如果使用php＆amp; Mysql的

Question

我需要你对PHP中if语句的一些帮助，因为我是新手

首先，我有这个问题：

$sqlCmd = "SELECT user.name,user.level 
    FROM user.user 
    WHERE user.id = '".$_GET['char']."' 
    LIMIT 1";

现在我有第二个查询

visible $sqlUpdate = "UPDATE user.user 
    SET level='1' 
    WHERE id='".$_GET['char']."' 
    LIMIT 1";

我想要做的是这样的：if level>1运行$sqlUpdate，否则会打印错误。

Answer 1

试试这个

visible $sqlUpdate = "UPDATE user.user 
                      SET level='1' 
                      WHERE id='".mysql_real_escape_string($_GET['char'])."' 
                      AND level >1 
                      LIMIT 1";

参考这些链接，他们可以帮助您学习 http://sqlzoo.net/，http://beginner-sql-tutorial.com/sql.htm

Answer 2

visible $sqlUpdate = "UPDATE user.user SET level='1' WHERE id='".mysql_real_escape_string($_GET['char'])."' AND level >1 LIMIT 1";

你的代码很容易被sql注入，你需要正确地转义所有请求

---

警告

您的代码容易受sql injection攻击，您需要撤消所有get，post和request，更好的方法是使用Prepared statement

好读

1. How to prevent SQL injection in PHP?
2. Are PDO prepared statements sufficient to prevent SQL injection?

---

注意

1. 整个 ext/mysql PHP扩展程序（提供名为mysql_的所有函数）为officially deprecated as of PHP v5.5.0，将来会被删除。因此，请使用PDO或MySQLi

好读

1. The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead
2. PDO Tutorial for MySQL Developers
3. Pdo Tutorial For Beginners

Answer 3

$sqlCmd = "SELECT user.name,user.level FROM user.user WHERE user.id = '".$_GET['char']."' AND level>1 LIMIT 1";
if(mysql_num_rows($mysql_con)===1){
//run second query
}else{echo'error:'}

Answer 4

我认为应该这样做......

$result = mysql_query($sqlCmd);
if($row = mysql_fetch_object($result)) {
    if ($row->level > 1)
    {
      mysql_query($sqlUpdate);
      //Might want to add a check here if it was successful... 
      echo "Done";
    }
    else
      echo 'error!!!';

}
mysql_free_result($result);

Answer 5

试试这个

$sqlCmd = "SELECT user.name,user.level 
FROM user.user 
WHERE user.id = '".$_GET['char']."' 
LIMIT 1";

$resource = mysql_query($sqlCmd);
$temp_arr = mysql_fetch_assoc($resource);
$level    = $temp_arr['level'];
if($level > 1)
{
    $sqlUpdate = "UPDATE user.user 
    SET level='1' 
    WHERE id='".mysql_real_escape_string($_GET['char'])."' 
    LIMIT 1";

    mysql_query($sqlUpdate);
 }

Answer 6

    mysql_query("UPDATE user.user SET level='1' WHERE id='".$_GET['char']."' and level>1 LIMIT 1");