Mysql Query + Variable

[Oni]

Heyas--

I know that almost all of the topics I start that ask for help, have to do with variables inside mysql queries.

This is another one of those topics 😉

$file = $ftp . $dir . $item['name'];  

$result = mysql_query("SELECT dir FROM ftp WHERE dir ='$file'")
$query = mysql_fetch_array( $result );  //Line 68
If($query != 0) {
mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES
('$file', '$item['type']', '$item['size']', 'date("d")') ")
or die(mysql_error()); 
}

Is what I have,

Parse error: syntax error, unexpected T_VARIABLE in C:\xampp\htdocs\ftp.php on line 68

Is the reply.

As you can see, the code tries to check to see if the string is already in the table, if it's not, it attempts to put it in the table.

But it doesn't work.. 🙁

[bastien]

missing the semi-colon at the end of the query line

[Oni]

ah!

Yea, that seems to be one problem...
Thanks!
However, I still get the same error?

$file = $ftp . $dir . $item['name'];  

$result = mysql_query("SELECT dir FROM ftp WHERE dir ='$file'")
$query = mysql_fetch_array( $result );  //Line 68
If($query != 0) {
	mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES
	('$file', '$item['type']', '$item['size']', 'date("d")') ");
	or die(mysql_error()); 
}

[bastien]

$result = mysql_query("SELECT dir FROM ftp WHERE dir ='$file'");

[Oni]

After I fix that, now I have a:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs\ftp.php on line 71

foreach($items as $line=>$item)
  {

$file = $ftp . $dir . $item['name'];  

$result = mysql_query("SELECT dir FROM ftp WHERE dir ='$file'");
$query = mysql_fetch_array( $result );  //Line 68
If($query != 0) {
	mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES
	('$file', '$item['type']', '$item['size']', 'date("d")') ") //line 71
	or die(mysql_error()); 
}
}

[bastien]

the date function is php and can't run inside the sql like you have it....change it to this anf give it a go

mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES 
    ('$file', '$item['type']', '$item['size']', '".date("d")."') ") //line 71 
    or die(mysql_error());

[Oni]

Thanks for the help...

But I still get the same problem?

foreach($items as $line=>$item)
  {

$file = $ftp . $dir . $item['name'];  

$result = mysql_query("SELECT dir FROM ftp WHERE dir ='$file'");
$query = mysql_fetch_array( $result );  //Line 68
If($query != 0) {
 	mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES
    ('$file', '$item['type']', '$item['size']', '".date("d")."') ") //line 71
    or die(mysql_error()); 
}
}

[bastien]

my bad, missed the array elements in there, those to be either

a. outside the query like the date bit

b. wrapped in {} braces to force it to evaluate

If($query != 0) { 
    mysql_query("INSERT INTO ftp(dir, klasor, boyut, gun) VALUES 
    ('$file', '{$item['type']}', '{$item['size']}', '".date("d")."') ") //line 71 
    or die(mysql_error());

[Oni]

heh--

Almost!

The foreach loop works, except I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\ftp.php on line 68

I had checked the value of $file before, and it's fine...

[bastien]

echo out the query and try it in a gui for mysql....its seems that your query is off somehow...test it

[Oni]

After I used some different commands that would have done the same thing, I found out that I was trying to use the wrong table... Now the table name is correct. But I seem to get a totally different problem.

$file = $ftp . $dir . $item['name'];  

$result = mysql_query("SELECT * FROM sunucu WHERE dir ='$file'");
$query = mysql_fetch_array( $result );  //Line 68
If($query != 0) {
    mysql_query("INSERT INTO sunucu(dir, klasor, boyut, gun) VALUES
    ('$file', '{$item['type']}', '{$item['size']}', '".date("d")."') ") //line 71
    or die(mysql_error()); 

}

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\ftp.php on line 68

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\ftp.php on line 69
Table 'ftp.ftp' doesn't exist

Where does it get the idea of a table named ftp.ftp? 🙁