Adding information from variables to database PHP - php

I'm trying to add $_SESSION['user_id'] into a database and, when I echo it in this function, it works fine. However, when I try to push it into my MySQL database, it adds the value 0. Really confused as to why. Thanks for any help!
function fill_team() {
$i = 1;
while ($i < 24) {
$first_name = first_name();
$last_name = second_name();
echo "<br>";
$add_names = mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`) VALUES ('$first_name', '$last_name', '.$_SESSION['user_id']'.)");
$i++;
}
}

Replace echo $row2['first_name']; with return $row2['first_name'];.
If you want to get some value from the function, you should use return operator to pass the value back. It has nothing to do with printing the value with print or echo.

There's some messy mixed quoting inside the SQL statement:
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '.$_SESSION['user_id']'.)");
Off the top of my head, I have no idea what SQL this will produce, but this should be written as:
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '$_SESSION[user_id]')");
or
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '".$_SESSION['user_id']."')");

Related

INSERT results in DUPLICATE ENTRY '0' FOR KEY PRIMARY

I am using a simple php script to insert data into database but it's failing. The query just doesn't become successful without showing a single error which is why I am unable to figure out the problem. Some expert here help me please.
echo $name." ".$email." ".$pass." ".$phone." ".$area." ".$specialization." ".$city." ".$latitude." ".$longitude;
The result of echo is normal - without any null elements.
$query = mysqli_query($conn, "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
if ($query) {
echo "Status: Registeration Successful!";
// creating directory for user and storing dummy profile picture
//mkdir('../profiles/'.$email_trim, 0777);
//$result_copy = copy("img/dp.jpg.jpg", "../profiles/".$email_trim."/dp.jpg.jpg");
} else {
echo "Status: Err";
}
This "Status: Err" is always printed. I don't know why.
P.S I have double checked the database the field labels are fine.
UPDATE 1:
I added the
die(mysqli_error($conn));
statement and it says "DUPLICATE ENTRY '0' FOR KEY PRIMARY'.
PROBLEM AND SOLUTION:
The issue was that I had an 'id' field which was primary key of the table but it was not set to AUTO_INCREMENT. So, whenever I tried to insert a new record, I was actually inserting entries with duplicate PKs which was the issue. I change it to AUTO_INCREMENT and it solved the problem.
It seems you try to insert a new element with a PK = 0, but there is already a record with this key !
What is the primary key of your table ? Do you use an "id" field which is not shown in your insert statement ? Is this field AUTO_INCREMENT ?
It would be helpful to see the structure of your 'users' table.
Wild guess: looks like you may have defined an "id" column (or with whatever other name) which is primary key with default value "0", but it's not auto increment. That way you can insert 1 row and it will get "0" as "id" column's value, but you cannot insert another row because it will also try to use default value "0", which cannot happen as primary key has to be unique.
If that is the case, then please alter users table and make sure that the primary key column is also 'auto increment'.
Please check the proper error by adding below mentioned code inside else:
echo mysqli_errno($conn) . '----' . mysqli_error($conn);
<?php
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ") or die(mysqli_error());
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
use ` Tick maybe because there's some reserved word in your fields.
I think you doubled your close parenthesis and do not put $conn inside the query..
$query = "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude')";
mysqli_query($query, $conn);
Something like this. I hope this helps

Why my database do not get vars sent from mysql_query();?

My connection to database works properly, but when I Send some information it does not update.
$name = mysql_escape_string($_POST['name']);
$mail = mysql_escape_string($_POST['mail1']);
$pass = mysql_escape_string($_POST['pass1']);
mysql_query("INSERT INTO `usrs` (`id`, `username`, `email`, `password`) VALUES (`NULL`, `$name`, `$mail`, `$pass`)");
I use to check data tables by:
MariaDB [BattleShip]> SELECT * FROM usrs;
output from this:
EMPTY set (0.00sec)
If your id is auto increment, don't include this column in query
...and replace value quote by '
mysql_query("INSERT INTO `usrs` (`username`, `email`, `password`) VALUES ('$name', '$mail', '$pass')");

