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Undefined index

 
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Hello

I working in a small project and i'm having an error Undefined index;



In line 34 and 35



What's wrong please?

Thank you
 
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Does this work?
 
Gil Carvalho
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Hi

No...

Now i have this in my logCat

02-20 14:43:43.704: E/Buffer Error(22965): Error converting result java.lang.NullPointerException: lock == null

Can't get what is wrong..

And this is my log from Android



 
Roel De Nijs
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Maybe you should first explain what you are trying to do... instead of posting some chuncks of code and let us guess about the purpose of the code
 
Gil Carvalho
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Hi

This is a login system for android.
It uses php, json e android.

Now the undefined index is resolved but i have other issues and i don't know why.

My php is this one after correcting the undefined index





And the error


The Json parse doesn't work....
 
Roel De Nijs
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But which json is invalid? I assume the var_dump of the user.
 
Gil Carvalho
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No

var_dump($user); prints the following data:



so the query works, the problem is here, response variable seem to be empty...because i don't receive any data in android app



so what i'm doing wrong?

Thank you
 
Roel De Nijs
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Based on the exception message Value array(6) of type java.lang.String cannot be converted to JSONObject, it seems to be about the var_dump of user, that's an array of 6 elements (and not a json object)
 
Gil Carvalho
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The var_dump was just to check the values, if i remove it i get this error in android app




org.json.JSONException: No value for user

So....i'm lost

 
Roel De Nijs
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And how does the json-object looks like that you are returning?
 
Gil Carvalho
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Json should return

 
Roel De Nijs
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Gil Carvalho wrote:Json should return


Does your php code returns this json?
 
Gil Carvalho
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No

Returns

{"tag":"login",
"success":1,
"error":0

}
 
Roel De Nijs
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That explains your error, because there is no user in the json-string and you try to retrieve the "user" key as a JSONObject.
 
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