How can i save today's day to a database with php? - php

How can i save today's day to a database with php?
I'm trying with:
$datenow=date('Y-m-d');
$stmt = $this->db->prepare("INSERT INTO user (age,wname,slid,time,sex) VALUES (?, ?, ?, ?, ?)");
$stmt->bind_param("isdds", $year, $reasons,$appy_level,$datenow,$sex);
this isn't working it writes 0000-00-00 to the database.

Depending on your database setup, remove the PHP date and use the mysql built in NOW() method. This will require your time field to have a data type of one of DATETIME or similiar.
eg:
$stmt = $this->db->prepare("INSERT INTO user (age,wname,slid,time,sex) VALUES (?, ?, ?, NOW(), ?)");
$stmt->bind_param("isdds", $year, $reasons, $appy_level, $sex);

you can use
INSERT INTO user (age,wname,slid,time,sex) VALUES (?, ?, ?, CURDATE(), ?)

Related

PHP mysqli won't insert data into table

I'm new to using mysqli_stmt statements in PHP. No matter, what I try, I can't insert data into my table. I have 2 tables, one, I can insert the data in but this one does not work
if ($new_profile_query = mysqli_prepare($db,
"INSERT INTO profile (
uid,
first_name,
last_name,
biography,
interests,
date_of_birth) VALUES (
?,
?,
?,
?,
?,
?)")) {
//VALUES ('$user_numberr', '$fname', '$lname', '$biography', '$interests', '$dob')";
mysqli_stmt_bind_param($new_profile_query,
"issbbs",
$user_numberr,
$fname,
$lname,
$biography,
$interests,
$dob);
mysqli_stmt_execute($new_profile_query);
mysqli_stmt_close($new_profile_query);
}

Update all values except auto-increment field using PHP / MySql / PDO

I have a quick question...I am updating all values in a row using a prepared statement and an array.
When initially inserting, my statement looks like this (and works perfect)
$sql="INSERT INTO $dbtable VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
The first and last values are NULL as the first is an auto increment ID field and last is a timestamp field.
Is there a way to keep my UPDATE statement as simple as my INSERT statement like this...
$sql="UPDATE $dbtable SET (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?) WHERE `announcements`.`id` = $id LIMIT 1";
I realize this does not work as is due to the first value being an auto increment field, is there a value I could put into my array to 'skip' this field?
This may not be the best way to describe my question but if you need more info, please let me know!
Thank you in advance!
UPDATE has no "implicit columns" syntax like INSERT does. You have to name all the columns that you want to change.
One alternative you can use in MySQL is REPLACE:
REPLACE INTO $dbtable VALUES (?, ?, ?, ?, ?, ...)
That way you can pass the current value for your primary key, and change the values of other columns.
Read more about REPLACE here: https://dev.mysql.com/doc/refman/5.6/en/replace.html
Note that this is internally very similar to #Devon's suggestion of using two statements, a DELETE followed by an INSERT. For example, when you run REPLACE, if you have triggers, both the ON DELETE triggers are activated, and then the ON INSERT triggers. It also has side-effects on foreign keys.
The solution I can think of doesn't involve an UPDATE at all.
DELETE FROM $dbtable WHERE id = $id;
INSERT INTO $dbtable VALUES ($id, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?);
Since you don't want to use the UPDATE syntax, this would delete the row and add a new row with the same id, essentially updating it. I would recommend wrapping it in a transaction so you don't lose your previous row if the insert fails for any reason.

Insert Statement won't work

I was always using normal querys for inserting data into the database but now I want to make it with prepared statements. I'm already using statements to select data in all my files but insert never worked... And now I ran out of ideas again. Maybe someone can see what I did wrong.
$animeId = $_POST['animeId'];
$username = $_POST['username'];
$rating = $_POST['rating'];
$story = $_POST['story'];
$genre = $_POST['genre'];
$animation = $_POST['animation'];
$characters = $_POST['characters'];
$music = $_POST['music'];
//Datum auslesen
$date = date("Y-m-d H:i:s");
if($insertRating = $con->prepare("INSERT INTO anime_rating (animeId, rating, story, genre, animation, characters, music, user, date) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?"))
{
$insertRating->bind_param("iiiiiiiss", $animeId, $rating, $story, $genre, $animation, $characters, $music, $username, $date);
$insertRating->execute();
$insertRating->close();
}
You have an errant comma in your query:
music, user,) VALUES (?, ?, ?, ?, ?, ?, ?
^^^
HERE
It should be
music, user) VALUES (?, ?, ?, ?, ?, ?, ?
In the statement:
INSERT INTO anime_rating (
animeId,
rating,
story,
genre,
animation,
characters,
music,
user /* 8 columns */)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?") /* 10 parameters */
There are 8 columns listed to insert values into and 10 parameters specified in the values section. Also as pointed out there is the extra comma in the list of values.
The number of columns must match the number of parameters and the number of parameters binding in the following statement:
`$insertRating->bind_param("iiiiiiiss", $animeId, $rating, $story, $genre, $animation, $characters, $music, $username, $date);`
Two errors in the statement:
INSERT INTO anime_rating (animeId, rating, story, genre, animation, characters, music, user,) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?"
^ here and ^ ^
remove the comma
add a closing parentheses before the end of the string.
remove one ,?
Furthermore you should chop one is from the binding:
$insertRating->bind_param("iiiiiiss", $animeId, $rating, $story, $genre, $animation, $characters, $music, $username, $date);
if($insertRating = $con->prepare("INSERT INTO anime_rating (animeId, rating, story, genre, animation, characters, music, user, date) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?"))
The last (") should be placed after the first ) at the end
New code:
if($insertRating = $con->prepare("INSERT INTO anime_rating (animeId, rating, story, genre, animation, characters, music, user, date) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)")

