dongwei4096 2012-09-04 21:35
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I'm using CodeIgniter for a project I'm working on.

I have an ajax call in a view like this:

    type: 'GET',
    url: 'extra/search/infojson/' + $(this).text().replace(/\s/g, "+");
    success: function(data) {
        /* Do something with that data */


infojson is a method in a controller that takes a parameter of 'username', does a search, and will return a JSON object. Is there any way I can return this data without having to create another view for it? This method will only be used to return such data from this one page, so I can't see why I need to create another view just for that. I've read about _output() but it didn't make any sense to me.

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3条回答 默认 最新

  • dotcraq3249 2012-09-04 21:45

    If you are compressing the output in your configuration file try to set $config[‘compress_output’] = FALSE;

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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