Problem with syntax error

Hi guys am fighting with a syntax error of my sql, saying exactly:

"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax"

Even though the code is working and doing what I wanted I still get the syntax error info! and here is the code:

  $person_id =mysql_query("SELECT person_id FROM person WHERE firstname='$array[0]'    AND lastname='$array[1]' AND city='$array[2]' ")
or die(mysql_error());
if (mysql_num_rows($person_id) )
{
     print 'user is already in table';
}
else
 {
mysql_query ("INSERT INTO person VALUES (NULL, '$a开发者_JAVA技巧rray[0]' ,'$array[1]' , '$array[2]' ")
 or die(mysql_error());  
 $person_id = mysql_insert_id();
}

$address_id =mysql_query("SELECT address_id FROM address WHERE street='$array[3]' AND city='$array[4]' AND region='$array[5]'")
or die(mysql_error());
if (mysql_num_rows($address_id) )
{
    print ' already in table';
}
else
{
mysql_query ("INSERT INTO address VALUES (NULL, '$array[3]', '$array[4]', '$array[5]'")
or die(mysql_error());  

$address_id = mysql_insert_id();
}

mysql_query ("INSERT INTO person_address VALUES($person_id, $address_id)")
or die(mysql_error());  

Thanks for any suggestions

It's probably because you haven't escaped your values...

Try:

$query = "SELECT age FROM person WHERE name='".mysql_real_escape_string($array[0])."' AND lastname='".mysql_real_escape_string($array[1])."' AND city='".mysql_real_escape_string($array[2])."'";

And read up on SQL injection.

EDIT

I think your problem is that you are trying to pass mysql result resources directly into a string, without fetching the actual values first.

Try this:

// Create an array of escaped values to use with DB queries
$escapedArray = array();
foreach ($array as $k => $v) $escapedArray[$k] = mysql_real_escape_string($v);

// See if the person already exists in the database, INSERT if not
$query = "SELECT person_id FROM person WHERE firstname='$escapedArray[0]' AND lastname='$escapedArray[1]' AND city='$escapedArray[2]' LIMIT 1";
$person = mysql_query($query) or die(mysql_error());
if ( mysql_num_rows($person) ) {
    print 'user is already in table';
    $person = mysql_fetch_assoc($person);
    $person_id = $person['person_id'];
} else {
    $query = "INSERT INTO person VALUES (NULL, '$escapedArray[0]', '$escapedArray[1]', '$escapedArray[2]')";
    mysql_query($query) or die(mysql_error());  
    $person_id = mysql_insert_id();
}

// See if the address already exists in the database, INSERT if not
$query = "SELECT address_id FROM address WHERE street='$escapedArray[3]' AND city='$escapedArray[4]' AND region='$escapedArray[5]'";
$address = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($address) ) {
    print 'address already in table';
    $address = mysql_fetch_assoc($address);
    $address_id = $person['address_id'];
} else {
    $query = "INSERT INTO address VALUES (NULL, '$escapedArray[3]', '$escapedArray[4]', '$escapedArray[5]')";
    mysql_query ($query) or die(mysql_error());  
    $address_id = mysql_insert_id();
}

// INSERT a record linking person and address
mysql_query ("INSERT INTO person_address VALUES($person_id, $address_id)") or die(mysql_error());  

ANOTHER EDIT

Firstly, I have modified the code above - added a couple of comments, corrected a couple of small errors where the wrong variable was referenced and re-spaced it to make it more readable.

Secondly...

You are getting that additional error because you are trying to insert a new row into your person_address table, which doesn't seem to have a sensibly configured primary key. The easy work around to the problem you currently have is to run a SELECT against this table to see if you have already got a record for that user, then if you have you can do an UPDATE instead of the INSERT to alter the existing record.

However, if I understand what your doing here correctly, you don't actually need the person_address table, you just need to add another integer column to the person table to hold the ID of the corresponding row in the address table. Doing this would make many of your future queries potentially much simpler and more efficient as it will be much easier to SELECT data from both tables at once (you could do it with your current structure but it would be much more confusing and inefficient).

