php\mysql not selecting all rows (php issue) - php

Question

I'm trying to SELECT all rows as specified in the query below, but I think the issue is related to mysql_fetch_row, because in MYSQL Workbench the MYSQL query (yes I know I should be using PDO) shows all desired rows.
$result = mysql_query("SELECT zebra_id FROM zebra WHERE user_id = '$id' AND pref = '1'")
or die(mysql_error());
$array = mysql_fetch_row($result);
echo json_encode($array);
JSON Output (only showing 1 row):
["986"]
I tried changing to mysql_fetch_array, but that doesn't output all rows in the JSON encoded $array.

Answer 1

mysql_fetch_row() only fetchs one row as the name implies. So your error is obvious.
mysql_fetch_array() won't work unless you iterrate through the entire result set*. You just don't call it once and expect to get the entire result set in an array.
$result = mysql_query("SELECT zebra_id FROM zebra WHERE user_id = '$id' AND pref = '1'")
or die(mysql_error());
$array = array();
while($row = mysql_fetch_assoc($result)) {
$array[] = $row;
}
echo json_encode($array);
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

Answer 2

You have to do this in a loop. See this:
$result = mysql_query("SELECT zebra_id FROM zebra WHERE user_id = '$id' AND pref = '1'")
or die(mysql_error());
$array = array();
while($row = mysql_fetch_row($result)){
$array[] = $row;
}
echo json_encode($array);
And also I have to say that, do not use mysql_* functions anymore. These are deprecated officially. You have to use mysqli_* or PDO instead.