I've been trying to insert a variable that has a comma in it to a SQL database for 30 minutes or so. I've echoed the variable, and the comma is there, but when it inserts, there's no comma!
Example (some code like mine):
$variable1 = "test";
$variable2 = "$variable1,";
$sql1 = "INSERT INTO table (`column`) VALUES ('$variable2')";
$query1 = mysqli_query($con,$sql1); //I dont think I need to put a con variable up there for an example code
And when I do:
echo $variable2;
The result is test, with the comma, but the data in the column is just test WITH NO COMMA.
Help please.
Thanks.
Edit:
Your Common Sense fixed it, apparently I needed brackets around '$variable2' so it's like:
$sql1 = "INSERT INTO table (`column`) VALUES (('$variable2'))";
Thanks Your Common Sense and everyone else who tried!
Well, the answer is simple.
It's your own code does remove this comma, either before insert or after fetch.
If you care to write a reproduceable test case, you will see that noone is taking your comma.
Test case means code that involves the behavior in question and nothing else. Not a single line of code beside insert and fetch:
$variable1 = "test";
$variable2 = "$variable1,";
$sql1 = "INSERT INTO users (username) VALUES ('$variable2')";
mysqli_query($db,$sql1);
$sql2 = "SELECT username FROM users WHERE username ='$variable2'";
$res = mysqli_query($db,$sql2);
$row = mysqli_fetch_row($res);
var_dump($variable1, $variable2, $sql1, $sql2, $row[0]);
run it, see it all with comma in place, and then search your own code for the comma trimming code
or may be you have just test without comma in your table, ans select this one all the time, instead of one with comma.
or whatever silly error of the like
Try it like this:
$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'world');
$stmt = $mysqli->prepare("INSERT INTO table ('column') VALUES (?)");
$stmt->bind_param($variable2);
/* execute prepared statement */
$stmt->execute();
This is more safe and will not trigger such strangeties. (Is that a word?)
What happens here is, that the query is send to the sql database and this returns a statement. The statement has some holes, these are the ?, in it.
When using bind_param you fill the holes and then you can execute.
This has a couple of advantages:
It is safe
You can reuse your statement
It is easier than string interpolation stuff
Try "INSERT INTO table ('column') VALUES ('" . $variable2 . "');"
Related
I've tried to follow several answers on this question but can't seem to get it to work for my specific problem.
I want to insert data but only if the flight_number doesn't exists already. How can I do that?
$sql = mysqli_query($con,
"INSERT INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`)
VALUES ('".$flight_number."', '".$mission_name."', '".$core_serial."', '".$payload_id."')"
);
Rob since you saying flight_number is a unique then you can use INSERT IGNORE
<?php
$sql = "INSERT IGNORE INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`) VALUES (?,?,?,?)";
$stmt = $con->prepare($sql);
$stmt->bind_param('isss',$flight_number,$mission_name,$core_serial,$payload_id);
if($stmt->execute()){
echo 'data inserted';
// INSERT YOUR DATA
}else{
echo $con->error;
}
?>
OR you could select any row from your database that equal to the provided flight number then if u getting results don't insert.
$sql = "SELECT mission_name WHERE flight_number = ? ";
$stmt = $con->prepare($sql);
$stmt->bind_param('i',$flight_number);
if(mysqli_num_rows($stmt) === 0){
// INSERT YOUR DATA
}
A unique index on flight number should do the trick.
CREATE UNIQUE INDEX flight_number_index
ON space (flight_number);
If you want to replace the existing row with the new one use the following:
$sql = mysqli_query($con,
"REPLACE INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`)
VALUES ('".$flight_number."', '".$mission_name."', '".$core_serial."', '".$payload_id."')"
);
Make note that I just copied your code and changed INSERT to REPLACE to make it easy to understand. PLEASE PLEASE PLEASE do not use this code in production because it is vulnerable to injection.
If you don't want to replace the existing row, run an insert and check for errors. If there is an error related to the index, the row already exists.
Disclaimer: I haven't tested any of this code, so there may be typos.
I am teaching myself php and MySQL, and right now I have a problem with MySQL.
I want to compare the phone number that the user put in with the phone number in MYSQL, and if it is in MYSQL to not register it again.
