Translating a database to JSON - php

Question

I have a method of creating JSON into an array that looks like this:
[{"date":"","name":"","image":"","genre":"","info":"","videocode":""},{...},{...}]
I first tried getting the data from a html page (not the database) like this:
$arr = array();
$info = linkExtractor($html);
$dates = linkExtractor2($html);
$names = linkExtractor3($html);
$images = linkExtractor4($html);
$genres = linkExtractor5($html);
$videocode = linkExtractor6($html);
for ($i=0; $i<count($images); $i++) {
$arr[] = array("date" => $dates[$i], "name" => $names[$i], "image" => $images[$i], "genre" => $genres[$i], "info" => $info[$i], "videocode" => $videocode[$i]);
}
echo json_encode($arr);
Where each linkExtractor looks a bit like this - where it grabs all the text within a class videocode.
function linkExtractor6($html){
$doc = new DOMDocument();
$last = libxml_use_internal_errors(TRUE);
$doc->loadHTML($html);
libxml_use_internal_errors($last);
$xp = new DOMXPath($doc);
$result = array();
foreach ($xp->query("//*[contains(concat(' ', normalize-space(#class), ' '), ' videocode ')]") as $node)
$result[] = trim($node->textContent); // Just push the result here, don't assign it to a key (as that's why you're overwriting)
// Now return the array, rather than extracting keys from it
return $result;
}
I now want to do this instead with a database.
So I have tried to replace each linkExtractor with this - and obviously the connection:
function linkExtractor6($html){
$genre = mysqli_query($con,"SELECT genre
FROM entries
ORDER BY date DESC");
foreach ($genre as $node)
$result[] = $node;
return $result;
}
But I am getting the error:
Invalid argument supplied for foreach()

Answer 1

Avoid redundancy and run a single SELECT
function create_json_db($con){
$result = mysqli_query($con,"SELECT date, name, image, genre, info, videocode
FROM entries
ORDER BY date DESC");
$items= array();
while ($row = mysqli_fetch_assoc($result)) {
$items[] = $row;
}
return $items ;
}

Answer 2

Try to use this. More info in the official PHP documentation:
function linkExtractor6($html){
$result = mysqli_query($con,"SELECT genre
FROM entries
ORDER BY date DESC");
$items = array();
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$items[] = $row;
}
return $items;
}

Answer 3

First, you are not iterating through your results via something like mysqli_fetch_array. So here is the function with mysqli_fetch_array in place. But there is a much larger issue. Read on.
function linkExtractor6($html){
$result = mysqli_query($con,"SELECT genre
FROM entries
ORDER BY date DESC");
$ret = array();
while ($row = mysqli_fetch_array($result)) {
$items[] = $row;
}
return $ret ;
}
Okay, with that done, it still won’t work. Why? Look at your function. Specifically this line:
$result = mysqli_query($con,"SELECT genre
But where is $con coming from? Without a database connection mysqli_query will not work at all. So if you somehow have $con set outside your function, you need to pass it into your function like this:
function linkExtractor6($con, $html){
So your full function would be:
function linkExtractor6($con, $html){
$result = mysqli_query($con,"SELECT genre
FROM entries
ORDER BY date DESC");
$ret = array();
while ($row = mysqli_fetch_array($result)) {
$items[] = $row;
}
return $ret ;
}
Remember, functions are self-contained & isolated from whatever happens outside of them unless you explicitly pass data into them.