JSON.parse: unexpected character - what character am I missing? - php

Question

Environment: PHP 5.3.5 MySQL Server 5.5.8, jquery version 1.6
Using Ajax to auto-populate a dropdown list of countries.
I keep getting this error and I have tried numerous things. Such as surround the $results with "'$results'" before encoding. The error still persists.
Here is the an example of output:
array(1) {
[0]=>
array(4) {
["id"]=>
string(2) "45"
[0]=>
string(2) "45"
["nicename"]=>
string(16) "Christmas Island"
[1]=>
string(16) "Christmas Island"
}
}
Here is the ajax (I even tried to change success to complete - the error code is just duplicated if I do that.
$.ajax({
type: "POST",
url: "models/ajaxHandler.php",
data: {handler:"getCountries", nli:"-1"},
dataType: "json",
success: function(results){
//results = $.parseJSON(results);
var resultStr = "";
for(var x in results)
resultStr = resultStr + results[x];
alert("RESULT" + resultStr);
//populateDropDown(results);
},
error: function(xhr, status, error){
alert(xhr+ "| ajax failure: could not populate list of countires | " + status + " | error:" + error);
var xhrStr = "";
for(var x in xhr)
xhrStr = xhrStr + xhr[x];
alert(xhrStr);
}
});
After I encode the json string in php I am escaping for special characters like so:
if (!empty($results)){
$json = json_encode($results);
//$json = form_safe_json($json);
echo $json;
}
function form_safe_json($json) {
$json = empty($json) ? '[]' : $json ;
$search = array('\\',"\n","\r","\f","\t","\b","'") ;
$replace = array('\\\\',"\\n", "\\r","\\f","\\t","\\b", "\'");
$json = str_replace($search,$replace,$json);
return $json;
}

Answer 1

After I encode the json string in php I am escaping for special characters
You don't need to do that -- json_encode() does all the necessary escaping for you, and in fact doing so is probably breaking the valid JSON that json_encode() has produced for you.
[EDIT]
To be clear: PHP's json_encode() function produces valid JSON from any input. (The only thing you need to test for is false if it fails, but even that will parse correctly in jQuery if you echo it, since an empty string is valid JSON).
If your program echos the output of json_encode(), and nothing else, then your program will be serving valid JSON and will not get the JSON parsing error in your JS code.
If your program echos anything else, or if you modify the JSON string before sending it, you may very well get errors.
Things to watch out for:
Don't try to send multiple JSON strings one after the other using multiple calls to json_encode(). This will be invalid JSON. Encode everything you want to send using a single call to json_encode().
Beware of PHP sending unwanted characters (particularly white space and UTF-8 BOM characters) which can cause errors in many situations.
If errors persist, load your JSON URL into the browser direct and view the source. You may see the error straight away. If not, copy and paste the JSON string into one of the JSON test sites on the web and see what it reports. This may help explain the problem.
If you're on PHP 5.4, you can use the PRETTY_PRINT option in json_encode(). This may help you with your debugging.

Answer 2

Perhaps json_encode() could be of some use? http://php.net/manual/en/function.json-encode.php

Answer 3

I am not sure what you try to achieve with the form_safe_json command.
The text string returned from:
$json = json_encode($result);
will contain correctly formated json and should not be further escaped in case you wish for the Javascript to parse it correctly. The escaping made by form_safe_json will break the json.