PHP Mysql Query (INSERT) Issue

I am having an issue with a MySQL query as follows:
My script generates this as an example query:
INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('Test2', '123-456-7890', 'test#test.com', 'mesa', 'az', '04-14-2013')
Which if I drop directly into PHPMyA, works fine. However, the PHP script I am trying to use to send the query from my website is not working and I can't get it figured out. Here it is:
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
mysql_query($sql);
$result = mysql_query($sql);
if($result)
{
echo("<br>Data Input OK");
}
else
{
echo("<br>Data Input Failed");
}
Nothing makes it to the MySQL DB and no PHP errors are displayed, however, if I echo $sql I get the exact query I posted previously.
Just remove the single line mysql_query($sql); on your code and you will be fine.. But you should better start practicing PHP MySQLi which stands for PHP MySQL Improved, such:
$con = mysqli_connect($host, $user, $password, $password);
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
$result = mysqli_query($con, $sql);
if($result) {
echo("<br>Data Input OK");
} else {
echo("<br>Data Input Failed");
}
$sql = 'INSERT INTO Table_name (`id`, `name`) VALUES ("1", "php");
You are executing $sql twice in your script wich is causing the error, please remove
mysql_query($sql);
And it will be ready to go
I would also suggest to stop using mysql_query please switch to mysqli or PDO
Are you sure there is a valid connection (..mysql_connect())? Try using the full syntax like so..
$conn = mysql_connect(...);
$result = mysql_query($query, $conn);
Also try forcing a commit after you execute the statement -
$mysql_query("COMMIT", $conn);
You are running mysql_query twice. Reason of the error. Try running the following code.
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
$result = mysql_query($sql) or die(mysql_error());
if($result){
echo("<br>Data Input OK");
} else{
echo("<br>Data Input Failed");
}
use this
"INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$_POST[name]', '$_POST[phone]', '$_POST[email]', '$_POST[city]', '$_POST[state]', '$_POST[date]')";
Try to use mysql_query($sql,$con); instead of mysql_query($sql);.
if(isset($_POST['submit']))
{
$name=$_POST['name'];
$age=$_POST['age'];
$address=$_POST['address'];
$ins="insert into table_name(`name`,`age`,`address`)values('".$name."','".$age."','".$address."')";
mysql_query($ins);
echo 'data inserted successfully';
}

