I getting error "Query error!" with this code:
$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1
ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1"));
echo "<br>".$result;die;
I also check different example like this example was running success but I found error in above code for my demo project.
The closing brace for VALUES is missing. There is also a non-needed brace at the end of your code line.
Please change
$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1 ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1"));
to
$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1) ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1");
Apparently you did not close the parentesis in the right place:
$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY,
UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1)
ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1");
Also... you should at least escape all the variables before concatenating in queries like this to avoid SQL injection
Related
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
I'm trying to grab SID from the insert into the first table (stories) so I can insert that SID into the writing table in my second insert.
I think the way to do this is with mysql_insert_id(); after the first query, but the primary key that auto-increments is called SID. If mysql_insert_id() could grab that value I'd be all set.
What I am finding from a var_dump is that the $SID = mysql_insert_id(); is just returning the value "0" and I'm not sure why.
There is a column called ID in stores, but if it was grabbing that, the value would be "1". Either way, I wish this method could be written as mysql_insert_SID();
Any idea what I am doing wrong or how I can fix this? And yes, I know this is a deprecated approach, but first I want to figure out how before I worry about converting to PDO.
// Get values from form
$category = $_POST['category'];
$genre = $_POST['genre'];
$story_name = $_POST['story_name'];
$text = $_POST['text'];
$query = "INSERT INTO stories (ID, category, genre, story_name, active) VALUES
('$user_ID', '$category', '$genre','$story_name', '1')";
$result = mysql_query($query);
$SID = mysql_insert_id();
$SID2 = "select stories.SID from stories where stories.SID=$SID";
$query2 = "INSERT INTO writing (ID, SID, text, position, approved)
VALUES('$user_ID', '$SID2', '$text', '1','N')";
$result = mysql_query($query2);
Retrieves the ID generated for an AUTO_INCREMENT column by the previous query (usually INSERT).
(http://php.net/manual/en/function.mysql-insert-id.php)
But you aren't executing any query (via mysql_query()). You're just assigning your query to a variable. Try following:
$query = "INSERT INTO stories (ID, category, genre, story_name, active) VALUES
('$user_ID', '$category', '$genre','$story_name', '1')";
mysql_query($query);
$SID = mysql_insert_id();
I think you've forgotten to execute the query most probably?
Try
$SID = mysql_insert_id();
after executing the query
$query = "INSERT INTO stories (ID, category, genre, story_name, active) VALUES
('$user_ID', '$category', '$genre','$story_name', '1')";
$result = mysql_query($query); // executing
$SID = mysql_insert_id(); // order of queries is important
If you cannot get the value through mysql_insert_id() then try SELECT LAST_INSERT_ID(). Of course there will be a value if you have executed an insert query with AUTOINCREMENT (which you haven't done yet)
I want to get the last id inserted but i don;t know how to get it in oracle. Any help will be appreciated. Below is my query.
$query = oci_parse($con,"INSERT INTO USER_LOGIN (USERNAME, PASSWORD, CNIC, ROLE_ID, PICTURE)
VALUES ('$Username', '$Password', '$CNIC', '$Role', '$Filename')");
Get above id: (don't know)
$OK = oci_parse($con,"SELECT LAST_INSERT_ID() FROM USER_LOGIN");
oci_execute($OK);
$row = oci_fetch_array($OK, OCI_ASSOC + OCI_RETURN_NULLS);
if($row)
{
$USERID = $row['USER_ID'];
}
It's not clear what you are trying to achieve (see Justin's questions). But if you want the trigger populated id of the user_login table back to your code you can use the returning clause.
$query = oci_parse($con,"INSERT INTO USER_LOGIN (USERNAME, PASSWORD, CNIC, ROLE_ID, PICTURE)
VALUES ('$Username', '$Password', '$CNIC', '$Role', '$Filename')
RETURNING YOUR_ID
INTO :LAST_INS_ID");
oci_bind_by_name($query, ':LAST_INS_ID', $theNewID, 8);
"INSERT INTO forum_topics (category_id, poster_id, poster_username, topic_title, topic_content, date) VALUES (".$category_id.", '$poster_id', '$topic_title', '$message', NOW()";
mysql_error() says that there is a problem with the syntax, however it might be something else. I'm gonna post the variables just so you know where they come from.
