[SOLVED] why is no one helping me????

[chishake]

i have another problem again. i am trying to check if an image i want to upload into my database already exist. i have used the

$result= SELECT  name from images WHERE name = $name,

($name = $_FILES['userfile']['name'])

$result = mysql_query($result);

$num=mysql_numrows($result);

if($num >0)

{

echo " there is a file in the database with name $name, upload a different file";

exit();

}

this piece of code gives me this error

 

Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource

it works on other pieces of data but not on my image table.

can anyone tell me why and how i can go about to sort this out since i want to make sure i am not uploading a file already in the database

[craygo]

the function is wrong 

 

mysql_num_rows not mysql_numrows

 

Ray

[phpSensei]

People seem to make this mistake alot. To prevent this, do not use notepad, because if you do, you can't tell if its a real function or not. In other programs, functions turn blue, or another color, so becareful.

[chishake]

that isnt the problem. i have tried mysql_num_rows but still getting the same warning

[phpSensei]

that isnt the problem. i have tried mysql_num_rows but still getting the same warning

 

try this, and tell us the error that the script spits out.

 

$result = mysql_query($result)or die(mysql_error());

[~n[EO]n~]

Is that all code you've got

Put your sql query inside double quote , like this

$result= "SELECT  name from images WHERE name = $name";

[trq]

If that is your actual code, it is well foobar'd. Post your actual code, within [ code ][/ code ] tags (no spaces).

 

[Goose87]

I dont really understand the ($name={FILES....  part as ive never come across it before, but i'd write it just like this..

 

($name = $_FILES['userfile']['name'])
$result=@mysql_query("SELECT name from images WHERE name = $name"); 

$num=@mysql_num_rows($result); 
if($num >0)
{
echo " there is a file in the database with name $name, upload a different file";
exit();
}

 

or maybe follow the way you did it and do :

 

$result=@mysql_query("SELECT name from images WHERE name = $name, ($name = $_FILES['userfile']['name'])"); 

$num=@mysql_num_rows($result); 
if($num >0)
{
echo " there is a file in the database with name $name, upload a different file";
exit();
}

 

Hope that works?

[craygo]

you will also get the error if there is an error with your sql statement. Use Sensei's suggestions below to help debug your sql statement.

 

You may want to start using some Quotes

 

<?php
$name = $_FILES['userfile']['name'];
$sql = "SELECT `name` from `images` WHERE name = '$name'";
$result = mysql_query($sql) or die(mysql_error());
$num=mysql_num_rows($result);
if($num >0)
{
echo " there is a file in the database with name $name, upload a different file";
exit();
}
?>

 

Ray

[zq29]

FYI: Probably a good idea to use a more descriptive topic title in future...

[chishake]

the problem get worst after adding the die(mysql_error())

this is what i get

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in

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 'JPG' at line 1

 

JPG is the type of file i am trying to upload

[craygo]

if you echo out $name what do you get???

[chishake]

this code has solved the problem.

<?php

$name = $_FILES['userfile']['name'];

$sql = "SELECT `name` from `images` WHERE name = '$name'";

$result = mysql_query($sql) or die(mysql_error());

$num=mysql_num_rows($result);

if($num >0)

{

echo " there is a file in the database with name $name, upload a different file";

exit();

}

?>

 

thanks you folks

[phpSensei]

Mark the topic solved.