Including $variable in MySQL

Here are my failed attempts to include a PHP variable in a MySQL expression. Replacing the variable with a 1 results in the results being printed. Any help will be appreciated.

 $query = "
 SELECT name FROM teams
 WHERE id = '$shooterID'";

$shooters = mysql_query($query)
 or die(mysql_error());

$i = 0;
while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[$i];
 $i++;
}

$shooters = mysql_query("
 SELECT name FROM teams
 WHERE id = '$shooterID'")
 or die(mysql_error());

$i = 0;
while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[$i];
 $i++;
}

Thanks

Attempting to utilize the methods here have not fully solved the problem (though thanks again). 开发者_Go百科Here are my revised efforts along with further context (I don't need to sanitize the data as it is coming directly from another query.

$shooters = mysql_query("
 SELECT * FROM events JOIN teams
 on events.shooter = teams.id
 ") or die(mysql_error());

$i = 0;
while($results = mysql_fetch_array( $shooters )) {
    $shooterIDs[$i] = $results[0];
    $i++;
}

//var_dump($shooterIDs); == array(1) { [0]=>  string(1) "1" } 

$query = "
 SELECT name FROM teams
 WHERE id = '".$shooterID[0]."'";

$shooters = mysql_query($query)
 or die(mysql_error());

while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[0];
}

Turns out my last attempt was missing a 's' in the variable namee $shooterIDs[0]. Stupid error. There were probably others as well that have been already solved with all of your help. Thanks!

The query is not your problem, the output is:

This is wrong:

$i = 0;
while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[$i];
 $i++;
}

This is correct:

while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[0];
}

Also

Just make sure you are properly sanitizing your input if you want to include the variable like that. For instance:

$shooterID = (int)$_GET['shooter_id'];

That forces the number to either be a 0 if it is not a number or a 1 if they pass in shooter_id[]=somthing, but it can never be a SQL injection string.

dont put the single quotes around $shooterID inside the query.

you'll probably also want something like:

while($shooter = mysql_fetch_array( $shooters )) {
 echo $shooter[0];
 $i++;
}

to print out the results.

Try something like this (comments added for clarity):

// Create the query, assuming $shooterID is an integer
$query = "SELECT name FROM teams WHERE id = '{$shooterID}'";

// Execute query
$shooters = mysql_query($query);

// Check result
if (!$shooters) { die(mysql_error()); }

// Iterate through rows
while ($shooter = mysql_fetch_array($shooters)) {
  // To display the entire $shooter array
  print_r($shooter);

  // To select the first item in $shooter array (no matter what it is)
  echo $shooter[0];

  // To specifically select the name field in $shooter array
  echo $shooter['name'];

  // To iterate over the $shooter array and display all fields
  // This will only be the name, unless you change the query to SELECT * FROM,
  // in which case this will return all fields in the table
  foreach ($shooter as $field) {
    echo $field;
  }
}

Have you tried:

 $query = "SELECT name FROM teams WHERE id = '" . $shooterID . "'";

Also, I don't see you defining $shooterID anywhere make sure you define it.
I.E.

$shooterID = 0;

Also,

$i = 0;  
while($shooter = mysql_fetch_array( $shooters )) {
     echo $shooter[$i];
     $i++;
}

should be

while($shooter = mysql_fetch_array( $shooters )) {
     echo $shooter[0];
}

or

while($shooter = mysql_fetch_array( $shooters )) {
     echo $shooter['name'];
}

or

while($shooter = mysql_fetch_object( $shooters )) {
     echo $shooter->name;
}

Also, you probably want some separation in your output:

while ($shooter = mysql_fetch_array( $shooters )) 
{
   echo $shooter[0], "\n";   //  or '<br>' if outputting to html
}