mdmartiny Posted March 25, 2012 Share Posted March 25, 2012 Hello Everyone, I am working on a project where I have to go in and grab the names of the fields in a table. I got it to work and for it to show me the field names but it also displays a lot more than what I would like for it to do. I keep getting the following error. It is repeated over my screen like 15 times. Warning: mysql_field_name() [function.mysql-field-name]: Field 5 is invalid for MySQL result index 7 in C:\xampp\htdocs\CMS\admin\includes\get_table.php on line 44 This is the section of code that it is referring to. Line 44 is $tablecolumn .= "<p>" . mysql_field_name($fieldnamesquery_result, $table_field) . "<input type='text' name='" . mysql_field_name($fieldnamesquery_result, $table_field) . "' id='" . mysql_field_name($fieldnamesquery_result, $table_field) . "'/>"; $tablecolumn = "<form name='insert_table' id='insert_table' action='get_database.php' method='post' > <fieldset> <legend>table information</legend>"; for ($table_field = 2; $table_field < $fieldnamesquery_result; $table_field++) { $tablecolumn .= "<p>" . mysql_field_name($fieldnamesquery_result, $table_field) . "<input type='text' name='" . mysql_field_name($fieldnamesquery_result, $table_field) . "' id='" . mysql_field_name($fieldnamesquery_result, $table_field) . "'/>"; <-----This is line 44 } $tablecolumn .= " <p> <input type='submit' name='submit_info' id='submit_info' /> </p> </fieldset> </form> "; echo $tablecolumn; Quote Link to comment https://forums.phpfreaks.com/topic/259672-for-loop-giving-to-much-out-put/ Share on other sites More sharing options...
Niixie Posted March 25, 2012 Share Posted March 25, 2012 Can you send your full code with database connection and query, it'll be easier to picture what happens then. Quote Link to comment https://forums.phpfreaks.com/topic/259672-for-loop-giving-to-much-out-put/#findComment-1330887 Share on other sites More sharing options...
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