Php 返回警告的Mysqli语句
这是我的密码:Php 返回警告的Mysqli语句,php,mysqli,Php,Mysqli,这是我的密码: 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"; } 它返回以下错误: Warning: mysqli_real_escape_string() expe
include "db_conx.php";
$sql = mysqli_query("INSERT INTO table(column) VALUES('" . mysqli_real_escape_string($var) . "')");
if ($sql) {echo "connection successful";
} else {
echo "failure";
}
它返回以下错误:
Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given
Warning: mysqli_query() expects at least 2 parameters, 1 given
我尝试使用PDO,但也没用…第一个参数是mysql连接链接标识符,第二个是字符串。有关详细信息,您可以访问此链接:
您在
mysqli\u real\u escape\u string()
和mysqli\u query()
将代码更改为
mysqli\u query($connection,$sql)
和mysqli\u real\u escape\u string($connection,$string)
两行都缺少连接变量,这就是您面临问题的原因:
尝试用以下内容替换代码:
//$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 ($sql) {echo "connection successful";
} else {
echo "failure";
}
只需在函数中传递连接标识符
//$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";
}