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So simple but just won't work

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So simple but just won't work General SQL help. Help with MySQL, MSSQL, Oracle, mSQL, PostgreSQL, etc.

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  #1  
Old July 26th, 2002, 06:45 AM
notbad notbad is offline
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So simple but just won't work
I just can't work out why this code:

<html>
<body>
<?PHP
$data_b = mysql_pconnect("dbs.server","root","");
mysql_select_db("databasename");
$query = "SELECT office_id, contact,site,street_add,Postal_add,city,state,pcod e,phone,fax,email,website FROM office_table WHERE site = ";
$query .= $HTTP_GET_VARS['of0'];
$result = mysql_query($query);
$row = mysql_fetch_array($result);
?>
<? PHP echo $row['contact'] ?>
</body>
</html>

Is giving me this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /web/htdocs/notbad/docs/Eplus_Beta/NSW/stationery2.php on line 4

If you can help it will make me very happy.
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  #2  
Old July 26th, 2002, 08:01 AM
EvilivE EvilivE is offline
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RE: So simple but just won't work
That error message is telling you that your query is wrong, either improperly formated, fields dont exist, etc.

The problem with your query is that there are no quotes around what site is suppose to be equal to. I find it helpful prior to putting my queries in code, I like to run them in my mysql client first. This way I see the result set and can verify that it returns what I wanted. You can test it by doing this:
mysql -u username -p dbname

One wat to fix your syntax problem is to use a query like this:
php Code:
Original - php Code 
  1. $query = "SELECT office_id, contact,site,street_add,Postal_add,city,state,pcod  e,phone,fax,email,website FROM office_table WHERE site='".$HTTP_GET_VARS['of0']."'";


Hopoe that helps.
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  #3  
Old August 3rd, 2002, 03:54 PM
notbad notbad is offline
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RE: So simple but just won't work
it seems that "$row = mysql_fetch_array($result); " has to be used as an argument like "if ($row = mysql_fetch_array($result)) { etc."
It's the only way I can assign the values to the array. I don't know why but it works.
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  #4  
Old August 3rd, 2002, 05:32 PM
EvilivE EvilivE is offline
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RE: So simple but just won't work
hmmm...

Very strange! I use code like this and it works fine:

$query = "select sount(*) from table";
$result = mysql_query($query);
$row = mysql_fetch_array($result);

Wierd.
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  #5  
Old August 3rd, 2002, 06:02 PM
EvilivE EvilivE is offline
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RE: So simple but just won't work
Let me rewrite that:

$query = "select count(*) as total from members";
$result = mysql_query($query);
$row = mysql_fetch_array($result);

echo "There are ".$row['total']." registered members";

Thats better.
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