Select specific column from specific row using old mysql_* API

Question

I've got a table in my SQL Database, and the table has two fields, one for an id, and the other for an integer. I want to grab the value in the integer field where the id is equal to a specific number I pass in. The id field is called "id" and the integer field is called "loglevel". This is the code I've got, but it doesn't give me the desired result.

$result = mysql_query("SELECT loglevel FROM Logs WHERE id='$number'");
echo "Pulled from SQL: " . $result;

The output for this is

Pulled from SQL: Resource id #2

Can you help me so the output is "2" if the value in the SQL table is 2?

Michael Berkowski · Accepted Answer · 2011-11-04 01:35:24Z

4

You need to fetch your result using mysql_fetch_assoc(), or mysql_fetch_array() (or others). $result as returned by mysql_query(), is a result resource, not an actual rowset:

$result = mysql_query("SELECT loglevel FROM Logs WHERE id='$number'");

// If the query completed without errors, fetch a result
if ($result) {
  $row = mysql_fetch_assoc($result);
  echo $row['loglevel'];
}
// Otherwise display the error
else echo "An error occurred: " . mysql_error();

When you are expecting multiple rows returned rather than just one, fetch them inside a while loop. There are many examples of this in the mysql_fetch_assoc() documentation .

edited Nov 4, 2011 at 1:35

answered Nov 4, 2011 at 1:29

Michael Berkowski

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sankar d. · Accepted Answer · 2011-11-04 03:44:48Z

0

$result = mysql_query("SELECT loglevel FROM Logs WHERE id='$number'");

here $result is a record set/ mysql resource not php array. you need to use mysql_fetch_assoc(), or mysql_fetch_array() to access $result.

if ($result) {
  while( $row = mysql_fetch_assoc($result)){
    $my_assoc[] = $row;
  }
}
var_dump($my_assoc);

this will print all data from the query.

answered Nov 4, 2011 at 3:44

sankar d.

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