hansman Posted July 7, 2008 Share Posted July 7, 2008 Hello, i currently have a for with a list of radio buttons. I would like the user to select once unique button, and that will determine what table the data is entered into. My radio buttons look like this: <input type="radio" name="Type" value="Crawling_Insects" id="Crawling_Insects" /> Crawling Insects</label> So, for this radio button it will submit date to Crawling_Insects table. Now on the submit to database page i do this. $Type=$_POST['Type']; ... $sql="INSERT INTO 'Type'(data form previous page)VALUES(Variables) Any idea on whats wrong here? Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/ Share on other sites More sharing options...
wildteen88 Posted July 7, 2008 Share Posted July 7, 2008 $sql="INSERT INTO 'Type'(data form previous page)VALUES(Variables) should be $sql="INSERT INTO $type(data form previous page)VALUES(Variables) Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/#findComment-583854 Share on other sites More sharing options...
hansman Posted July 7, 2008 Author Share Posted July 7, 2008 still does not work $sql="INSERT INTO $Type( i put that in and i still get unsuccessful insert Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/#findComment-583866 Share on other sites More sharing options...
wildteen88 Posted July 7, 2008 Share Posted July 7, 2008 You might want to use the following instead. $sql = 'INSERT INTO `'.$type.'`(data form previous page)VALUES(Variables)'; Before using $sql in mysql_query, echo $sql to see if it is being populated properly. Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/#findComment-583867 Share on other sites More sharing options...
mbeals Posted July 7, 2008 Share Posted July 7, 2008 do what wildteen said with the echo command, but you may also need to encapsulate the table name. Try: $sql="INSERT INTO `$type` (data form previous page)VALUES(Variables)"; [\code] and make sure the sql in the 'data from previous page' and 'variables' is correct. Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/#findComment-583871 Share on other sites More sharing options...
hansman Posted July 7, 2008 Author Share Posted July 7, 2008 thanks alot, echoing $SQL made me find my error, i had another form using the "Type" variable SOLVED Quote Link to comment https://forums.phpfreaks.com/topic/113620-solved-using-a-form-to-insert-data-into-a-database/#findComment-583879 Share on other sites More sharing options...
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