PHP mysql_result错误[重复]
This question already has an answer here:
My SESSION username is Hassan
This is my PHP code:
$user = $_SESSION['username'];
echo $user;
$sql = mysql_query('SELECT `full_name` FROM `users` WHERE username="$user"');
$full_name = mysql_result($sql,0);
echo $full_name;
To make sure there is a $_SESSION['username'] I echo the $user and the result I get is perfect, this is what I get when I goto home.php:
Hassan
Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 5 in C:\xampp\htdocs\PHP\Projects\UserSys\home.php on line 15
I have now ticked the correct answer, question solved close post.
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此问题已经存在 这里有一个答案： p>
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警告：mysql_result（）[function.mysql-result]：无法跳转到profile.php上MySQL结果索引5的第0行 第11行
1回答
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My SESSION用户名是Hassan p>
这是我的PHP代码： p>
$ user = $ _SESSION ['username']; echo $ user; $ sql = mysql_query（'SELECT`full_name` FROM`users` WHERE username =“$ user”'）; $ full_name = mysql_result（$ sql，0）; echo $ full_name; code> pre>确保有
$ _ SESSION ['username'] code>我回显$ user code>和r 我得到的是完美的，这是我在转到home.php时获得的： p>Hassan p>
警告：mysql_result（ ）：无法跳转到第15行的C：\ xampp \ htdocs \ PHP \ Projects \ UserSys \ home.php中的MySQL结果索引5的第0行 p> blockquote>
我现在已经勾选了正确答案，问题已经解决了。 p> div>
You are trying to stick your $user variable in a single-quote string. It won't be evaluated.
Simply change your query to use double quotes:
$sql = mysql_query("SELECT `full_name` FROM `users` WHERE username='$user'");
Though while this should fix your immediate issue, please realize you are using deprecated mysql_* functions, and potentially opening yourself to SQL injection. You should be escaping your query params at least, ideally using prepared statements.