MYSQL On Duplicate Error - php

Question

My problem:
Error: Duplicate entry '2' for key 'cnt_2'
My table:
CREATE TABLE UploadLog(
id int(11) NOT NULL AUTO_INCREMENT,
date text NOT NULL DEFAULT '',
cnt int(11) NOT NULL DEFAULT 0,
UNIQUE INDEX (id, cnt),
PRIMARY KEY (id)
)
My php code:
$date1=date("Y-m-d");
$sql = "SELECT * FROM UploadLog WHERE date='$date1'";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
$dateindb=$row["cnt"];
if ($dateindb==0) {$dateindb=1;};
$sql = "INSERT INTO UploadLog (date, cnt) VALUES ('$date1', '$dateindb') ON DUPLICATE KEY UPDATE date=date, cnt=cnt+1";
if (mysqli_query($conn, $sql)) {} else {echo mysqli_error($conn)};
Data in MYSQL table:
+------+------------+-----+
| id | date | cnt |
+------+------------+-----+
| 1 | 2015-04-16 | 3 |
| 2 | 2015-04-17 | 2 |
| 3 | 2015-04-18 | 1 |
+------+------------+-----+
Current date = 2015-04-18.
Edited:
Command:
DESCRIBE DateLog
+-------+---------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------+---------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| date | text | NO | | NULL | |
| cnt | int(11) | NO | UNI | 0 | |
+-------+---------+------+-----+---------+----------------+
Command:
SHOW CREATE TABLE UploadLog
UploadLog | CREATE TABLE `UploadLog` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`date` text NOT NULL,
`cnt` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`id`),
UNIQUE KEY `id` (`id`),
UNIQUE KEY `cnt_2` (`cnt`),
KEY `cnt` (`cnt`),
KEY `id_2` (`id`),
KEY `cnt_3` (`cnt`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=latin1

Answer 1

I cannot explain where the error come from, but I expect, it is caused by different MySQL versions.
But first off all I would recommend to change the date column from text to char(10). This is because the content have a very defined length.
After that the table definition will look like this:
CREATE TABLE DateLog
(
id INT(11) NOT NULL AUTO_INCREMENT,
date CHAR(10) NOT NULL,
cnt INT DEFAULT 0 NOT NULL,
UNIQUE INDEX (id, date, cnt),
PRIMARY KEY (id)
);
To get your code working you have just to add the id into the insert query. For that your code will change a little bit to:
$date1 = date("Y-m-d");
$sql = "SELECT * FROM DateLog WHERE date='$date1'";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
// Check if entry for the current date already exists
if (null === $row) {
// If not, set id to NULL
$id = 'NULL';
$dateindb = 1;
} else {
// If it exist, use it
$id = $row["id"];
$dateindb = $row["cnt"];
}
// Insert with id
$sql = "INSERT INTO DateLog (id, date, cnt) VALUES ($id,'$date1', '$dateindb') ON DUPLICATE KEY UPDATE cnt=cnt+1";
if (mysqli_query($conn, $sql)) {
} else {
echo mysqli_error($conn);
}
To simplify the situation, there is no need for an auto increment primary key, because you can use the date column as one. From this it follows that you also don’t need an unique index, too.
So simply change your Table definition to:
CREATE TABLE DateLog
(
date CHAR(10) PRIMARY KEY NOT NULL,
cnt INT DEFAULT 1
);
After you can use the this to count the uploads:
$date1 = date("Y-m-d");
$sql = "INSERT INTO DateLog (date) VALUES ('$date1') ON DUPLICATE KEY UPDATE cnt=cnt+1";
if (mysqli_query($conn, $sql)) {
} else {
echo mysqli_error($conn);
}
Simple, or not?
Hope this helps.

Answer 2

The error is pretty obvious... you are trying to insert duplicate values in a column, defined as unique...
cnt=cnt+21 gives you 2, which you already have in your table.

Answer 3

The problem is the field cnt_2, which is UNIQUE. That means it can't have the same value for two different registers. You aren't setting it a value in the INSERT query, so it takes the default value, which already exist in another register.