exception in sql syntax

sladepowers

i hope i'm in the right place

hi, my name is carlton. i'm having a situation with
this error message:

" 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 '', '', '', '', '', '', '', '', '', '', '')' at line 3 ".

i know hardly any about sql and know not where to find where this error is but am eager to go forward. i access the database
via dreamweaver or my hosting service through php myadmin. but i am
unable to view posted web form data in the database.

do i need both dreamweaver and my phpadmin to work out this issue?

please help

laserlight

You should post the relevant SQL statement along with your relevant database schema.

Roger_Ramjet

Single quotes within single quotes, so the sql command is broken.

If you are single-quoting the string values then double-quote the sql command

"INSERT INTO table(col1,col2) VALUES('val1','val2')"

sladepowers

laserlight, how would i go about getting the sql statement you mentioned? i use my phpadmin via my hosting service so i never actually read the the sql statements because they do it all for you. so please tell me how to obtain the statement you are refering to.
thank you so much.

sladepowers

roger ramjet, thanks for the reply. here is the sql command i believe may be in question?

<?php

FileName="Connection_php_mysql.htm"

Type="MYSQL"

HTTP="true"

$hostname_babiesnmovies = "XXX";
$database_babiesnmovies = "XXX";
$username_babiesnmovies = "XXX";
$password_babiesnmovies = "XXX";
$con = mysql_pconnect($hostname_babiesnmovies, $username_babiesnmovies, $password_babiesnmovies)
or trigger_error(mysql_error(),E_USER_ERROR);

mysql_select_db("db337879506", $con);

$yourfirstname = $POST['YourFirstName'];
$yourlastname = $POST['YourLastName'];
$yourphonenumber = $POST['YourPhoneNumber'];
$youremailaddress = $POST['YourE-mailAddress'];
$babysfirstname = $POST['BabysFirstName'];
$babysmiddlename = $POST['BabysMiddleName'];
$babyslastname = $POST['BabysLastName'];
$babysbirthdate = $POST['BabysBirthdate'];
$babysage = $POST['BabysAge'];
$babysheight = $POST['BabysHeight'];
$babysweight = $POST['BabysWeight'];
$babyseyecolor = $POST['BabysEyeColor'];
$babyshaircolor = $POST['BabysHairColor'];
$babysethicity = $POST['Babys Ethnicity'];

$sql="INSERT INTO register (id, YourFirstName, YourLastName, YourPhoneNumber, YourEmailAddress, BabysFirstName, BabysMiddleName,
BabysLastName, BabysBirthdate, BabysAge, BabysHeight, BabysWeight, BabysEyeColor, BabysHairColor, BabysEthnicity) values ( 'NULL',
'$yourfirstname','$yourlastname', '$yourphonenumber', '$youremailaddress', '$babysfirstname', '$babysmiddlename', '$babyslastname',
'$babysbirthdate', '$babysage', '$babysheight', '$babysweight', '$babyseyecolor', '$babyshaircolor', '$babysethnicity')";

mysql_query($sql, $con) or die (mysql_error());

also can you tell me how to echo my php/sql statements? i'm so green.

Roger_Ramjet

echo $sql;

Now looking at your code it is apparent that one error is the use of 'NULL'
If the id field is an autoinc column then you can not supply a value, not even a NULL value. What you do instead is to just omit that column from the list. (and NULL is a PHP constant anyway so you don't need quotes around it which turn it into a string)

$sql="INSERT INTO register (id, YourFirstName, YourLastName, YourPhoneNumber, YourEmailAddress, BabysFirstName, BabysMiddleName,
BabysLastName, BabysBirthdate, BabysAge, BabysHeight, BabysWeight, BabysEyeColor, BabysHairColor, BabysEthnicity) values ( 'NULL',
'$yourfirstname','$yourlastname', '$yourphonenumber', '$youremailaddress', '$babysfirstname', '$babysmiddlename', '$babyslastname',
'$babysbirthdate', '$babysage', '$babysheight', '$babysweight', '$babyseyecolor', '$babyshaircolor', '$babysethnicity')";

becomes 

$sql="INSERT INTO register (YourFirstName, YourLastName, YourPhoneNumber, YourEmailAddress, BabysFirstName, BabysMiddleName,
BabysLastName, BabysBirthdate, BabysAge, BabysHeight, BabysWeight, BabysEyeColor, BabysHairColor, BabysEthnicity) values ( '$yourfirstname','$yourlastname', '$yourphonenumber', '$youremailaddress', '$babysfirstname', '$babysmiddlename', '$babyslastname',
'$babysbirthdate', '$babysage', '$babysheight', '$babysweight', '$babyseyecolor', '$babyshaircolor', '$babysethnicity')";

then 

echo $sql;

Finally, the use of mysql_query is nearlt a decade out of date. I guess you are using some book or titorial written for PHP 4 and not upgraded since.
Learn how to use PDO now instead of mysql_query

sladepowers

i followed the instructions. thank you. i submitted the form with the new code and this is the error message i received:

"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 '', '', '', '', '', '', '', '', '', '', '')' at line 3"

this is code as revised:

<?php

FileName="Connection_php_mysql.htm"

Type="MYSQL"

HTTP="true"

mysql_select_db("db337879506", $con);

$sql="INSERT INTO register (id, YourFirstName, YourLastName, YourPhoneNumber, YourEmailAddress, BabysFirstName, BabysMiddleName,
BabysLastName, BabysBirthdate, BabysAge, BabysHeight, BabysWeight, BabysEyeColor, BabysHairColor, BabysEthnicity) values ('$yourfirstname',
'$yourlastname', '$yourphonenumber', '$youremailaddress', '$babysfirstname', '$babysmiddlename', '$babyslastname',
'$babysbirthdate', '$babysage', '$babysheight', '$babysweight', '$babyseyecolor', '$babyshaircolor', '$babysethnicity')";

mysql_query($sql, $con) or die (mysql_error());

and i will be sure to get a look at PDO...maybe not this instant but i will review the link. thanks. so can you give me a clue as to where my error is? if i knew how to get to it i may be able to see it through but as for now i can't see where it's at.

Roger_Ramjet

You still have the id field in the query so the values do not match the fields: 15 fields and 14 values.

As to the rest of it, it looks like none of the vars have any data as well, '','','' etc

Have you echoed the string? Be nice to see what the query really looks like when you've finished building it.