JSON error

Question

Hi I am working on a maintenance project and they have used JSON in that project. I am getting the following error Microsoft JScript compilation error: Syntax error in the code snippet. I am new to JSON and after adding watch window I got to know the error and that is expected '(' .But I am not getting the correct syntax. Following is the code snippet where I am getting the error: parse : function(jsonString) { // filter out while statement var js = jsonString; if (js.substr(09) == while(1);) { js = js.substr(9); } if (js.substr(02) == /*) { js = js.substr(2js.length-4); } return eval('('+js+')'); } Can anyone help me? Thanks Sheetal

sheetal · Answer

Hi whole code is : var JSON = { stringify: function stringify(arg) { var c i l s = '' v; switch (typeof arg) { case 'object': if (arg) { if (arg.constructor == Array) { for (i = 0; i v = stringify(arg[i]); if (s) { s += ''; } s += v; } return '[' + s + ']'; } else if (typeof arg.toString != 'undefined') { for (i in arg) { v = stringify(arg[i]); if (typeof v != 'function') { if (s) { s += ''; } s += stringify(i) + ':' + v; } } return '{' + s + '}'; } } return 'null'; case 'number': return isFinite(arg) ? String(arg) : 'null'; case 'string': l = arg.length; s = ''; for (i = 0; i c = arg.charAt(i); if (c >= ' ') { if (c == '\' || c == '') { s += '\'; } s += c; } else { switch (c) { case '\b': s += '\b'; break; case '\f': s += '\f'; break; case '
': s += '\n'; break; case '': s += '\r'; break; case '	': s += '\t'; break; default: c = c.charCodeAt(); s += '\u00' + Math.floor(c / 16).toString(16) + (c % 16).toString(16); } } } return s + ''; case 'boolean': return String(arg); case 'function' : // Added for use of jqGrid T. Tomov return arg.toString(); default: return 'null'; } } parse : function(jsonString) { // filter out while statement var js = jsonString; if (js.substr(09) == while(1);) { js = js.substr(9); } if (js.substr(02) == /*) { js = js.substr(2js.length-4); } return eval('('+js+')'); } } and I am getting error on last line: return eval('('+js+')'); as soon as I enter the any number in the textbox.