Object of class CI_DB_mysql_result could not be converted to string |
[eluser]Mitja B.[/eluser]
Code: $data['title'] = ":: BlueTraker :: Home ::"; and i get Quote:A PHP Error was encountered What is wrong here?
[eluser]xwero[/eluser]
You can't add the query object as content. You have to use the result method before you pass it on to the table library.
[eluser]Lotus18[/eluser]
[quote author="xwero" date="1223917015"]You can't add the query object as content. You have to use the result method before you pass it on to the table library.[/quote] What do you mean by result method? I also have same problem with the OP. I read this article but I guess this is wrong or maybe I don't really get it. Please give us some example on how to this or give me some sites regarding this problem. So far I have this code: Code: function comments(){ Any suggestions will be greatly appreciated ![]() Rey Sean
[eluser]Lotus18[/eluser]
As I woke up early this morning and after a series of trial and error at last I got it ![]()
[eluser]Jan_1[/eluser]
Could you please explain how you solved it? Got the same - Need a hint. Thank you!!!
[eluser]andrewtheandroid[/eluser]
[quote author="Jan from Hamburg" date="1235317167"]Could you please explain how you solved it? Got the same - Need a hint. Thank you!!![/quote] I know this is an old post but for anyone else wondering how to solve this.. What the first reply meant was that you can't convert the query object to a result. To do this you need to use the result() or result_array() method in the Generating Query Results userguide. Example: Where you are getting your result (eg. a model) Code: function getProfile($staffid) { Where you are parsing the results (eg. controller). and sending to view. Code: // Obtaining result
[eluser]seismic[/eluser]
This is how i make it: Controller Code: <?php Model: Code: <?php View: Code: <HTML><head></head> |
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