My code:
<?php
require_once 'connection/connection.php';
// Variables from HTML to php
$worker_Name = $_POST['workerNameFromHtml']; // worker Name
$worker_City = $_POST['workerCityFromHtml']; // workerCity
$worker_career = $_POST['workerCareerFromHtml']; // worker career
$worker_PhoneNumber = $_POST['workerPhonNumberFromHtml']; // worker Phone Number
$worker_SecondPhoneNumber = $_POST['workerSecondPhoneNumberFromHtml']; // worker Second Phone Number
$submt=$_POST['submitFromHtml'];
if($submt){
$qry = ( "SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '$worker_PhoneNumber'") or die(mysql_error());
$result = $connect->query($qry);
$num = $result->num_rows;
if ($num == 1) {
$here = "INSERT INTO workersTable VALUES('','$worker_Name','$worker_City','$worker_career','$worker_PhoneNumber','$worker_SecondPhoneNumber')";
$query = $connect->query($here);
print "Successfully added!";
}
else {print "This number has already been entered Thank you for your cooperation!";}}
$connect->close();
So far I have not found a solution to this problem.
your biggest problem here is that you are trying to include variables inside of a string.
"SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '$worker_PhoneNumber'"
If you want to do it this way, you need to concatenate your variables with your string.
"SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '".$worker_PhoneNumber."'"
Keep in mind if you do this you will want to sanitize your variables first to prevent SQL injections. Also, when you INSERT variables, you will actually want to use a prepared statement like this:
"INSERT INTO table_name (column1, column2, column3,...) VALUES (value1, value2, value3,...)"
where the 1st set of values are the names of your columns in the database and the second set are your PHP variables you are putting into it.
I need to insert encrypted values in mysql table, but when I use traditional pdo method to insert its inserting the data in wrong format. ex: I insert aes_encrypt(value, key) in place of inserting encrypted value its inserting this as string.
Following is the code :
$update = "insert into `$table` $cols values ".$values;
$dbh = $this->pdo->prepare($update);
$dbh->execute($colVals);
$arr = array("col"=>"aes_encrypt ($val, $DBKey)");
I know i am doing it wrong, but not able to find correct way.
You are almost there, here is a simplified version:
<?php
$sql = "insert into `users` (`username`,`password`) values (?, aes_encrypt(?, ?))";
$stmt = $this->pdo->prepare($sql);
// Do not use associative array
// Just set values in the order of the question marks in $sql
// $fill_array[0] = $_POST['username'] gets assigned to first ? mark
// $fill_array[1] = $_POST['password'] gets assigned to second ? mark
// $fill_array[2] = $DBKey gets assigned to third ? mark
$fill_array = array($_POST['username'], $_POST['password'], $DBKey); // Three values for 3 question marks
// Put your array of values into the execute
// MySQL will do all the escaping for you
// Your SQL will be compiled by MySQL itself (not PHP) and render something like this:
// insert into `users` (`username`,`password`) values ('a_username', aes_encrypt('my_password', 'SupersecretDBKey45368857'))
// If any single quotes, backslashes, double-dashes, etc are encountered then they get handled automatically
$stmt->execute($fill_array); // Returns boolean TRUE/FALSE
// Errors?
echo $stmt->errorCode().'<br><br>'; // Five zeros are good like this 00000 but HY001 is a common error
// How many inserted?
echo $stmt->rowCount();
?>
you can try it like this.
$sql = "INSERT INTO $table (col) VALUES (:col1)";
$q = $conn->prepare($sql);
$q->execute(array(':cols' => AES_ENCRYPT($val, $DBKey)));
i'am beginner in php and i have problem in insertion query
if(isset($id)){
$qry = "insert into user_to_birds(user_id,tax_id)values( 1 ,'.$id .') ";
$result = mysql_query($qry);
}
I'am connected to the database but the query didn't work.
Why it is not working? how can i correct it?
Don't create queries this way. It is very vulnerable to SQL injection.
Use a prepared statement instead. A prepared statement is precompiled, hence will not be subject to SQL injection.
$id = 99;
$tax = 8;
$stmt = $mysqli->prepare("insert into user_to_birds(user_id,tax_id)values(?,?)"));
$stmt->bind_param("ii", $user, $tax);
$stmt->execute();
.. work on it ..
$stmt->close();
ii stands for two integers. After that first part of the binding, telling which type of variables you use in which order, can you add the values of those variables to the statement. The values will be escaped automatically using this method.
if(isset($id)){
$qry = "insert into user_to_birds(user_id, tax_id)values('1','$id') ";
$result = mysql_query($qry);
}
Work like a charm.
I think your single quotes should be double quotes:
$qry = "insert into user_to_birds(user_id,tax_id )values( 1 ,".$id .") ";
You are confusing strings in PHP with strings in SQL (which is, admittedly, easy to do).
For how to insert into there's a nice article here
http://www.w3schools.com/php/php_mysql_insert.asp
INSERT INTO table_name (column1, column2, column3,...)
VALUES (value1, value2, value3,...)
