CODEWITHSUNDEEP
jquery
Delirium tremens
Delirium tremens

Reputation: 4739

beginning to use jquery

The code

uri = 'http://www.test.com/';
$.ajax({
    url: uri,
    success: function(msg){
        sc.searchcontents = msg;
        alert(sc.searchcontents);
    }
});
alert(sc.searchcontents);

alerts undefined, then alerts the page's source code.

Upvotes: 0

Views: 92

Answers (3)

Scott
Scott

Reputation:

The second alert box is being called before the ajax is done because the ajax call is async by default. Add the async: false clause and that should resolve your issue.

Upvotes: 1

Mutation Person
Mutation Person

Reputation: 30498

How are you determining that it works or fails?

What are the error messages you are getting?

Have you read the jQuery Ajax Documentation?

To help you deterine what is going on, try adding an error function:

uri = 'http://www.test.com/'; 
$.ajax({ 
    url: uri, 
    success: function(msg){ 
        sc.searchcontents = msg; 
        alert(sc.searchcontents); 
    }
    error: function(XMLHttpRequest, textStatus, errorThrown) {
        alert(errorThrown);
    }
});

Upvotes: 0

Brandon Henry
Brandon Henry

Reputation: 3720

the success function of ajax runs after it loads. however, the javascript on your page will keep going while ajax loads its url. the alert you are seeing first is the one outside of your ajax call (undefined) and then the one within the ajax call (source code).

for testing, try putting a delay in the one at the end of your code sample. you'll see what i mean.

Upvotes: 1

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