mysql query result in php variable

Question

Is there any way to store mysql result in php variable? thanks

$query = "SELECT username,userid FROM user WHERE username = 'admin' ";
$result=$conn->query($query);

then I want to print selected userid from query.

Answer 1

Of course there is. Check out mysql_query, and mysql_fetch_row if you use MySQL.
Example from PHP manual:

<?php
$result = mysql_query("SELECT id,email FROM people WHERE id = '42'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

echo $row[0]; // 42
echo $row[1]; // the email value
?>

Answer 2

There are a couple of mysql functions you need to look into.

• mysql_query("query string here") : returns a resource

• mysql_fetch_array(resource obtained above) : fetches a row and return as an array with numerical and associative(with column name as key) indices. Typically, you need to iterate through the results till expression evaluates to false value. Like the below:

  while ($row = mysql_fetch_array($query)){
      print_r $row;
  }

  Consult the manual, the links to which are provided below, they have more options to specify the format in which the array is requested. Like, you could use mysql_fetch_assoc(..) to get the row in an associative array.

Links:

• http://php.net/manual/en/function.mysql-query.php
• http://php.net/manual/en/function.mysql-fetch-array.php

In your case,

$query = "SELECT username,userid FROM user WHERE username = 'admin' ";
$result=mysql_query($query);
if (!$result){
    die("BAD!");
}
if (mysql_num_rows($result)==1){
    $row = mysql_fetch_array($result);
    echo "user Id: " . $row['userid'];
}
else{
    echo "not found!";
}