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向前台ajax返回错误的一种方法 #17

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sfPPP opened this issue May 4, 2018 · 0 comments

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sfPPP commented May 4, 2018

在异常处写：
return response.Response({'error_code': 'xxx'})
直接返回给前台，前台在接受时进行判断：

 success: function (res) {
                if (res.error_code == 'xxx')
                    $('#元素').html("{% trans 'x_xxx_error' %}");

用某元素显示异常

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