build an array of an array from SQL query with php - php

Question

I have the following working array of an array in PHP:
$explore1 = array
(
array("a.html", "A"),
array("b.html", "B"),
array("c","C")
);
$arrlength = count($explore1);
for($x = 0; $x < $arrlength; $x++) {
echo '<li>'.$explore1[$x][1].'</li>';
} }
I want to populate the explore1 array from SQL. If i simply change the code like below it has errors but I don't know what I'm suppose to do instead?
$sql = 'SELECT url, name FROM explore_items WHERE menuID="item1"';
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
$explore1 = array //IT DOESN'T LIKE THIS LINE
(
while($row = $result->fetch_assoc()) {
// BUILD SAME LINES AS ABOVE WITH ECHO
echo "array('" . $row["url"]. ", '" . $row["name"]. "'),";
}
);
Can anybody help?

Answer 1

Either you don't need $explore1 at all:
$sql = 'SELECT url, name FROM explore_items WHERE menuID="item1"';
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
//echo "array('" . $row["url"]. ", '" . $row["name"]. "'),";
echo '<li>'.$row["name"].'</li>';
}
}
or if you need you can fetch_all():
$sql = 'SELECT url, name FROM explore_items WHERE menuID="item1"';
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$explore1=$result->fetch_all(MYSQLI_ASSOC);
foreach($explore1 as $row ) {
//echo "array('" . $row["url"]. ", '" . $row["name"]. "'),";
echo '<li>'.$row["name"].'</li>';
}
}

Answer 2

Please learn some php basics.
Defining an array of arrays is something like:
$explore1 = array();
while($row = $result->fetch_assoc()) {
$explore1[] = array($row["url"], $row["name"]);
}

Answer 3

You must populate the array inside the loop:
while($row = $result->fetch_assoc()) {
$explore1[$row["url"]] = $row["name"];
}