我试图从数据库中显示数据的输出,但它显示错误。我想创建将在同一页面上显示消息的论坛
我的错误发生在:
$ mysqli_query($ SQL);
注意:未定义的变量:mysqli_query
致命错误:函数名称必须是字符串
我的编码是
` if(trim($ _ POST [' forum_numberUser'])== NULL || trim($ _ POST [' forum_text'])== NULL)
{ // If the user didn't fill in all fields
echo "<h3>PLEASE COMPLETE THE FORM TO REGISTER</h3>";
}
else {
include('connect.php');
$_SESSION['FILLED'] = TRUE;
// username and password sent from form
$mynumberUser=$_POST['forum_numberUser'];
$mytext=$_POST['forum_text'];
// $mydate=$_POST['forum_date'];
// To protect MySQL injection (more detail about MySQL injection)
$mynumberUser = stripslashes($mynumberUser);
$mytext = stripslashes($mytext);
$mynumberUser = mysql_real_escape_string($mynumberUser);
$mytext = mysql_real_escape_string($mytext);
$mydate=date('Y-m-d H:i:s');
//$mytime=date('H:i:s');
$sql = "INSERT INTO forum VALUES(null, '$mynumberUser', '$mytext', '$mydate')";
$mysqli_query($sql); //HERE IS MY ERROR...
if(mysqli_query($conn, $sql) === TRUE) {
//&['forum_text']
echo "Successfully registered ";
}
else {
echo "<h4>Unable to register.</h4>";
}
}
}`
答案 0 :(得分:0)
在函数调用$
之前有mysqli_query()
,这是变量的语法,而不是函数。