The following code example could be used if you add another integer column on the end of your person, and call that column address_id. You will notice it's very similar to the above, but there are two key differences:

• We do the address stuff first, since we will keep track of the relation in the person record
• We do an UPDATE only if we find a person, otherwise we just INSERT a new person as before

// Create an array of escaped values to use with DB queries
$escapedArray = array();
foreach ($array as $k => $v) $escapedArray[$k] = mysql_real_escape_string($v);

// See if the address already exists in the database, INSERT if not
$query = "SELECT address_id FROM address WHERE street='$escapedArray[3]' AND city='$escapedArray[4]' AND region='$escapedArray[5]'";
$address = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($address) ) {
    print 'address already in table';
    $address = mysql_fetch_assoc($address);
    $address_id = $person['address_id'];
} else {
    $query = "INSERT INTO address VALUES (NULL, '$escapedArray[3]', '$escapedArray[4]', '$escapedArray[5]')";
    mysql_query ($query) or die(mysql_error());  
    $address_id = mysql_insert_id();
}

// See if the person already exists in the database, UPDATE if he does, INSERT if not
$query = "SELECT person_id FROM person WHERE firstname='$escapedArray[0]' AND lastname='$escapedArray[1]' AND city='$escapedArray[2]' LIMIT 1";
$person = mysql_query($query) or die(mysql_error());
if ( mysql_num_rows($person) ) {
    print 'user is already in table';
    $person = mysql_fetch_assoc($person);
    $person_id = $person['person_id'];
    $query = "UPDATE person SET address_id = '$address_id' WHERE person_id = '$person_id'";
    mysql_query($query) or die(mysql_error());
} else {
    $query = "INSERT INTO person VALUES (NULL, '$escapedArray[0]', '$escapedArray[1]', '$escapedArray[2]', '$address_id')";
    mysql_query($query) or die(mysql_error());
}

If we structure the database in this way, it allows us to do this:

SELECT person.*, address.* FROM person, address WHERE person.address_id = address.address_id AND [some other set of conditions]

Which will return the person record, and the address record, in the same result set, all nicely matched up for you by the database.

YET ANOTHER EDIT

You need to add an auto-increment primary key to the person_address table, and perform a SELECT on it to make sure you are not adding duplicate records.

You should replace the final INSERT statement with the following code segment. This code assumes that you have a primary key in the person_address table called relation_id. It also assumes that the id field names in this table are named in the same way as they are in the other two tables.

// See if a relation record already exists for this user
// If it does, UPDATE it if the address is different
// If it doesn't, INSERT an new relation record
$query = "SELECT relation_id, address_id FROM person_address WHERE person_id = '$person_id' LIMIT 1";
$relation = mysql_query($query);
if ( mysql_num_rows($relation) ) {
    $relation = mysql_fetch_assoc($relation);
    if ($relation['address_id'] == $address_id) {
        print 'The record is identical to an existing record and was not changed';
    } else {
        $relation_id = $relation['relation_id'];
        $query = "UPDATE person_address SET address_id = '$address_id' WHERE relation_id = '$relation_id'";
        mysql_query($query) or die(mysql_error());  
    }
} else {
    $query = "INSERT INTO person_address VALUES(NULL, '$person_id', '$address_id')";
    mysql_query($query) or die(mysql_error());  
}

EVEN MORE EDITING

Try this to replace the code from above:

// See if a relation record already exists for this user
// If it doesn't, INSERT an new relation record
$query = "SELECT person_id FROM person_address WHERE person_id = '$person_id' AND address_id = '$address_id' LIMIT 1";
$relation = mysql_query($query);
if ( !mysql_num_rows($relation) ) {
    $query = "INSERT INTO person_address VALUES('$person_id', '$address_id')";
    mysql_query($query) or die(mysql_error());  
}

You cannot use array values like that inside of quotes - instead you could, for example, separate the values from the query using dots.

$query = "SELECT age FROM person WHERE name='".$array[0]."' AND lastname='".$array[1]."' AND city='".$array[2]."'";

the second and fourth query do not have an ending ')' at the end of the values