Warning: Wrong parameter count for mysql_query()

astoller

When running this script I get wrong parameter count for mysql_query.
The database and the table all exist ok in mysql.

How can I fix this?

$Host= "localhost";
$User= "root";
$Password = "123";
$DBName = "NewDatabase";
$TableName = "feedback2";

$Link = mysql_connect($Host, $User, $Password);

$Query= "INSERT into '$TableName'
(company,logo,strapline,activity,contact,phone,mobile,email,add1,add2,add3,add4,postcode,text)
values ('0',
'$Array[company]',
'$Array[Logo]',
'$Array[Strapline]',
'$Array[Activity]',
'$Array[Contact]',
'$Array[Phone]',
'$Array[Mobile]',
'$Array[Email]',
'$Array[Add1]',
'$Array[Add2]',
'$Array[Add3]',
'$Array[Add4]',
'$Array[Postcode]',
'$Array[Text]')";

//The following line is a test for developemnt only. Then remove it.
print ("The query is : $Query \n");

if (mysql_query ($DBName,$Query,$Link)) {
print ("The query was successfully executed \n");
} else {
print ("The query coud not be executed \n");
}

mysql_close ($Link);
?>
</body>
</html>

laserlight

you should select the database with mysql_select_db()
as such, the $DBName argument should not be there.

actually, if you have questions about native php functions, just search the php manual, it should be able to explain in detail.

superwormy

www.php.net/mysql_query

astoller

I changed as you said, but stil get the error message

The query is :
INSERT into 'feedback2' (company,logo,strapline,activity,contact,phone,mobile,email,add1,add2,add3,add4,postcode,text) values ('0', '', '', 'wer', '', 'werwer', '', '', '', '', '', '', '', 'werwer', '')
The query coud not be executed

Here is my changed code. Any ideas?

<?php
/this page receives and handles the data frm form.html/
//trim the incoming data

$Array ["Company"] = trim ($Array["Company"]);
$Array ["Logo"] = trim ($Array["Logo"]);
$Array ["Strapline"] = trim ($Array["Strapline"]);
$Array ["Activity"] = trim ($Array["Activity"]);

$Array ["Contact"] = trim ($Array["Contact"]);
$Array ["Phone"] = trim ($Array["Phone"]);
$Array ["Mobile"] = trim ($Array["Mobile"]);
$Array ["Email"] = trim ($Array["Email"]);

$Array ["Add1"] = trim ($Array["Add1"]);
$Array ["Add2"] = trim ($Array["Add2"]);
$Array ["Add3"] = trim ($Array["Add3"]);
$Array ["Add4"] = trim ($Array["Add4"]);

$Array ["Postcode"] = trim ($Array["Postcode"]);
$Array ["Text"] = trim ($Array["Text"]);

//Set the variables for the database access

$Host= "localhost";
$User= "root";
$Password = "123";
$DBName = "NewDatabase";
$TableName = "feedback2";

$Link = mysql_connect($Host, $User, $Password);

//The following line is a test for developemnt only. Then remove it.
print ("The query is : $Query \n");
mysql_select_db($DBName);
if (mysql_query ($Query,$Link)) {
print ("The query was successfully executed \n");
} else {
print ("The query coud not be executed \n");
}

mysql_close ($Link);

superwormy

Try this for your query... see if it returns an error.

mysql_query ($Query,$Link) or die (mysql_error());

astoller

I am not sure I know what you mean.
Here is the exisitng query:

if (mysql_query ($Query,$Link)) {
print ("The query was successfully executed \n");
} else {
print ("The query coud not be executed \n");
}

What should I change it to?

superwormy

mysql_query ($Query,$Link) or die (mysql_error());

:-)

astoller

Hereis my changed code with your suggestion:

//The following line is a test for developemnt only. Then remove it.
print ("The query is : $Query \n");
mysql_select_db($DBName);

mysql_query ($Query,$Link) or die (mysql_error());

if (mysql_query ($Query,$Link)) {
print ("The query was successfully executed \n");
} else {
print ("The query coud not be executed \n");
}

mysql_close ($Link);
?>

The result is:

INSERT into feedback2 values ('0', '', 'dfg', 'dsfgsdfg', '', '', '', '', '', '', '', '', '', '', '')
The query coud not be executed

superwormy

You misunderstand... take out your bit... and put in my bit. Replace yours with mine.

