[RESOLVED] pass a name into a function

[jasondavis]

Below is my function which will get
- a users image
- a users name
- a users online status

Some of my quiries already pull in a users name so instead of and extra query for name is there a way to pass the name from my page into this function when I call it

here is how I call it now

<?

show_user_photo($line['auto_id'],$photo_type='small',$user_name='yes')

//could I do something like this

show_user_photo($line['auto_id'],$photo_type='small',$user_nameSET='JasonDavis')

 ?>

function show_user_photo($user_id, $photo_type = 'big', $user_name = 'yes', $online = 'yes') {
	$LOCALIP = $_SERVER['REMOTE_ADDR'];
	$sql_check = "select auto_id from friend_reg_user where auto_id='$user_id'";
	$result_check = executeQuery($sql_check);
	if ($line_check = mysql_fetch_array($result_check)) {
		// checking for user exist
		$sql_photo = "select photo_name  from friend_user_photo  where userid='$user_id' and defaultphoto='yes' order by auto_id desc";
		$result_photo = executeQuery($sql_photo);
		$pic1 = '';
		$pic = '';
		$file = '';

	if ($photo_type == 'big') {
		$noimg = 'nopic.gif';
	} else {
		$noimg = 'noimagesmall.jpg';
	}
	if ($line_photo = mysql_fetch_array($result_photo)) {
		$file = "uploded_files/userphoto/" . $line_photo['photo_name'];
		if ($line_photo['photo_name'] != '' && file_exists($file)) {
			$fil_ext1 = pathinfo($file);
			$fil_ext = $fil_ext1["extension"];
			$fil_explode = '.' . $fil_ext;
			$arr = explode($fil_explode, $line_photo['photo_name']);
			if ($photo_type == 'big') {
				$pic1 = $arr[0] . "_thumb" . $fil_explode;
				$height = 130;
				$width = 160;

			} else {
				$pic1 = $arr[0] . "_thumb1" . $fil_explode;
				$height = 80;
				$width = 90;
			}
			if (file_exists("uploded_files/userphoto/" . $pic1)) {
				$img_name = $pic1;
			} else {
				$img_name = $noimg;
			}
		} else {
			$img_name = $noimg;
		}
	} else {
		$img_name = $noimg;
	}
	echo "<div align='center'><a href='?p=profile&userid=$user_id'><img src='uploded_files/userphoto/$img_name' border='0'></a></div>";
	if ($user_name == 'yes') {
		$sql = "select disp_name from friend_reg_user where auto_id='$user_id'";
		$name = getSingleResult($sql);
		echo "<div align='center' class='boldblue' valign='middle'>";
		$trunc_name = preg_replace("/(\S{15,}?)/", "$1<br />", $name);
		echo $trunc_name;
		if (check_online_status($user_id, $LOCALIP) && $online == 'yes') {
			echo "<BR><img src='images/online.gif' >";
		}

		echo "</div>";
	} else {
		if (check_online_status($user_id, $LOCALIP) && $online == 'yes') {
			echo "<div align='center' class='boldblue' valign='middle'><img src='images/online.gif' ></div>";
		}
	}
} else {
	if ($photo_type == 'big') {
		$noimg = 'noexist.jpg';
	} else {
		$noimg = 'noexist1.jpg';
	}
	echo "<div style='height:$height;width:$width; overflow:hidden' align='center'><img src='uploded_files/userphoto/$noimg'    border=0></div>";
}
}

[wilku]

call your function this way:

show_user_photo($line['auto_id'],'small','JasonDavis') ;

And change that if structure where the function checks $user_name for yes/no to check if $user_name is empty or not. If not use $user_name as user's name

[jasondavis]

yeah got it thanks