How do I separate output from database? - php

Question

I tried to get multiple rows out from my database with PHP but all I get is one line of text like: "8910"
My code is following:
$sql = "SELECT * FROM posts WHERE idUsers=$id";
$sth = $conn->query($sql);
if(!$sth) {
echo("Error description: " . mysqli_error($conn));
die();
}
$rows = mysqli_num_rows($sth);
for ($x = 0; $x <= $rows; $x++) {
$sql = "SELECT idPosts FROM posts WHERE idUsers=$id";
if(!$sth) {
echo("Error description: " . mysqli_error($conn));
die();
}
$result = mysqli_fetch_array($sth);
$postId = $result['idPosts'];
echo $postId;
}
And then I edit this: echo $postId." ";
And get a space between the id's like this: 8 9 10.
I tried to do $postIds = explode(" ", $postId);
And then echoing out for example $postIds[0] but I get all the id's once again
Now I do not know what to do so I need help ^^

Answer 1

Replace $postId = $result['idPosts']; with $postId[] = $result['idPosts'];. That way you create an array an you can just access it like $postId[0].
You also forgot to query again.
...
$sql = "SELECT idPosts FROM posts WHERE idUsers=$id";
$sth = $conn->query($sql);
if(!$sth) {
...

Answer 2

Your second query is unnecessary, you already have all the data from your first query. Just replace your code with this:
$sql = "SELECT * FROM posts WHERE idUsers=$id";
$sth = $conn->query($sql);
if (!$sth) {
echo("Error description: " . mysqli_error($conn));
die();
}
$postIds = array();
while ($row = mysqli_fetch_array($sth)) {
$postIds[] = $result['idPosts'];
}
After this loop you will have an array of all the idPosts values. You can then process them as you need to. Or you can process them in the loop:
while ($row = mysqli_fetch_array($sth)) {
$postId = $result['idPosts'];
// code to process postId e.g.
echo "$postId<br/>";
}