If mine doesn't die, then the query seems to have successfully executed.

Check the database and see if the query inserted the record.

The_Chancer

You'll also need to select the database like :

$Link = mysql_connect($Host, $User, $Password); 

//insert this next line 
mysql_select_db($dbname);

$Query= "INSERT into '$TableName' 
(company,logo,strapline,activity,contact,phone,mobile,email,add1,add2,add3,add4,postcode,text) 
values ('0', 
'$Array[company]', 
'$Array[Logo]', 
'$Array[Strapline]', 
'$Array[Activity]', 
'$Array[Contact]', 
'$Array[Phone]', 
'$Array[Mobile]', 
'$Array[Email]', 
'$Array[Add1]', 
'$Array[Add2]', 
'$Array[Add3]', 
'$Array[Add4]', 
'$Array[Postcode]', 
'$Array[Text]')";

So you should be able to insert...

superwormy


<?php

$Host= "localhost"; 
$User= "root"; 
$Password = "123"; 
$DBName = "NewDatabase"; 
$TableName = "feedback2"; 

$Link = mysql_connect($Host, $User, $Password) or die (mysql_error());

mysql_select_db($DBName) or die (mysql_error());

mysql_query ($Query,$Link) or die (mysql_error()); 

mysql_close ($Link); 

?>

The_Chancer

Always a good one...

I don't know how or why, but I tend to use it in an include file (which can help keep the passwords etc out of the web directory...)

Such as

<?php
global $db_user;
global $db_pass;
global $db_host;
global $db_name;
$db_user	= 'root';
$db_pass	= '123';
$db_host	= 'localhost';
$db_name= 'dbname';
function do_query ( $sql ) 
{
	global $db_user;
	global $db_pass;
	global $db_host;
	global $db_name;
	$headers .= "MIME-Version: 1.0\r\n"; 
	$headers .= "Content-type: text/html; charset=iso-8859-1\r\n"; 
	$headers .= "From: Some Website <here@thissite.com>\r\n"; 
	$headers .= "To: Me cos I made a mistake<me@home.com>\r\n"; 
	$headers .= "Reply-To: That site<there@thatsite.com>\r\n"; 
	$headers .= "X-Priority: 1\r\n"; 
	$headers .= "X-MSMail-Priority: High\r\n"; 
	$headers .= "X-Mailer: Just That Server";
	mysql_connect( $db_host, $db_user, $db_pass )or die("Cant connect to $db_host: ");
	mysql_select_db( $db_name )or die("Cant select $db_name: ");
	$res = mysql_query( $sql )or die("You messed up in your sql Using <b>$sql</b>\n\r" . mysql_error() . mail("thechancer99@hotmail.com","Error message in  SQL - Error <br>".mysql_error()."<br> Occurred with $sql",$headers));
	return $res;
}
?>

Then all you do is make your query and do_query($thisquery)
If it works, then all OK - if not, you get the error message and a mailback as well (if you want it to...)

astoller

Yes, that worked.
I got no error message.
Thanks.
But how could I incorporate the usefuel line of code
that lets me know whether or not the query was executed?

Also, I get "sorry duplicate key with value "0" already exists.
How could I fix this?

superwormy

Is that first field set to UNIQUE, is it AUTO_INCREMENT...?

You can incorporate that useful line of code anyway you wish to :-)

I showed you the basics of it, The Chancer showed you more, figure the rest our yourself :-)

The_Chancer

Make sure your id row is an auto increment one... should work

As for the "Success" code - you have made that already 🙂

http://www.php.net
http://www.mysql.com

What did my mother tell me ?
Read learn and inwardly digest.....

Dunno what it means but it sounds good...

The_Chancer

Oops - didn't nick your idea Superwormy....

Just took too long in answering it...