[RESOLVED] need a pair of eyes

[dugindog]

running into an issue with this query. db connection is fine, selects the database, but errors on query itself. commenting it out everything runs smooth.

help

$sqlquery = "INSERT INTO dogs ('Name', 'Owner', 'Age', 'Breed', 'Colour', 'Genotype', 'Sire', 'Dam', 'Hips', 'Elbows', 'Altered', 'Images', 'temperment', 'trainability', 'other') VALUES('$name', '$owner', '$dog[8]', '$dog[1]', '$dog[14]', '$genotype', '$sire', '$dam', '$dog[3]', '$elbows', '$altered', '$images', '$dog[10]','$dog[12]', '$other' ) ";

mysql_query($sqlquery) or die( "Unable to complete the query <HR>" . $sqlquery);

this is what I get returned to me

[code=html]Unable to complete the query

INSERT INTO dogs ('Name', 'Owner', 'Age', 'Breed', 'Colour', 'Genotype', 'Sire', 'Dam', 'Hips', 'Elbows', 'Altered', 'Images', 'temperment', 'trainability', 'other') VALUES('sdfg', 'owner', '4', 'Affenpinscher', 'Silver', 'genotype', 'sire', 'dam', 'Good', 'elbows', 'alt', 'images', 'Friendly','Stubborn', 'other' ) [/code]

also here's the entire php page:

<?php


$database = "changed on purpose";
$owner = "owner";
$genotype = "genotype";
$sire = "sire";
$dam = "dam";
$elbows = "elbows";
$altered = "alt";
$images = "images";
$other = "other";


$link = mysql_connect('localhost', 'changed on purpose', 'password');
if (!$link) {
   die('Could not connect: ' . mysql_error());
}


@mysql_select_db($database) or die( "Unable to select database");


#$sqlquery = "INSERT INTO dogs ('Name', 'Owner', 'Age', 'Breed', 'Colour', 'Genotype', 'Sire', 'Dam', 'Hips', 'Elbows', 'Altered', 'Images', 'temperment', 'trainability', 'other') VALUES('$name', '$owner', '$dog[8]', '$dog[1]', '$dog[14]', '$genotype', '$sire', '$dam', '$dog[3]', '$elbows', '$altered', '$images', '$dog[10]','$dog[12]', '$other' ) ";

#$result = mysql_query($sqlquery, $link) or die( "Unable to complete the query <HR>" . $sqlquery);
mysql_close($link);


?> 

[dugindog]

I have also checked by the following and it outputted exactly as I expected (correctly).

the error must be somewhere within the string.

$result = mysql_list_tables("$database");
$num_rows = mysql_num_rows($result);
if ($num_rows >= 1) {echo "<HR>Connected and read $num_rows tables in the current database.<HR>";}

mysql_free_result($result);

$result = mysql_query("SHOW COLUMNS FROM dogs");
if (mysql_num_rows($result) > 0) {
   while ($row = mysql_fetch_assoc($result)) {
       $qt++;
   }
}
echo "$qt fields in the table dogs<HR>";

[rincewind456]

What error message do you get when using mysql_error() like so:

$result = mysql_query($sqlquery, $link) or die(mysql_error(). "Unable to complete the query <HR>" . $sqlquery); 

[dugindog]

Here's the error I get when I add that in.

[code=html]You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''dogs' ('Name', 'Owner', 'Age', 'Breed', 'Colour', 'Genotype', Unable to complete the query

INSERT INTO 'dogs' ('Name', 'Owner', 'Age', 'Breed', 'Colour', 'Genotype', 'Sire', 'Dam', 'Hips', 'Elbows', 'Altered', 'Images', 'temperment', 'trainability', 'other') VALUES('joe', 'owner', '3', 'Affenpinscher', 'Silver', 'genotype', 'sire', 'dam', 'Good', 'elbows', 'alt', 'images', 'Friendly','Stubborn', 'other' ) [/code]

I have stripped the insert down to only 1 set and it still errors here's what I stripped it to:

$sqlquery = "INSERT INTO dogs ('Name') VALUES('$name') ";

heres the error I am recieving.:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''Name') VALUES('joe')' at line 1Unable to complete the query 

[rincewind456]

You don't want or need quotes around the field names such as Name, you do want and need them around the values such as 'joe', to be inserted.

[dugindog]

I was being stupid, and not thinking. I appreciate all your help. I will mark this as resolved now.