Re: parsing out quoted text

sam a écrit :

$str='bass "electric organ" bagpipes'; $parser($str); $query="SELECT * FROM table WHERE tb_instr = "bass" AND tb_instr = "electric organ" //<<quoted phrase AND tb_instr = "bagpipes"; Anybody know where I can just copy code that will do the above? you wish to split $str and keep what's between the " intact ?

you can split with spaces, then parse the array and treat difefrently if the first character of your element is a ". $query = "SELECT *FROM table WHERE "; $array = split ($str," "); foreach ($array as $element) { // "element if ( substr($element,0,1) == "\"" ) { $quoted_element = $element; } // element" elseif ( substr($element,strlen($element)-1,1) == "\"" ) { $query .= "tb_instr = \"".$quoted_element." ".$element."\" AND ";

      $quoted_element = "";

} // element after "element elseif ($quoted_element != "" ) { $quoted_element .= " ".$element; } else { $query .= "tb_instr = \"".$element."\" AND "; } } // remove the last AND // please check if 4 or 5 characters to remove, 5 should be fine $query = substr ($query,0,strlen($query)-5); print $query; N F

thanks