[SQL] Duplicate field '0' for key 1 Solved

Question

Hello, as mentioned in the title, I have this message ( Duplicate of field '0' for key 1 ) that appears after executing my query, but I can't see where the error is coming from because I don't believe there is a duplicate, and especially what key 1 refers to..

Thank you for your help

Configuration: Windows XP Firefox 3.0.10

dam75 · Accepted Answer

Hello,
you must NOT set the auto_increment field...
so
-> either you set it to NULL
-> or you use the syntax:
INSERT INTO Clients SET nom="$nom", prenom="$prenom", ...
WITHOUT setting the Id...

PS Also be careful to properly escape quotes and others if you are in php: see mysql_real_escape_string, mysql_escape_string, addslashes, ...

Good luck!

Absot · Answer

Does anyone know, please?

NitroGen® · Answer

I'm trying to see your problem, sorry maybe I'm wrong but well it's better than nothing lol ...
Is that field set to auto-increment and is the key set to primary mode?
The key 1 is related to the auto-increment field but I'm not sure!!!

Hoping this can give you a little help lol

Absot · Answer

Here is how I created my database:

CREATE TABLE Product (
idProduct int(11) NOT NULL PRIMARY KEY auto_increment,
name varchar(30) NOT NULL default '',
category varchar(30) NOT NULL default '',
reference varchar(30) NOT NULL default '',
photo varchar(50) NOT NULL default 'non_available.PNG',
priceHT int(11) NOT NULL default '0'
) ;

NitroGen® · Answer

Do you have Wamp server? It will give you the code in PHPMyAdmin, and why don't you do your database manually with PHPMyAdmin? It will create your MySQL database.

Absot · Answer

I'm used to using easyphp at my high school and I have successfully created this database with phpmyadmin but without results.. :/

I even have the problem that it doesn't display the list in the database even though it's the same adapted code as the one from my high school that works.. '-_-

NitroGen® · Answer

??? That's weird... are you sure the passwords, username, etc. are correct? does this database exist? is the name correct?  I can't say much, I've never used this function in a script...

Absot · Answer

Yes, normally everything is fine because in my code, I execute the command to connect to my database with a login and if it can't, it gives me an error message and it doesn't do it, so it should be fine logically..

Here is my code for help:
------------------------------------------------------------------------------------------------------

/* connection to the MySql server
replace localhost with the server's IP
then provide the login and password*/

$connexion = mysql_connect("localhost","root") or die("Unable to connect to the server");

/* connection to your database indicate your login name*/
mysql_select_db("Site") or die("Unable to connect to the database");

/* query to execute */
$query = "SELECT nom,categorie,reference,photo FROM Produit";

/* execution of the query */
$result = mysql_query($query) or die("Query failed");

echo "<table border='2' class='tableau_produit'>";
echo "<tr>
<th class='tableau_nom'>
Product name
</th>
<th class='tableau_nom'>
Product category
</th>
<th class='tableau_nom'>
Product reference
</th>
<th class='tableau_photo'>
Product photo
</th>
</tr>";

// as long as there are rows returned, we display the products
while ($line = mysql_fetch_array($result)) {

/* the mysql_fetch_array function returns the results in the form of an associative array in which we access the different elements as follows
*/

echo "<tr>";
echo "<td>$line[nom]</td>";
echo "<td>$line[categorie]</td>";
echo "<td>$line[reference]</td>";
echo "<td><img src=images/$line[photo] width=150px height=100px></td>";
echo "</tr>";

}

echo "</table>";

------------------------------------------------------------------------------------------------------

I put my code in bold and the rest is comments..

Product name	Product category	Product reference	Product photo
$line[nom]	$line[categorie]	$line[reference]

"; ------------------------------------------------------------------------------------------------------ I put my code in bold and the rest is comments..

Absot · Answer

Does anyone else know, please?

NitroGen® · Answer

Try putting a star for your query like this "SELECT * FROM Product" to test if it’s not the names that are wrong..

dam75 · Answer

Hello,

It's not a SELECT query that can return this kind of message (unless MySQL is seriously sick!), but an INSERT query...
In this case, your IdProduit table is a primary key, hence unique.

Since it is auto-increment, you shouldn't have to worry about it, but if you include it in an INSERT (manually or due to a poorly designed tool :), it will indeed cause a "duplicate entry ... key 1" error...
Key 1 refers to the first found entry, and there is only one in your table :)

Example of such a query:
INSERT INTO Produit SET IdProduit=1, nom="test";
--> run it twice in a row (assuming your table was empty before you started), and the second will return your error message...

Good luck!

Absot · Answer

Donc ma requête $query = "INSERT INTO Client values('', '$sexe', '$nom', '$prenom', '$mail','$mdp', '$naissance', '$adr', $telephone);"; is not correct and should be replaced by $query = "INSERT INTO Client values('$sexe', '$nom', '$prenom', '$mail','$mdp', '$naissance', '$adr', $telephone);";?

le père · Answer

Hello  No, the change to be made is not that one; it is sure not to work. Can you provide the definition of the Client table?

Absot · Answer

Here is my client table:
CREATE TABLE Client (
idClient int(11) NOT NULL PRIMARY KEY auto_increment,
civility varchar(4) default NULL '',
lastName varchar(50) default NULL '',
firstName varchar(50) default NULL '',
email varchar(50) NOT NULL default '',
password varchar(50) NOT NULL default '',
birthdate varchar(10) default NULL,
address varchar(200) default NULL,
phone int(10) default NULL
) ;

dam75 · Answer

INSERT INTO Clients (title, last_name, first_name, email, password, birth_date, address, phone) VALUES ('$gender', '$lastName', '$firstName', '$email','$password', '$birthDate', '$address', $phone);

Absot · Answer

I’m not sure what the difference is between your request and mine, but I’ll never get used to it.. :(

It seems to be working only now, but now it tells me that my table does not exist even though it successfully connected to it normally since it doesn’t give me an error message.. ?

Here’s the error message: The table 'site.clients' does not exist

When I right-click on the easyphp icon in the bottom right corner, go to "administration" and then "database management", I select my database "site" and I can see my tables "client" and "produit" listed below, but written in lowercase; maybe that’s the error, does easyphp take case into account?

le père · Answer

Client or Clients?

dam75 · Answer

What changes: I list the fields I want to update, and I have REMOVED your idClient from the list :)
As for case sensitivity: it actually depends on a lot of things (OS like Win vs Linux, config, ...) but it's better to assume that YES MySQL is case sensitive ...

Best of luck!

Absot · Answer

Client.. ^^

My request seems to have executed since I am not getting any error messages..

And given my code, I think that’s normal?

-----------------------------------------------------------------------------------------------

$query = "INSERT INTO Client (civilite, nom, prenom, mail, mdp, naissance, adresse, telephone) VALUES ('$sexe', '$nom', '$prenom', '$mail','$mdp', '$naissance', '$adr', $telephone);";

/* executing the query */
$result = mysql_query($query) or die(mysql_error());
}

-----------------------------------------------------------------------------------------------

dam75 · Answer

yes ;) just check (via phpMyAdmin or other) if your record has been properly inserted?

Absot · Answer

It was displayed correctly, I went to "show" in phpMyAdmin and I found my client, I'm proud of myself^^  Thank you for your help :)  Now all I have left to do is display the list of clients...

[SQL] Duplicate field '0' for key 1

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