What does this error message mean?

[judacoor]

Look, i'm getting this error message after performing a query with php:Error: Column count doesn't match value count at row 1and i have no idea what it is......Please help me! i'm stuck!!

[vijay]

Hi.. May I have your 'query' and 'table structure'?Regards,Vijay

[Synook]

Are you trying to perform an INSERT query? If so, then that means that the number of columns you have specified for insertion is not the same as the count of the list of values you gave.For example, if I tried to perform the query INSERT INTO table (name, email) VALUES ('John'); then the error would be returned because the number of columns specified for insertion (2) is not the same as the number of values given (1).Hope that helped

[judacoor]

Uhmmmm right!And I certainly am missing one field.....the first one, wich is 'id_user' but I used the "auto-increment" option, so I thought I didn't have to specify a value for mysql would automatically generate it....So what should I do?

[aspnetguy]

put in an empty string for the auto_increment column

VALUES ('','John')

[judacoor]

Thanks!it's working now!!!thanks a lot!!!

[aspnetguy]

no problem

[Synook]

You could also, if you wanted to, only specify certain columns. So, if you had your `users` table

+--------+----------+---------+|user_id |user_name |user_age |+--------+----------+---------+

and only wanted to insert the user's name you could do

INSERT INTO users (user_name) VALUES ('John')

[judacoor]

Hey guys.....Now I'm getting the following error:Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\Program......and I haven't been able to figure out what it is.....Here's the function that's causing the error:function getConsult(){$id_user = $_SESSION['id_user];$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error());return $result;}Since the error mentions something about a not-valid Link resource........I thought it could probably be the $link variable, which is:function getLink(){$link=mysql_connect("localhost","root","");$selectdb=mysql_select_db("mydb",$link);if($link==false){echo "Error connecting to DB";exit();}else if($selectdb==false){echo "Error selecting DB";exit();}else{return $link;}}but this $link variable works perfectly in the other querys I've made....so I don't know what the ###### is going on...Every functions is in a diferent file, actually in a different folder.........but I used the "include (file.php)" in every case so...........really.............I need some help!!!!Please help me out!!!!

[justsomeguy]

You need to specify global variables in functions in order to use the global version of the variable. Otherwise it is trying to use a variable called $link that was defined in the function, and obviously there is no $link variable defined in the function. Add this line at the top of the function:global $link;

[smartalco]

$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error());

you have an open apostrophe, change it to this

$result = mysql_query("SELECT * FROM `consult` WHERE `id_user` = '$id_user'",$link) or die(mysql_error());

[judacoor]

Oh....but the $link variable is defined above the function.........like this:<?php$link=Conn();function getConsult(){$id_user = $_SESSION['id_user];$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error());return $result;}?>And the thing is that $link works perfectly with another function that i've been using it with.........I changed it anyways.......alas it still doesn't work:<?phpglobal $link;$link=Conn();function getConsult(){$id_user = $_SESSION['id_user];$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error());return $result;}?>Any ideas??Please!

[Synook]

No, the problem is in this line:

$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error());

See this: $result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user",$link) or die(mysql_error()); ?There is no closing apostrophe for the id_user value. It needs to be like this: $result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user'",$link) or die(mysql_error());See, or else you open the value for id_user but never close it

[smartalco]

thanks for reading my post judacoor /sarcasm (if you don't know what im talking about, look at post 11 in this thread, or the post right above this (Synooks))

[justsomeguy]

Instead of this:

<?phpglobal $link;$link=Conn();function getConsult(){

I meant for you to do this:

<?php$link=Conn();function getConsult(){  global $link;

[judacoor]

smartalco and synook: I'm terrybly sorry....I forgot to mention earlier that it was actually a copy-paste mistake........the code is as follows:$result = mysql_query("SELECT * FROM `consult` WHERE `consult`.`id_user` = '$id_user'",$link) or die(mysql_error());so I don't see any errors at first sight, but I forgot to to tell you guys in the last post, for I was in a hurry when I posted it.I'd appreciate any further help!!!

[judacoor]

I'll try using the global attribute like you say Justsomeguy, and I will tell you guys later what happened