }, 100);
		  if(userdata[0].confirmed = 1){
			  userdata[0].confirmed = "¿¿¿¿";
			  }else{
				userdata[0].confirmed = "¿¿ ¿¿¿¿";  
			  }
			  
		if(userdata[0].role = 1){
			userdata[0].role = "¿¿¿¿";
		}
		if(userdata[0].role = 2){
			userdata[0].role = "¿¿¿¿";
		}
		
		
		
          
       $('table.table > tbody').prepend("<tr  id='record- "+ userdata[0].id +" '><td class='a-center'>"+ userdata[0].id +"</td><td><a href='http://mymvc104/admin/users/details/"+ userdata[0].id +"'>"+ userdata[0].first_name + userdata[0].last_name +"</a></td><td>"+ userdata[0].email +"</td><td>"+ userdata[0].role +"</td><td>"+ userdata[0].confirmed +"</td><td>"+ userdata[0].registration_date +"</td><td class='action'><a href='http://mymvc104/admin/users/details/"+ userdata[0].id +"' rel='"+ userdata[0].id +"' id='userdetails'><img width='16' height='16' title='¿¿¿ ' src='/assets/img/icons/magnifier.png'></a> <a href='http://mymvc104/admin/users/update/"+ userdata[0].id +"' id='userupdate'><img width='16' height='16' title='¿¿¿¿' src='/assets/img/icons/edit_default.png'></a> <a href='http://mymvc104/admin/users/delete/"+ userdata[0].id +"' rel='"+ userdata[0].id +"' id='userdelete'><img width='16' height='16' title='¿¿¿ ' src='/assets/img/icons/delete.png'></a></td></tr>"); 
		  
          },
        complete: function() {

          $("div#adduserform").dialog("close");

        },
          error:function(xhr, status, error){
             console.log("problem");
          }
        });

the relevant php code:

                 if ((!$posterrors && !$postmissing) )
                  {
                       //insert the post data alone
                             
                             $insert = $db->Insert('users',$data);

                            $theid = $db->lastId;
         
           
                               $fields = array('id','first_name','last_name','email','registration_date','role','confirmed');
                              $userdetails = $db->selectMultiFieldsById($fields, 'users', 'id', $theid, 'last_name', 'DESC');
                              
                              echo json_encode($userdetails);
                            

 
                  }
                  elseif (isset($posterrors) && isset($postmissing))
                  {
                     
                       echo json_encode($postmissing);
                       echo json_encode($posterrors);
                  
                  }

Open in new window

Gurvinder Pal Singh

where did you assigned the onclick event handler to the new links?

leakim971

it may be good to split your question...

you did not post the part open the dialogs (view, update, delete)

derrida

ASKER

hi

as i said i use jquery dialogs. this is the delete code:

//user delete
    $("a#userdelete").click(function (e) {
        e.preventDefault();
               
        var uid = $(this).attr('rel');
        //alert(uid);
        var parent = $(this).parent("td").parent("tr");
         //alert(parent);

        
        $("div#userdeleteConfirm").dialog({
            title : ' ¿¿¿ ¿¿¿¿¿?',
            autoOpen: false,
            width: '40%',
            height: 'auto',
            modal: true,
            draggable: false,
            buttons: {
                "¿¿¿¿" : function (){
                    $(this).dialog("close");	
                },
                "¿¿¿" : function (){
                  deleteuser(uid,parent);
                   $(this).dialog("close");
                }
            }
		
        });
        $("div#userdeleteConfirm").dialog('open');
		
        return false;
    });

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SOLUTION

Gurvinder Pal Singh

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ASKER CERTIFIED SOLUTION

leakim971

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SOLUTION

hielo

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derrida

ASKER

really thank you i had no knowledge about live function.

do i need to publish the second question seperatly?

best regards

Gurvinder Pal Singh

which one?

derrida

ASKER

2- i have client side validation and a server side validation. but i cannot display the server side validation .

Gurvinder Pal Singh

Didn't get that.
<<but i cannot display the server side validation>>
what does it mean?

derrida

ASKER

if you`ll take a look at my php code.

i have my validation check and then i use if else: if all required fields are filled and they are acceptable the i insertit into the database but if some required fields are empty or if they are filled incorrectly i echo the missing fields and errors.

the thing is that in firebug (when i delibratly leave some fields empty and some errors) i do get the the right response but when i try to alert them it does not work.
the bizzare thing is that when there are no errors or missing required fields i can use the returned data.

so i am confused. i am not that great with ajax yet.
in the code i published in the original you can see that i use the returned data to generate the tr of the table.

hielo

>>but i cannot display the server side validation .
It would best if you post that as a new question in the PHP code AND provide code + some description as what exactly is the problem. Simply saying " but i cannot display the server side validation ." is a "useful" as saying "My program doesn't work. Please help me". The more descriptive you are, the faster others will be able to help you.

hielo

Do you have a link to this page?

derrida

ASKER

thank all you saved me again. hope to get better with ajax soon