mysqli prepared issue Number of variables doesn't match

I am receiving error 'Number of variables doesn't match number of parameters in prepared statement'
$stmt = $this->conn->prepare( "INSERT INTO user
(
st1, u1, e1, sa1,
h1, roles_id, name_titles_id, first_name,
last_name, phone, mobile, address_road,
address_area, address_region, post_code, city,
country_id, creation_date, activated_at, modified_date_time,
created_by, referred_by, gender, ad1, status
)
VALUES
(
?, ?, ?, ?,
?, ?, ?, ?,
?, ?, ?, ?,
?, ?, ?, ?,
?, ?, ?, ?,
?, ?, ?, ?, ?
)"
)
$stmt->bind_param('i',$st1);
$stmt->bind_param('s',$u1);
$stmt->bind_param('s',$e1);
$stmt->bind_param('s',$sa1);
$stmt->bind_param('s',$h1);
$stmt->bind_param('i',$roles_id);
$stmt->bind_param('i',$name_titles_id);
$stmt->bind_param('s',$first_name);
$stmt->bind_param('s',$last_name);
$stmt->bind_param('s',$phone);
$stmt->bind_param('s',$mobile);
$stmt->bind_param('s',$address_road);
$stmt->bind_param('s',$address_area);
$stmt->bind_param('s',$address_region);
$stmt->bind_param('s',$post_code);
$stmt->bind_param('s',$city);
$stmt->bind_param('i',$country_id);
$stmt->bind_param('s',$creation_date);
$stmt->bind_param('s',$activated_at);
$stmt->bind_param('s',$modified_date_time);
$stmt->bind_param('i',$created_by);
$stmt->bind_param('i',$referred_by);
$stmt->bind_param('s',$gender);
$stmt->bind_param('s',$ad1);
$stmt->bind_param('i',$status);
Edit:
Just make a small test and it confirms, we can't use multiple bind_param with mysqli.
Not work:
$stmt->bind_param('s',$a);
$stmt->bind_param('s',$b);
Work:
$stmt->bind_param('ss',$a, $b);
Hopefully it'll be useful for future searches.
Your problem is simple. You are trying to do the thing manually, while the number of data asks for the automated process. You have to make a program to create a query for you.
Suppose You have an array with data already. All you need is to define the list of fields to insert
$fields = "st1,u1,e1,sa1,h1,roles_id,name_titles_id,first_name,last_name,phone,";
$fields .= "mobile,address_road,address_area,address_region,post_code,city,";
$fields .= "country_id,creation_date,activated_at,modified_date_time,";
$fields .= "created_by,referred_by,gender,ad1,status" ;
$fields = explode(",",$fields);
and then use some programming. Luckily, it's already done:
include 'safemysql.class.php';
$db = new safeMysql();
$insert = $db->filterArray($_POST,$fields);
$db->query("INSERT INTO user SET ?u", $insert);
And yeah, you are using bind_param wrong way. Correct usage can be seen in the manual page.

properly sanitize multiple user inputs with mysqli

Im currently using mysqli, and I want a way to properly sanitize every single user input. Im looking for the most simple lightweight way to do this, as I understand that Im NOT supposed to use mysql_real_escape....
my query is like so
$stmt = $sql->prepare("INSERT INTO Persons (msg, ip, time, main, twit, city, lat, lon, lang)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)");
as i understand i'm supposed to use the function bindParam... If i use it like so, am i completley securing my user inputs?
$stmt->bind_param('sssssssss', $_POST[msg], ('$ip'), ('$date'), '$_POST[main]', '$_POST[twit]', ('$cit'), ('$lat'), ('$lon'), '$_POST[lang]');
$stmt->execute();
$stmt->close();
If this isn't securing my user inputs how do i properly do so?
You need to prepare the statement to be safe. Something like below (its probably not 100% but gives you an idea)
$sql = new mysqli("localhost", "my_user", "my_password", "world");
$stmt = $sql->prepare("INSERT INTO Persons (msg, ip, time, main, twit, city, lat, lon, lang)
VALUES
(?, ?, ?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param("sssssssss",$_POST[msg], $ip, $date, $_POST[main], $_POST[twit], $cit, $lat, $lon, $_POST[lang]);
$stmt->execute();
First of all you have to follow basic PHP syntax
'$_POST[msg]' would be inserted as a literal $_POST[msg] string, while you expecting a value for $_POST['msg'] variable.

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