Question

Ok在试图将这一错误插入数据库时

"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near @email.com, UT, 84505, NOW(), 69.169.186.192) at line 1"

我可以指出这个问题。下面是我的发言守则。

$insert_query = sprintf("INSERT INTO contacts (first_name, last_name, email, state, zip, date, ip) VALUES (%s, %s, %s, %s, %s, NOW(), %s)",
                        $fname,
                        $lname,
                        $email,
                        $state,
                        $zip,
                        $ip);


$result = mysql_query($insert_query, $connection) or die(mysql_error());

我的议席结构如下:

    id int(11)                              
    first_name  varchar(100)                                 
    last_name   varchar(100)                             
    email   varchar(100)                                 
    state   varchar(3)                               
    zip int(10)                             
    date    datetime                                
    ip  varchar(255)

Answer 1

您需要引用说明中所有类型的栏目。将<代码>%<>/code>改为 %s in the sprintf Format.

Please read about SQL Injection if you haven t done so already.

Answer 2

这可能有助于你。

$insert_query = "INSERT INTO contacts set first_name =  $fname , last_name =  $lname , email =  $email , state =  $state , zip =  $zip , date = ". time() .", ip =  $ip )";


$result = mysql_query($insert_query, $connection) or die(mysql_error());

如果你想要检查询问

echo $insert_query;

Answer 3

如果你能够回过头来,那将有所帮助,但是,它希望你不要在var的参数上引用。

$insert_query = sprintf("INSERT INTO contacts (first_name, last_name, email, state, zip, date, ip) VALUES ( %s ,  %s ,  %s ,  %s ,  %s , NOW(),  %s )",
                        $fname,
                        $lname,
                        $email,
                        $state,
                        $zip,
                        $ip);

By the way, you have an extra column in your insert - NOW doesn t appear related to a column. I m assuming ZIP is a varchar column, not a number, by the way.

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