Xtremer360 Posted March 17, 2010 Share Posted March 17, 2010 I have a field in my database called height and obviously an example would look like 5'6". I am calling that row and it shows up as 5'6 and doesn't include that following quotation mark. I'm supposed to escape it somehow I think but not sure how. Any tips? Link to comment https://forums.phpfreaks.com/topic/195597-escaping-a-quotation/ Share on other sites More sharing options...
premiso Posted March 17, 2010 Share Posted March 17, 2010 Look into mysql_real_escape_string as that will escape any information going into the database properly. Link to comment https://forums.phpfreaks.com/topic/195597-escaping-a-quotation/#findComment-1027753 Share on other sites More sharing options...
Xtremer360 Posted March 17, 2010 Author Share Posted March 17, 2010 Still doesn't show that following quotation. Am I dong something wrong still? $query = "SELECT s.hometown, s.height, s.weight, h.kowtitles, h.kowawards, w.nicknames, w.finisher, w.setup, w.music FROM `efed_bio_singles` AS s, `efed_bio_history` AS h, `efed_bio_wrestling` AS w WHERE s.bio_id = '$defaultcharacterid' AND h.bio_id = '$defaultcharacterid' AND w.bio_id = '$defaultcharacterid'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $height = mysql_real_escape_string($height); And then this in the form: <?php echo $row['height']; ?> Link to comment https://forums.phpfreaks.com/topic/195597-escaping-a-quotation/#findComment-1027760 Share on other sites More sharing options...
Xtremer360 Posted March 17, 2010 Author Share Posted March 17, 2010 Did some research and came up with <?php echo htmlspecialchars($row['height']); ?> and it works. Link to comment https://forums.phpfreaks.com/topic/195597-escaping-a-quotation/#findComment-1027773 Share on other sites More sharing options...
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