Problem with ajax/jquery and django

Question

I want to send a value to the server.I send it using:

$.post("http://127.0.0.1:8000/search/",{'long':"30"});

and my views.py is:

def search(request):   
    lon = request.POST.get('long','')  
    place = Places.objects.filter(longtitude__icontains=lon)  
    if place is not None:  
        return render_to_response('sr.html', {'place': place, 'query': lon})  
    else:  
        return HttpResponse("NOTHING FOUND")

But i can not get my database searched for places with longitude containing 30!!!!

Please post the relevant bits of your actual error page, or specify what happens when you have this failure. — Paul McMillan
– Paul McMillan, Commented Oct 11, 2009 at 17:03

Daniel Roseman · Accepted Answer · 2009-10-11 17:33:40Z

2

filter() returns a queryset, not an individual item. So the variable you have called place will not be a Places object, but a queryset containing one or more places.

If for example your template has something like {{ place.name }}, that will print nothing because a queryset does not have a name attribute.

Assuming that there will end up being several Places with the same longitude, you will need to iterate through the queryset in your template with a {% for place in places %} loop.

answered Oct 11, 2009 at 17:33

Daniel Roseman

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1 Comment

thikonom Over a year ago

I have done it as you said .See my comment below. The thing is that i can not get the 'query'printed meaning lon has no value!

emeryc · Accepted Answer · 2009-10-12 04:29:40Z

0

Have you made sure you are even calling the function?

I would try printing out each value, including place. Since Place.objects.filter can't return None, you should be getting some kind of query set, even if it's empty.

Honestly this code is pretty buggy in general...It would help to know if Places.longtitude is really a string, if it is a string why did you decide to do that?

since place can never be None, did you want to check for an empty queryset in which case you would write

if not place:

I think we need more information then you've given us to help you at all.

answered Oct 12, 2009 at 4:29

emeryc

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Comments

googletorp · Accepted Answer · 2009-10-12 07:41:45Z

DR is spot on, but is missing a little thing. When you do this call in jQuery, you get one huge disadvantage, you can't use the Django template language. When you write a normal view, you would do exactly like DR explained. However since you are doing this with javascript, there are two things you want to do.

Since you can't rely on Django's auto escaping you need to manually escpape user inputted values like place.name.
The data you return, will need to be returned in JSON. This is not stricly necessary, but it will make it a lot easier for you to handle the data in your javascript. Django comes with the simplejson library, that you can fall back on, but I would recommend using the json library, which you can use to easily convert the python objects to json.

Brandon Henry · Accepted Answer · 2009-10-12 21:30:55Z

0

remove the ' around long in the jquery

$.post("http://127.0.0.1:8000/search/",{'long':"30"});

should be

$.post("http://127.0.0.1:8000/search/",{long:"30"});

your view is never seeing that 'long' is in the request. using the jquery post method and django together can get confusing, because the stuff between {} is structured differently.

answered Oct 12, 2009 at 21:30

Brandon Henry

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3 Comments

jb. Over a year ago

Nope! In JavaScript, {"long": "30"} means create an object with the key "long" with a value 30. { long : "30" } means create an object with key value of the variable long with a value of 30.

Brandon Henry Over a year ago

if that were the case this would not work. i have no variables named url, title, or tags. var item = $(this).parent(); var data = { url: item.find("#id_url").val(), title: item.find("#id_title").val(), tags: item.find("#id_tags").val() }; $.post("/save/?ajax", data, function (result) { if (result != "failure") { item.before($("li", result).get(0)); item.remove(); $("ul.bookmarks .edit").click(bookmark_edit); } else { alert("Failed to validate bookmark before saving."); } });

jb. Over a year ago

Embarrassing! Disregard everything I said.

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