How to retrieve one row/one field using PHP mysql_fetch_array - php

Question

Just started learning PHP, Angular and mySQL and am trying to retrieve just one field value from one single row. The same code below works for a query that returns multiples rows:
$qry = "SELECT ID FROM SharePoint001 ORDER BY DownloadedTimeStamp DESC LIMIT 1"; //returns a5f415a7-3d4f-11e5-b52f-b82a72d52c35
$data = array();
while($rows = mysql_fetch_array($qry))
{
$data[] = array("ID" => $rows['ID']);
}
print_r(json_encode($data[0]));

Answer 1

I highly recommend switching to the mysqli extension. It's a much better way of doing things and you will probably find it much easier.
Here's a simple mysqli solution for you:
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
$row = $resource->fetch_assoc();
echo "{$row['field']}";
$resource->free();
$db->close();
If you're grabbing more than one row, I do it like this:
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
With Error Handling: If there is a fatal error the script will terminate with an error message.
// ini_set('display_errors',1); // Uncomment to show errors to the end user.
if ( $db->connect_errno ) die("Database Connection Failed: ".$db->connect_error);
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
if ( !$resource ) die('Database Error: '.$db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
With try/catch exception handling: This lets you deal with any errors all in one place and possibly continue execution when something fails, if that's desired.
try {
if ( $db->connect_errno ) throw new Exception("Connection Failed: ".$db->connect_error);
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
if ( !$resource ) throw new Exception($db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
} catch (Exception $e) {
echo "DB Exception: ",$e->getMessage(),"\n";
}
The MySQL extension is:
Officially deprecated (as of PHP 5.5. Will be removed in PHP 7.)
Lacks an object-oriented interface
Much slower than mysqli
Not under active development
Does not support:
Non-blocking, asynchronous queries
Transactions
Stored procedures
Multiple Statements
Prepared statements or parameterized queries
The "new" password authentication method (on by default in MySQL 5.6; required in 5.7+)
Since it is deprecated, using it makes your code less future proof. See the comparison of SQL extensions.

Answer 2

You have not executed the query, so you'll never get results.
First you have to prepare query and place it into a variable lets say $qry.
Then execute that query by using mysql_query() function and capture result into a variable lets say $result
Now you can iterate that $result with loop to fetch rows and cells (what you call fields).
Below is code I have corrected:
$qry = "SELECT ID FROM SharePoint001 ORDER BY DownloadedTimeStamp DESC LIMIT 1";
$result = mysql_query($qry); // You are missing this line
$data = array();
while($rows = mysql_fetch_array($result)) // You have to pass result into mysql_fetch_array not query string
{
$data[] = array("ID" => $rows['ID']);
}
print_r(json_encode($data[0]));