I'm getting this error:
Column count doesn't match value count at row 1
From the following code:
$name = $_GET['name'];
$description = $_GET['description'];
$shortDescription = $_GET['shortDescription'];
$ingredients = $_GET['ingredients'];
$method = $_GET['method'];
//$image = $_GET['image'];
$username = $_GET['username'];
$length = $_GET['length'];
$dateAdded = uk_date();
$conn = mysql_connect('localhost', 'dbname', 'pass');
mysql_select_db('dbname');
$query = sprintf("INSERT INTO dbname (id, Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')",
mysql_real_escape_string($name),
mysql_real_escape_string($description),
mysql_real_escape_string($shortDescription),
mysql_real_escape_string($ingredients),
//mysql_real_escape_string($image),
mysql_real_escape_string($length),
mysql_real_escape_string($dateAdded),
mysql_real_escape_string($username));
$result = mysql_query($query) or die(mysql_error());
What does the error mean?
You have 9 fields listed, but only 8 values. Try adding the method.
The number of column parameters in your insert query is 9, but you've only provided 8 values.
INSERT INTO dbname (id, Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')
The query should omit the "id" parameter, because it is auto-generated (or should be anyway):
INSERT INTO dbname (Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')
Your query has 8 or possibly even 9 variables, ie. Name, Description etc. But the values, these things ---> '', '%s', '%s', '%s', '%s', '%s', '%s', '%s')", only total 7, the number of variables have to be the same as the values.
I had the same problem but I figured it out. Hopefully it will also work for you.
Related
require('config.php');
$sql = sprintf(
"INSERT INTO users (fname,lname,email,contact,pwd,isTeacher ) VALUES ('%s', '%s', '%s', '%s', '%s', '%s')",
$conn->real_escape_string($fname),
$conn->real_escape_string($lname),
$conn->real_escape_string($email),
$conn->real_escape_string($contact),
$conn->real_escape_string($pwd),
$isTeacher );
$conn->query($sql);
// $sql = "INSERT INTO users (fname, lname, email, contact, pwd, isTeacher) VALUES ('$fname, '$lname', '$email', '$contact, '$pwd', '$isTeacher')";
// $conn->query($sql);
if($conn=='true') { echo "Registered successfully";}
else{ echo "Issue entereing data" . $conn->error; }
}
I am trying to add data in users table. if I add data using sprintf, it works and adds just fine. but when I add using commented out syntax, it says check MariaDb syntax to use near '$fname'. What am I doing wrong in the second syntax. why do i have to use sprintf always.
I'm trying to insert into an SQLite database file using PHP. I have an existing database file that I can verify is working with SQLite Database Browser. This is the error I get:
Warning: sqlite_query() [function.sqlite-query]: no such table: players in
C:\wamp\www\espnapi\getPlayers.php on line 17
if ($db = sqlite_open("nhl.db", 0666, $sqliteerror)) {
$sql = sprintf("INSERT INTO players (url, team, number, firstname,
lastname, position)
VALUES ('%s', '%s', '%s', '%s', '%s', '%s')", $aTempPlyr[0],
$aTempPlyr[1], $aTempPlyr[2], $aTempPlyr[3], $aTempPlyr[4], $aTempPlyr[5]);
sqlite_query($db, $sql);
}
I'm trying to make a registration page but PHP is telling me that I have the wrong parameters, which doesn't make sense unless I need to add a parameter for the auto-incremental primary ID key.
Here's my SQL query call:
mysql_query("INSERT INTO Users (username, password, fname, lname, email) VALUES ('%s', '%s', '%s', '%s, '%s')",
mysql_real_escape_string($username),
mysql_real_escape_string($password),
mysql_real_escape_string($first),
mysql_real_escape_string($last),
mysql_real_escape_string($email)) or die(mysql_error());
It gives me the wrong paramater count on the last line in this code block. Any ideas? I copied and pasted the row-names straight from my database.
my table is as follows:
id - int(11) - auto-incrementing
username - varchar(20)
password - varchar(20)
fname - varchar(35)
lname - varchar(35)
email - varchar(254)
You have formatted the SQL query as a sprintf() call, but don't call sprintf()
mysql_query(sprintf("INSERT INTO Users (username, password, fname, lname, email) VALUES ('%s', '%s', '%s', '%s', '%s')",
mysql_real_escape_string($username),
mysql_real_escape_string($password),
mysql_real_escape_string($first),
mysql_real_escape_string($last),
mysql_real_escape_string($email))) or die(mysql_error());
// also note some parentheses out of place ^^^^^^^^^^^^^^^^^^^^^^^^
It's PHP that's telling you off about parameters, not MySQL.