PHP foreach issue with empty fields

I have a big form
This form is processed by a PHP file called by a serialize jQuery function
foreach($_GET['claimant'] as $k=>$v) {
$insClaim = "INSERT INTO `cR_Claimants` (`memberID`, `ParentSubmission`, `Name`, `DOB`, `Company`, `Email`, `MainPhone`, `OtherPhone`, `MobilePhone`, `OwnershipPercentage`, `Address`, `ZIPcode`, `Country`) VALUES ('".$memberID."', '".$refNumb."', '".mysql_real_escape_string($v['name'])."', '".$v['DOB']."', '".mysql_real_escape_string($v['company'])."', '".$v['email']."', '".$v['mainPhone']."', '".$v['alternatePhone']."', '".$v['mobilePhone']."', '".mysql_real_escape_string($v['percentage'])."', '".mysql_real_escape_string($v['address'])."', '".$v['ZIP']."', '".$v['country']."')";
$resultinsClaim=mysql_query($insClaim) or die("Error insert Claimants: ".mysql_error());
}
The problem is that $_GET['claimant'] in certain cases can be empty. I mean that the relative field has not been entered at all.
When this happens obviously the Insert should not run when that specific $_GET['claimant'] is empty.
I tried the two following solutions, but they do not work, the Insert runs anyway, putting in my DB empty strings.
Please help.
foreach($_GET['claimant'] as $k=>$v) {
if($_GET['claimant'] != "") {
$insClaim = "INSERT INTO `cR_Claimants` (`memberID`, `ParentSubmission`, `Name`, `DOB`, `Company`, `Email`, `MainPhone`, `OtherPhone`, `MobilePhone`, `OwnershipPercentage`, `Address`, `ZIPcode`, `Country`) VALUES ('".$memberID."', '".$refNumb."', '".mysql_real_escape_string($v['name'])."', '".$v['DOB']."', '".mysql_real_escape_string($v['company'])."', '".$v['email']."', '".$v['mainPhone']."', '".$v['alternatePhone']."', '".$v['mobilePhone']."', '".mysql_real_escape_string($v['percentage'])."', '".mysql_real_escape_string($v['address'])."', '".$v['ZIP']."', '".$v['country']."')";
$resultinsClaim=mysql_query($insClaim) or die("Error insert Claimants: ".mysql_error());
}
}
AND
foreach($_GET['claimant'] as $k=>$v) {
if(!empty($_GET['claimant'])) {
$insClaim = "INSERT INTO `cR_Claimants` (`memberID`, `ParentSubmission`, `Name`, `DOB`, `Company`, `Email`, `MainPhone`, `OtherPhone`, `MobilePhone`, `OwnershipPercentage`, `Address`, `ZIPcode`, `Country`) VALUES ('".$memberID."', '".$refNumb."', '".mysql_real_escape_string($v['name'])."', '".$v['DOB']."', '".mysql_real_escape_string($v['company'])."', '".$v['email']."', '".$v['mainPhone']."', '".$v['alternatePhone']."', '".$v['mobilePhone']."', '".mysql_real_escape_string($v['percentage'])."', '".mysql_real_escape_string($v['address'])."', '".$v['ZIP']."', '".$v['country']."')";
$resultinsClaim=mysql_query($insClaim) or die("Error insert Claimants: ".mysql_error());
}
}
If $_GET['claimant'] is an array, you should ask for its length:
if (count($_GET['claimant']) > 0) { ... }
The check should be:
if(!empty($v)) {
// Stuff here
}
This is assuming that the GET variable actually contains an array of arrays.
Most likely you don't need the foreach.
This code is also vulnerable to SQL injection, all parameters needs to be escaped before entered into a SQL query
Try this instead:
$vals = $_GET['claimant'];
if(!empty($vals)) {
$query = "INSERT INTO `cR_Claimants` (`memberID`, `ParentSubmission`, `Name`, `DOB`, `Company`, `Email`, `MainPhone`, `OtherPhone`, `MobilePhone`, `OwnershipPercentage`, `Address`, `ZIPcode`, `Country`) VALUES ('".$memberID."', '".$refNumb."', '".mysql_real_escape_string($vals['name'])."', '".mysql_real_escape_string($vals['DOB'])."', '".mysql_real_escape_string($vals['company'])."', '".mysql_real_escape_string($vals['email'])."', '".mysql_real_escape_string($vals['mainPhone'])."', '".mysql_real_escape_string($vals['alternatePhone'])."', '".mysql_real_escape_string($vals['mobilePhone'])."', '".mysql_real_escape_string($vals['percentage'])."', '".mysql_real_escape_string($vals['address'])."', '".mysql_real_escape_string($vals['ZIP'])."', '".mysql_real_escape_string($vals['country'])."')";
$resultinsClaim=mysql_query($insClaim) or die("Error insert Claimants: ".mysql_error());
}
Not sure why you're using a foreach() loop here..._GET['claimant'] is probably not an array of values unless you have multiple fields on your form called claimant[].
Just do this:
$claimant = $_GET['claimant'];
if( $claimant != ""){
$insClaim = "YOUR REALLY LONG QUERY";
// etc.
}
ALSO: please, please, please use mysql_real_escape_string() on all incoming request parameters.

mysql query not inserting to database

i have a db query in php that is not inserting into database. Have used this format lots of times but for some reason its not working now. any ideas please
$query = "INSERT INTO `databasename`.`member_users` (`id`, `first_name`, `last_name`, `username`, `password`, `address1`, `address2`, `postcode`, `access`, `expires`) VALUES (NULL, '$fname', '$lname', '$email', '', '$add1', '$add2', '$postcode', '0', '')";
$result = mysql_query($query);
if($result){
echo"query inserted";
}else{
echo "nope";
}
Instead of echo "nope"; I suggest something like :
echo 'error while inserting : ['.mysql_errno().'] '.mysql_error();
echo 'query : '.$query;
This way you will be able to see the exact error and the query that was executed.
It can be a lot of things :
Constraint error with a foreign key
Data type error
Non-existent field
Wrong database or table name
Instead of...
$query = "INSERT INTO `databasename`.`member_users` ..."
do
$query = "INSERT INTO member_users ..."
Hope it works. :)
If databasename and member_users are variables then,
Instead of
$query = "INSERT INTO databasename.member_users...
do
$query = "INSERT INTO $databasename.$member_users...

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