MySQL problem with php ^^ helpz!

[antoha]

well im using a Denwer program on wich i created MySql database,
now, when im using this code:

<?php
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_close($link);
?> 

im getting "connected successfully" so its all should be good
but when im tryin to extract somethin from the database like:

<?php
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$query = 'SELECT password From ".au_server_ls.".account_data';
$result = mysql_query($query);
$array = mysql_fetch_lengths($result);


echo $array;
mysql_close($link);
?> 

it gives me the following error:

Connected successfully
Warning: mysql_fetch_lengths() expects parameter 1 to be resource, boolean given in Z:\denwer\www\denwer\index.php on line 9

cant some1 pliz help me with this issue?

[laserlight]

Change this:

$result = mysql_query($query);

to this, for debugging:

$result = mysql_query($query) or die('Query error: ' . mysql_error());

[dagon]

what's au_server_ls suppose to be ? if its just the db name

$query = 'SELECT password From au_server_ls.account_data'; 

will work

[antoha]

hmmmm.... after playing with it abit i got to this problem:
the code is:

<?php
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$query = 'select password From au_server_ls.account_data where name = "alex"';
$result = mysql_query($query) or die('Query error: ' . mysql_error()); 



echo mysql_fetch_row($result);;
mysql_close($link);
?>

result:

Connected successfullyArray

what is should display is:

Connected successfullyjLIjfQZ5yojbZGTqxg2pY0VROWQ=

[laserlight]

You might want to read the PHP manual on [man]mysql_fetch_row/man.

By the way, this is the legacy MySQL extension. You should be using the PDO extension or MySQLi extension instead.

[antoha]

hmmmm yes u were right, i kinda used the wrong command line, using it properly works just fine, anyway, i got the answer i needed, thnx alot
also... another question.
im not used to working with MySql been working with msSql(odbc) generaly, so, how do i get
numerous results?
i have tryed to use this:

 <?php
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$query = 'select id from au_server_ls.account_data where last_ip = "127.0.0.1"';
$result = mysql_query($query) or die('Query error: ' . mysql_error()); 
while (mysql_fetch_row($result)){
echo mysql_result($result, 1);
}


mysql_close($link);
?>

wich should have worked with ODBC but it dosent work with mysql, it just displays 1 result, also it should display 2.

[laserlight]

I suggest that you do this:

<?php
$link = mysql_connect('localhost', 'root', '')
    or die('Could not connect: ' . mysql_error()); // debugging

$query = "SELECT id FROM au_server_ls.account_data WHERE last_ip = '127.0.0.1'";
$result = mysql_query($query)
    or die('Query error: ' . mysql_error()); // debugging
while ($row = mysql_fetch_assoc($result)) {
    echo $row['id'];
}

mysql_close($link);

Note that the mysql_close is optional. Strings in SQL should be delimited by single quotes, not double quotes, hence I change "127.0.0.1" to '127.0.0.1'.

[antoha]

ty, all problems are solved thnx again to every1 who helped!

[laserlight]

You're welcome 🙂

You can mark this thread as resolved using the thread tools. Oh, and remember to consider using the PDO extension or MySQLi extension 🙂