You've tried to use mysql_query like sprintf, which it is not. mysql_query accepts an optional database resource identifier, and the query string. Two parameters. That is all.
If you do want to use sprintf, then go for it:
mysql_query(
sprintf(
"INSERT INTO Users (username, password, fname, lname, email) VALUES ('%s', '%s', '%s', '%s, '%s')",
mysql_real_escape_string($username),
mysql_real_escape_string($password),
mysql_real_escape_string($first),
mysql_real_escape_string($last),
mysql_real_escape_string($email)
)
) or die(mysql_error());
But remember that the first argument to mysql_query is just a string. No magic.
mysql_query(
sprintf("INSERT INTO Users (username, password, fname, lname, email) VALUES ('%s', '%s', '%s', '%s, '%s')",
mysql_real_escape_string($username),
mysql_real_escape_string($password),
mysql_real_escape_string($first),
mysql_real_escape_string($last),
mysql_real_escape_string($email)))
or die(mysql_error()); // sprintf to build a final string of your query by given format and "or die statement" is outside the mysql_query function call.
I'm getting this error:
Column count doesn't match value count at row 1
From the following code:
$name = $_GET['name'];
$description = $_GET['description'];
$shortDescription = $_GET['shortDescription'];
$ingredients = $_GET['ingredients'];
$method = $_GET['method'];
//$image = $_GET['image'];
$username = $_GET['username'];
$length = $_GET['length'];
$dateAdded = uk_date();
$conn = mysql_connect('localhost', 'dbname', 'pass');
mysql_select_db('dbname');
$query = sprintf("INSERT INTO dbname (id, Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')",
mysql_real_escape_string($name),
mysql_real_escape_string($description),
mysql_real_escape_string($shortDescription),
mysql_real_escape_string($ingredients),
//mysql_real_escape_string($image),
mysql_real_escape_string($length),
mysql_real_escape_string($dateAdded),
mysql_real_escape_string($username));
$result = mysql_query($query) or die(mysql_error());
What does the error mean?
You have 9 fields listed, but only 8 values. Try adding the method.
The number of column parameters in your insert query is 9, but you've only provided 8 values.
INSERT INTO dbname (id, Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')
The query should omit the "id" parameter, because it is auto-generated (or should be anyway):
INSERT INTO dbname (Name, Description, shortDescription, Ingredients, Method, Length, dateAdded, Username) VALUES ('', '%s', '%s', '%s', '%s', '%s', '%s', '%s')
Your query has 8 or possibly even 9 variables, ie. Name, Description etc. But the values, these things ---> '', '%s', '%s', '%s', '%s', '%s', '%s', '%s')", only total 7, the number of variables have to be the same as the values.
I had the same problem but I figured it out. Hopefully it will also work for you.
I am creating an application where I am generating pins dynamically based on user's input and storing them into mySql database.
$sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status')
VALUES
for($i=0;$i<$npin;$i++)
{
('$pin[$i]','$ownerid', 'Free', '1');
}
;";
how can I do that?
$s = $pdo->prepare("INSERT INTO xy (a,b,c,d) VALUES (?,?,?,?)");
foreach ($pins as $i) {
$s->execute($i,$ownerID,"free",1);
}
Try this:
$sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ";
for($i=0; $i<sizeof($pin); $i++) {
if ($i>0)
$sql .= ", ";
$sql .= "('$pin[$i]', '$ownerid', 'Free', '1')";
}
Of course you need to escape the values of $pin in case they contain any characters which could mess with the SQL query.
Something like
$sql = sprintf( "INSERT INTO `tblpin` (`pinId`, `ownerId`, `usedby`, `status`) VALUES ('%s', '%s', '%s', '%s')",
generatePIN($pin),
mysql_real_escape_string($ownerId),
mysql_real_escape_string($usedBy),
mysql_real_escape_string( $status) );
or (edited for Conspicuous Compiler)
$pins = generatePINS($user); // ? however they're generated
foreach( $pins as $pin) {
$sql = sprintf( "INSERT INTO `tblpin` (`pinId`, `ownerId`, `usedby`, `status`) VALUES ('%s', '%s', '%s', '%s')",
$pin,
mysql_real_escape_string($ownerId),
mysql_real_escape_string($usedBy),
mysql_real_escape_string( $status) );
$result = mysql_query($sql);
}
where generatePIN is your function to make your pin based on whatever the heck you're basing it off of. or generatePINS returns an array of them