PHP sqlite3不显示行结果

Question

I have the following code that runs ok but if I want to show the records from the fetched array it doesn't and no error is displayed.

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$db = new SQLite3('db/ZIPDB.sl3');

if(!$db) {
  echo $db->lastErrorMsg();
}

$sql =<<<EOF
  SELECT id, zip, country FROM ZIPDB WHERE zip LIKE '%32038%';
EOF;

$ret = $db->query($sql);
$row = $ret->fetchArray(SQLITE3_ASSOC);

echo "The entire array: \n";
var_dump($row);
//echo "ID: $row['id']\n";
//echo "ZIP: $row['zip']\n";
//echo "Country: $row['country']\n";

$db->close();
?>

The array is displayed correctly as:

The entire array: array(3) { ["id"]=> string(6) "213" ["zip"]=> string(4) "32038" ["country"]=> string(13) "UNITED STATES" }

If I uncomment the lines after var_dump() I get no output and no error either. I must specify the output is a single row not multiple so I would use a foreach loop. Any idea what's wrong?

EDIT:

Sorry, I think I forgot to:

echo "ID: " .$row['id']. "\n";
echo "ZIP: " .$row['zip']. "\n";
echo "Country: " .$row['country']. "\n";

Now works... I think I should close the question.

Answer 1

Since no answer, I will answer myself with a working version of the little snippet:

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$db = new SQLite3('db/ZIPDB.sl3');

if(!$db) {
  echo $db->lastErrorMsg();
}

$sql =<<<EOF
  SELECT id, zip, country FROM ZIPDB WHERE zip LIKE '%32038%';
EOF;

$ret = $db->query($sql);
$row = $ret->fetchArray(SQLITE3_ASSOC);

echo "The entire array: \n";
var_dump($row);
echo "ID: " .$row['id']. "\n";
echo "ZIP: " .$row['zip']. "\n";
echo "Country: " .$row['country']. "\n";

$db->close();
?>