//not sure if this will make a difference buy i would try a space between tax_id) and values(
also, im not sure if the way youve done it is wrong but i would have written like this
if(isset($id))
{
$qry = "insert into user_to_birds (user_id, tax_id)
values( '1' ,'".$id ."') ";
$result = mysql_query($qry);
}
look at string concatination aswell either have
" ' ' ".$variable." ' ' ";
in that fashion
As others have said, it looks like you're not using string concatenation correctly in your query. Try changing your query to something like:
$qry = "INSERT INTO user_to_birds (user_id,tax_id) VALUES ( 1 ,'$id') ";
Another possibility is that your $id variable isn't set. Try printing out the variale before doing the isset() check and that will tell you if you need to look at an earlier point in your code.
Finally, I'd recommend you look at mysqli functions rather than mysql.
http://php.net/manual/en/book.mysqli.php
You have some confusion in quotes: your string in " ", your sql value in ' ', but when you concatenate you need to close your string and write dot and variable, after this you need write dot, open string quotes again and write text if it needed. Your mistake - you didn't close string (") before concatenation and this leads to misinterpretation of the code. In this case your code will look like:
$qry = "insert into user_to_birds(user_id,tax_id)values( 1 ,'" .$id ."') ";
But you can not use concatenation,you can do it simply: PHP allows write your variable $id in string, without use concatenation:
$qry = "insert into user_to_birds(user_id,tax_id)values( 1 ,'$id') ";
<?php
$username = $_POST['username'];
$password = $_POST['password'];
if($username&&$password)
{
$connect = mysql_connect("CiniCraftData.db.55555555.hostedresource.com", "CiniCraftData", "*********") or die("Couldn't Connect");
mysql_select_db("CiniCraftData") or die ("Couldn't Find Database");
$query = "INSERT INTO CiniUsers ('username.CINIDAT') VALUES('$username')";
$result = mysql_query($query) or die("Error occurred.");
}
else die("Please enter a username and password.");
?>
For this part of the code:
$query = "INSERT INTO CiniUsers ('username.CINIDAT') VALUES('$username')";
The VALUES seem to not be working properly, I need whatever the string value of $username is to be inserted into my CiniUsers database. What do I need to do to make the code above work? I'm very new to php and sql syntax and the guides I'm finding online are all completely different from each other as if they keep updating php.
Try reviewing this part:
$query = "INSERT INTO CiniUsers ('username.CINIDAT') VALUES('$username')";
The syntax is:
$query = "INSERT INTO table (column) VALUES ('$strvar')";
What is the column name you wanted to insert into?
If it is username.CINIDAT then try removing the qoutes.
Like this:
$query = "INSERT INTO CiniUsers (username.CINIDAT) VALUES ('$username')";
or maybe your column is named username so:
$query = "INSERT INTO CiniUsers (username) VALUES ('$username')";
UPDATE
The query from your comment, change it to this:
$query = "INSERT INTO CiniUsers (username.CINIDAT) VALUES ('$username')";
The format for the SQL statement is as so:
INSERT INTO nameOfTable (column1, column2, column3, etc) VALUES ('column1', 'column2', 'column3', 'etc')
You MUST make sure that you are using the field names exactly as they are stored in MySQL.
Your SQL could appear like so:
$query = "INSERT INTO CiniUsers (username) VALUES('$username')";
OR
$query = "INSERT INTO CiniUsers (username) VALUES('{$username}')";
Another thing that may help is that your die() statement is not very helpful. Yes, it is a bummer when your php program quits early, but it will save you a lot of time and frustration if you know why it quit. Although you may still be learning PHP and MySQL and may not know what the errors mean, they will start to make sense the more you see them and can tell you whether your query was bad, the connection failed or many more things. Change to something like this:
$connect = mysql_connect("CiniCraftData.db.55555555.hostedresource.com", "CiniCraftData", "*********") or die("Couldn't Connect: mysql_error()");
mysql_select_db("CiniCraftData") or die ("Couldn't Find Database: mysql_error()");
...
$result = mysql_query($query) or die("Some kind of error occurred...Query failed: mysql_error()");
You find that seeing the mysql_error() will help you solve problems like this much faster.
USE phpMyAdmin to test your query out, your query may be working perfectly. It is really the only way to know for sure. Use the suggested SQL and replace the PHP variable with some dummy data like "testUsername_1". If the query works, you will have manually added the username to the db, if not, the problem lies in SQL statement.
Here is some documentation on SQL INSERT INTO statements if you need more details:
http://www.w3schools.com/sql/sql_insert.asp
I think you should use mysqli or pdo. This liberary you are using is deprecated.
That said, what is username.CINIDAT? I think this is where your problem is. It should be something like this
$query = "INSERT INTO CiniUsers (username) VALUES('$username')";
I am assuming that CiniUsers is the table name and username is the column name.
The simplest way is to build the query by concatenating the statement with the value.
$query = "INSERT INTO CiniUsers ('username.CINIDAT') VALUES('".$username."')";
Without validation, this is not a very good idea, or something like this is very easy.