Question

$sql = "UPDATE 'biz_feedback'
    SET
        advice1 = '".mysql_real_escape_string($advice1)."',
        advice2 = '".mysql_real_escape_string($advice2)."'
    WHERE
        review_id = '".$_POST["review_id"]."'
        AND biz_id = '".$_POST["biz_id"]."'
        AND username = '".$_POST["post_review_username"]."'";

$result = mysql_query($sql) or die('Error, query'.$sql.'failed'.mysql_error());

带来此错误

Error, query

UPDATE 'biz_feedback'
SET
    advice1 = 'No issues but would be good to have more help to hand so person doesn\'t have to go hunting for help',
    advice2 = ''
WHERE
    review_id = '1117'
    AND biz_id = '2371881'
    AND username = 'user'

failed

错误，查询UPDATE＆＃39; biz_feedback＆＃39; SET advice1 =＆＃39;没有任何问题，但最好有更多的帮助，所以人们不必去寻求帮助＆＃39;，advice2 =＆＃39;没有更多建议谢谢＆＃39 ;在哪里review_id =＆＃39; 1117＆＃39; AND biz_id =＆＃39; 2371881＆＃39; AND username =＆＃39;＆＃39; failed您的SQL语法错误;查看与您的服务器版本对应的手册，以便在＆＃39;＆quot; biz_feedback＆＃39;附近使用正确的语法。 SET advice1 =＆＃39;没有任何问题但最好有更多的帮助＆＃39;在第1行

Answer 1

你在这里错位了你的点.mysql_real_escape_string (.$advice2)。最后一个应该在(之后，所以它看起来像.mysql_real_escape_string($advice2).

您的查询的下一个问题是'在表名周围签名。你应该跳过它或使用反引号`而不是。

语法无效的问题my_real_escape_string

1 个答案: