MySQLi Prepared Statement Query Results in an Array - php

Question

Trying to transition over my old mysql queries to mysqli prepared statements. I've got everything figured out except for one thing. How can I get the query results stored as an array? I used to do this like this:
$sql = "SELECT * FROM Users";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result) {
// do stuff
}
Now I have the following code. In this case, my array is a single record, so I don't need to iterate over it, but I want to hold it as an array so that I can refer to its field names. Also, I will have other queries that will return multiple records, so I'll need to iterat then.
$sql = "SELECT * FROM Users
WHERE (LOWER(first_name)=LOWER(?) && LOWER(last_name)=LOWER(?))";
$stmt = mysqli_stmt_init($link);
$this_user;
if (mysqli_stmt_prepare($stmt, $sql)) {
/* Bind the input parameters to the query */
mysqli_stmt_bind_param($stmt, 'ss', $first_name, $last_name);
/* execute query, store results in an array */
mysqli_stmt_execute($stmt);
$result = mysqli_fetch_array($stmt);
if (mysqli_num_rows($result) == 0) {
mysqli_stmt_close($stmt);
mysqli_close($link);
$tag_result = "failure";
$tag_message = "No matching user found";
echo encodeJSONObj($tag_result, $tag_message);
die();
}
if (mysqli_num_rows($result) > 1) {
mysqli_close($link);
$tag_result = "failure";
$tag_message = "Multiple records found for this user.";
echo encodeJSONObj($tag_result, $tag_message);
die();
}
$this_user = mysqli_fetch_array($result);
/* close statement */
mysqli_stmt_close($stmt);
}
$id = $this_user['id'];
$first_name = $this_user['first_name'];
$last_name = $this_user['last_name'];
// and so on...
Can somebody tell me what I am doing wrong? Thanks!
EDIT: With big thanks to Phil, I've modified my code. However, I still seem to be returning 0 rows even though my input parameters should return exactly 1 row. Here is what I have:
$sql = "SELECT id, first_name, last_name, group_id, email, cell
FROM Users
WHERE (first_name=? && last_name=?)";
$stmt = mysqli_stmt_init($link);
if (mysqli_stmt_prepare($stmt, $sql)) {
/* Bind the input parameters to the query */
mysqli_stmt_bind_param($stmt, 'ss', $first_name, $last_name);
/* execute query, bind result, and fetch value */
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $id, $first_name, $last_name, $group_id, $email, $cell);
mysqli_stmt_fetch($stmt);
if (mysqli_stmt_num_rows($stmt) == 0) {
mysqli_stmt_close($stmt);
mysqli_close($link);
echo "No results returned";
die();
}
...
}
This always outputs No results returned when it should find 1 row and skip right past that block. I've been staring at this for a long time, but I just can't see what I am doing wrong.

Answer 1

Your script contains numerous errors (as mentioned in comments above). Here's a simple step-by-step...
Prepare a statement and bind parameters
$stmt = $link->prepare($sql);
if (!$stmt) {
throw new Exception($link->error, $link->errno);
}
// you can error check this too but it rarely goes wrong
$stmt->bind_param('ss', $first_name, $last_name);
Execute the statement and store the result
if (!$stmt->execute()) {
throw new Exception($stmt->error, $stmt->errno);
}
$stmt->store_result();
Do your number of row checks against $stmt->num_rows...
if ($stmt->num_rows == 0) {
// ...
}
if ($stmt->num_rows > 1) {
// ...
}
Bind and fetch the result
// This relies on the SELECT column ordering.
// You should probably change your SELECT statement to
// SELECT id, first_name, last_name FROM Users...
$stmt->bind_result($id, $first_name, $last_name);
$stmt->fetch();
$stmt->close();
$link->close();
If you want to fetch the single result row as an associative array, try this instead
$result = $stmt->get_result(); // note - this requires the mysqlnd driver
$this_user = $result->fetch_array(MYSQLI_ASSOC);
$result->free();
$stmt->close();
$link->close();