CODEWITHSUNDEEP
javascriptjqueryjqgridfree-jqgrid
jfprevot
jfprevot

Reputation: 49

jqgrid 4.13.1 search filter error

I use jqgrid 4.13.1. In the jqfiddle example http://jsfiddle.net/9ezy09ep/162/, when I try to filter the list like this (important : replace %tab% with the tab char, ascii value=9):

Customer ID > Contains > %tab%HANAR

I get the error in the file jquery-1.12.0.min.js : 

SyntaxError: JSON.parse: bad control character in string literal at line 1 column 67 of the JSON data

How to escape the tab char in the value field of the jqgrid filter ?

Upvotes: 1

Views: 211

Answers (1)

Oleg
Oleg

Reputation: 221997

I analysed the problem. The problem exist because of usage xmlJsonClass.toJson in the line of code of free jqGrid 4.13.1. The line come from old jqGrid (see here). I thought already to remove the usage of xmlJsonClass.toJson and to use only JSON.stringify, but I hold the usage of old methods hoping to hold better compatibility with old versions of jqGrid.

The problem is that JSON standard (see here) require to escape only " and \ symbols and some other symbols, like tab, could be escaped:

enter image description here

The method xmlJsonClass.toJson generates the string, which escapes tab (converts to two characters \t) and JSON.stringify don't do this.

The problem come later after one deserializes postData.filters using $.parseJSON, which calls JSON.parse internally.

I made modification of the code of the Searching Dialog to use JSON.stringify as the first choice (which exists in all modern web browsers and can be included in old web browsers by including json2.js). I will continue to use xmlJsonClass.toJson for the fallback scenario only.

I committed the fix to GitHub, which fixes the problem. See http://jsfiddle.net/OlegKi/9ezy09ep/163/, which uses the latest sources from GitHub.

Upvotes: 2

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