mysqli的語句，返回警告

Question

這裏是我的代碼：

include "db_conx.php"; 
$sql = mysqli_query("INSERT INTO table(column) VALUES('" . mysqli_real_escape_string($var) . "')"); 
if ($sql) {echo "connection successful"; 
} else { 
echo "failure"; 
} 

它返回這些錯誤：

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given 
Warning: mysqli_query() expects at least 2 parameters, 1 given 

我嘗試使用PDO，但沒有工作，要麼...

Answer 1

第一參數是mysql連接鏈接標識符，其次是字符串有關更多詳細信息，請訪問此鏈接：http://in2.php.net/manual/en/mysqli.real-escape-string.php。

include "db_conx.php"; 
$sql = mysqli_query(pass_your_connection_identifier_here ,"INSERT INTO table(column) VALUES('" . mysqli_real_escape_string(pass_your_connection_identifier_here, $var) . "')"); 
if ($sql) {echo "connection successful"; 
} else { 
echo "failure"; 
} 

Answer 2

你缺少連接標識符都mysqli_real_escape_string()和mysqli_query() 改變你的代碼，mysqli_query($connection,$sql)和mysqli_real_escape_string($connection,$string)

Answer 3

你缺少在兩個線路連接變量，這就是爲什麼你所面對的問題：

嘗試用此代替您的代碼：

//$con // it is your connection variable 
include "db_conx.php"; 
$sql = mysqli_query($con,"INSERT INTO table(column) VALUES('" . mysqli_real_escape_string($con,$var) . "')"); 
if ($sql) {echo "connection successful"; 
} else { 
echo "failure"; 
}