Not sure how to parse this PHP/JSON object in Javascript/jQuery

Question

So from what I understand, because I created a JSON object in PHP using json_encode and then used echo to display it, I can access it directly as an object in JS, like this

.done(function(response) {
    var result = response;
    $(result).hide().prependTo('#messages').fadeIn('slow');
});`

However, how do I access the data within the object? My object contains error which will be either true or false, and error_message while contains the errors formatted as <li>error</li>

PHP returns this - {"error":true,"error_messages":" <li>Name too short (minimum of 4 characters)<\/li> <li>Name too short (minimum of 4 characters)<\/li>"}

How does your JSON output look like? Install Firebug and paste a sample output... — user1386320
– user1386320, Commented Jul 17, 2013 at 21:39
response.error and response.error_messages will store these values — apartridge
– apartridge, Commented Jul 17, 2013 at 21:40
@cantsay: To improve your question, use the "edit" link under the question, not comments. — T.J. Crowder
– T.J. Crowder, Commented Jul 17, 2013 at 21:42

T.J. Crowder · Accepted Answer · 2013-07-17 22:00:48Z

3

If your server returns the correct Content-Type header (application/json), jQuery will parse the response for you and give you an object, which you can use like this:

console.log(response.error_messages); // "<li>Name too short...

If your server does not return the correct Content-Type header, you can force the issue by supplying

dataType: "json"

in your $.ajax call.

Either way, the JSON you've quoted in the comment on your question is valid.

So it may be that you want:

if (response.error) {
    $(response.error_messages).hide().prependTo('#messages').fadeIn('slow');
}
else {
    // whatever you show when it's successful
}

But note that #messages must be a menu, ul, or ol element, since your error_message defines an li element.

edited Jul 17, 2013 at 22:00

answered Jul 17, 2013 at 21:41

T.J. Crowder

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7 Comments

Ian Over a year ago

I don't know if anyone agrees, but I think always explicitly setting dataType to what you expect it to be is a good idea, so that it's always parsed correctly. Then again, the better way may be to never do that, so you ensure your server is configured correctly (but then you're leaving possible failures)

T.J. Crowder Over a year ago

@Ian: Yeah, it's a judgment call. I lean toward making sure the server is sending back the right information, so all links in the chain (proxies and whatnot) have good information to work with. This is really easy with most server-side technologies (including PHP).

cantsay Over a year ago

thanks, but I tried alert(result.error_messages) and it says undefined

T.J. Crowder Over a year ago

@cantsay: Okay. So what do you see when you walk through this with the debugger? alert is not a very useful debugging tool. Your browser has a debugger built into it, with powerful tools for stepping through your code statement by statement, using breakpoints, etc. What do you see when you do that?

cantsay Over a year ago

okay, did console.log(response) and it shows TypeError: e is undefined (jquery.min line 4) then below it shows the response from PHP which I added to my original question with () around it. I tried changing error_messages to "test" as well

|

Ryan Endacott · Accepted Answer · 2013-07-17 21:41:32Z

1

You should just be able to use it as an object like normal. E.g:

result.error; // true or false
result.error_messages; // the error messages

You may need to parse it, but jQuery (which it looks like you're using) may do that for you. If it doesn't, use JSON.parse:

var myobj = JSON.parse(result);

answered Jul 17, 2013 at 21:41

Ryan Endacott

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