Error on constructing JSONArray from String

Question

I am trying to construct a JSONArray from a String, and extract the JSONObjects from inside it. Here is my code:

String jsonStr = "[{\"name\" : \"John Doe\",\"gender\":\"male\",\"age\":40},{\"name\" : \"Jane Doe\",\"gender\":\"female\",\"age\":30}]";

JSONArray jsonArr = new JSONArray(jsonStr);

I get the following error on the line JSONArray is declared:

Exception in thread "main" java.lang.RuntimeException: Uncompilable source code
        at org.json.JSONTokener.nextValue(JSONTokener.java:319)
        at org.json.JSONArray.<init>(JSONArray.java:119)
        at org.json.JSONArray.<init>(JSONArray.java:146)

I am thinking it should be possible to have an array of JSONObjects. Is there something wrong with the way I have constructed the string ? Any help is appreciated.

Thanks!

I get the same error, if the JSON source code is in a separate JAVA library project. I moved the source code for the JSON classes to the same project as the main/calling code and I no longer get the error? — simon
– simon, Commented Sep 5, 2012 at 21:37

Joe · Accepted Answer · 2011-09-10 01:49:05Z

2

You're missing the trailing ]

answered Sep 10, 2011 at 1:49

Joe

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4 Comments

Joe Over a year ago

+1, but he's using java not javascript. single quotes are chars

Joe Over a year ago

Oh, good call. It wasn't tagged Java/Javascript, and I've always used JSON in JS, so I assumed JS heh. Is it possible to do it the other way round then, or will Java try and parse the keys as chars there? And totally totaly unrelated, but nice name :)

ds13 Over a year ago

Sorry about the typo with the missing ]. I am getting the same error with it as well.

Joe Over a year ago

That's a good question, beyond my limited knowledge. I would assume it would error because double quotes are the standard for JSON.

confucius · Accepted Answer · 2011-09-10 02:05:57Z

2

 String jsonStr =" {\"array\": [{\"name\": \"John Doe\",\"gender\": \"male\",\"age\":40},{\"name\": \"Jane Doe\",\"gender\": \"female\",\"age\": 30 }]}";

this string is fine

if you want to validate any JSON syntax data this can help you

edited Sep 10, 2011 at 2:05

answered Sep 10, 2011 at 1:50

confucius

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5 Comments

Joe Over a year ago

According to the docs, the string should start with [ and end with ]. If the code you posted works, it's probably worth submitting a bug report or whatever :)

ds13 Over a year ago

Its funny how its possible to initialize a JSONObject from a String, but not a JSONArray of JSONObject.

ds13 Over a year ago

Thanks Nammari, for the link. If I remove the escape character "\" from the string, JSONLint says its a valid JSON string.

confucius Over a year ago

if i answer you check my answer in the voting area :) this way how you can thank people in stackoverflow

ds13 Over a year ago

I am new to stackoverflow. Apparently you need to garner 15 reputation points to vote up. Will do so as soon as I have them!

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