$message = $_POST['topic_message'];
$topic_title = $_POST['topic_title'];
$category_id = $_GET['id'];
EDIT
Changed it to
$topic_sql = "INSERT INTO forum_topics (category_id, poster_id, poster_username, topic_title, topic_content, date) VALUES (".$category_id.", '$poster_id', '$username', '$topic_title', '$message', NOW())";
However it still doesn't work...
You're missing the closing paren for VALUES:
... NOW())";
There are other issues:
The parameter count is incorrect
Your query is vulnerable to injection since you are not using parameterized queries with PDO/mysqli
Maybe you list 6 columns but only give data for 5? And missing closing ).
Looks like you're missing a closing parenthesis and only inserting 5 values into 6 columns...
INSERT INTO forum_topics (category_id, poster_id, poster_username, topic_title, topic_content, date)
VALUES (".$category_id.", '$poster_id', '$username', '$topic_title', '$message', NOW())
You missing the user name?
I'm trying to get the last inserted id of multiple inserted rows.
record_id is auto increment
$sql = "INSERT INTO records (record_id, user_id, status, x) values ";
$varray = array();
$rid = $row['record_id'];
$uid = $row['user_name'];
$status = $row['status'];
$x = $row['x'];
$varray[] = "('$rid', '$uid', '$status', '$x')";
$sql .= implode(',', $varray);
mysql_query($sql);
$sql2 = "INSERT INTO status_logs (id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES";
$varray2[] = "(' ', mysql_insert_id(), '$status', '$uid', '$x')";
$sql2 .= implode(',', $varray2);
mysql_query($sql2);
This is the result:
INSERT INTO records (record_id, user_id, notes, x) values ('', '1237615', 'this is a note', 'active')
INSERT INTO status_logs (log_id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES('', INSERT INTO records (record_id, user_id, notes, x) values ('', '1237615', 'this is a note', 'active')
INSERT INTO status_logs (log_id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES('', mysql_insert_id(), '1', '2013:05:16 00:00:01', '', this is a note'', '1237615', 'active'), '1', '2013:05:16 00:00:01', '', this is a note'', '1237615', 'active')
There is no value for mysql_insert_id().
You're mixing php function mysql_insert_id() and SQL INSERT statement syntax.
Either use MySQL function LAST_INSERT_ID() in VALUES clause of INSERT statement
INSERT INTO records (user_id, notes, x) VALUES('1237615', 'this is a note', 'active');
INSERT INTO status_logs (record_id, status_id, date, timestamp, notes, user_id, x)
VALUES(LAST_INSERT_ID(), '1', ...);
^^^^^^^^^^^^^^^^^
or retrieve the last inserted id by making a separate call to mysql_insert_id() right after first mysql_query(). And then use that value when you as a parameter to your second query.
$sql = "INSERT INTO records (user_id, ...)
VALUES(...)";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error()); //TODO beter error handling
}
$last_id = mysql_insert_id();
// ^^^^^^^^^^^^^^^^^^
$sql2 = "INSERT INTO status_logs (record_id, ...)
VALUES $last_id, ...)";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error()); //TODO beter error handling
}
Note:
You don't need to specify auto_incremented column in column list. Just omit it.
Use at least some sort of error handling in your code
On a side note: Instead of interpolating query strings and leaving it wide open to sql-injections consider to use prepared statements with either mysqli_* or PDO.
Unless I mis-reading your code, you're calling the PHP function mysql_insert_id from within the SQL?
What you need to do is grab that into a PHP variable first, then use the variable in the SQL. Something like this:
// Run the first query
mysql_query($sql);
// Grab the newly created record_id
$recordid= mysql_insert_id();
Then in the second INSERTs just use:
$varray2[] = "(' ', $recordid, '$status', '$uid', '$x')";