jmogera jmogera - 1 month ago 17
Javascript Question

How to return success from ajax post in Node.js

I have a function like this:

exports.saveAction = function (req, res) {
var conn = mysql.createConnection({
host : nconf.get("database:host"),
//port: 3306,
user : nconf.get("database:username"),
password : nconf.get("database:password"),
database : nconf.get("database:database"),
multipleStatements: true,
//ssl: 'Amazon RDS'
});
var action = req.body;
conn.query('UPDATE actions SET ? WHERE Id = ?',
[action, action.Id], function (err, result) {
conn.end();
if (err) throw err;
res.writeHead(200, { "Content-Type": "application/json" });
res.end("Updated Successfully");
});
};


and I am return "200", but it always returns in the error clause shown below:

$.ajax({
url: "/api/action/SaveAction",
type: "PUT",
data: ko.toJSON(self.stripDownObj()),
datatype: "json",
contentType: "application/json; charset=utf-8",
success: function (result) {
console.log(result);
if(result.status == 200){
self.isEditMode(!self.isEditMode());
}
},
error: function(result){
console.log(result);
}
});


Note: the sql query is successful and does save the data.

Answer

By returning JSON when you're expecting JSON

res.end('{"success" : "Updated Successfully", "status" : 200}');

and then

$.ajax({
     ....
    datatype: "json", // expecting JSON to be returned

    success: function (result) {
        console.log(result);
        if(result.status == 200){
            self.isEditMode(!self.isEditMode());
        }
    },
    error: function(result){
        console.log(result);
    }
});

In Node you could always use JSON.stringify to get valid JSON as well

var response = {
    status  : 200,
    success : 'Updated Successfully'
}

res.end(JSON.stringify(response));

Express also supports doing

res.json({success : "Updated Successfully", status : 200});

where it will convert the object to JSON and pass the appropriate headers for you automatically.

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