Like this my query is working fine:
$pdo = $db->prepare('SELECT *
FROM projects WHERE project_id = :project_id ');
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "ok";
}
But when I add a subselection then I get a blank page:
$pdo = $db->prepare('SELECT *
(SELECT * FROM animals WHERE projects.animal=animals.id) AS animal
FROM projects WHERE project_id = :project_id ');
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "ok";
}
What did I do wrong?
You can use INNER JOIN to have both tables.
$pdo = $db->prepare('SELECT *
FROM projects
INNER JOIN animals ON projects.animal=animals.id
WHERE project_id = :project_id ');
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "ok";
}
Try this query beacuse this query will return all the projects that have an animal associated in table animals.
I think is this that you want
Found this solution:
$pdo = $db->prepare('SELECT *
FROM projects
LEFT JOIN animals ON projects.animal=animals.id
WHERE project_id = :project_id ');
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "ok";
}
Related
I got issue with nested array which seems are not Json object for some reason. When i try for e.g access jsonData["user"] it works, but when i try go deeper such as jsonData["user"]["photo_url"] then it's treated like a string and i cant access the value.
Current code:
<?php
require_once("db_connection.php");
$userId = $_GET["user_id"];
$query = "WITH user AS (SELECT id, JSON_OBJECT('display_name', u.display_name, 'photo_url', u.photo_url) AS user FROM users u WHERE id = :userId), info AS (SELECT id, JSON_ARRAYAGG(JSON_OBJECT('text', text, 'start_at', start_at, 'end_at', end_at, 'status', status)) AS information FROM report GROUP BY id), img AS (SELECT report_id, JSON_ARRAYAGG(JSON_OBJECT('name', name)) AS images FROM report_images GROUP BY report_id), cmt AS (SELECT report_id, COUNT(*) AS totalcomments FROM report_comments rc JOIN users u ON rc.user_id = u.id GROUP BY report_id) SELECT u.user, info.information, img.images, cmt.totalcomments FROM report r JOIN user u ON u.id = r.user_id LEFT JOIN info ON info.id = r.id LEFT JOIN img ON img.report_id = r.id LEFT JOIN cmt ON cmt.report_id = r.id";
$stmt = $db->prepare($query);
// Bind our variables.
$stmt->bindValue(":userId", $userId);
// Execute.
$stmt->execute();
$result = $stmt->fetchAll();
if (count($result) > 0) {
$toJson = json_encode($result);
echo $toJson;
} else {
$toJson = json_encode("Error");
echo $toJson;
}
?>
Old code which worked:
<?php
require_once("db_connection.php");
require_once("functions.php");
$userId = $_GET["user_id"];
$query = "SELECT * FROM report WHERE `user_id` = :userId";
$stmt = $db->prepare($query);
// Bind our variables.
$stmt->bindValue(":userId", $userId);
// Execute.
$stmt->execute();
$result = $stmt->fetchAll();
if (count($result) > 0) {
foreach ($result as $key => $value) {
$result[$key]["user"] = getUserById($db, $value["user_id"]);
}
$toJson = json_encode($result);
echo $toJson;
} else {
$toJson = json_encode("Error");
echo $toJson;
}
?>
Because you are aggregating some of the values as JSON in your query, you need to decode those first before attempting to use the values and then JSON encode the whole result set. You need to do that as you fetch the data, so replace:
$result = $stmt->fetchAll();
with:
$result = array();
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
$row['user'] = json_decode($row['user']);
$row['images'] = json_decode($row['images']);
$row['comments'] = json_decode($row['comments']);
$result[] = $row;
}
I created a SELECT to get my communities.
And create two SELECTs to get the communities I'm following.
But I get just my communities.
I do not get the communities I'm following.
$user_id = $_GET["id"];
$row1 = array();
$row2 = array();
// get my communities
$res1 = mysql_query("SELECT * FROM communities where user_id = '$user_id'");
while($r1 = mysql_fetch_assoc($res1)) {
$row1[] = $r1;
}
// get "id" of my communities I'm following
$res = mysql_query("SELECT * FROM communities_follow where user_id = '$user_id'");
while($r = mysql_fetch_assoc($res)) {
$coid = $r["coid"];
// get my communities I'm following
$res2 = mysql_query("SELECT * FROM communities where id = '$coid'");
while($r2 = mysql_fetch_assoc($res2)) {
$row2[] = $r2;
}
}
$resp = array_replace_recursive($row1, $row2);
print json_encode( $resp );
The inner join will get you those communities only, where you are following:
SELECT c.* FROM communities c
INNER JOIN communities_follow cf ON c.id = cf.coid
WHERE cf.user_id = '$user_id';
Or, without a JOIN:
SELECT * FROM communities
WHERE EXISTS (SELECT 1 FROM communities_follow cf
WHERE c.id = cf.coid AND cf.user_id = '$user_id')
Try this sql.
SELECT * FROM communities c LEFT JOIN communities_follow cf ON c.user_id = cf.user_id where
cf.user_id = '$user_id';
hello I have this variable $company->id
and I have function like this
<?php
function GetEmploees(){
$db = JFactory::getDBO();
$query = 'SELECT * FROM #__jbusinessdirectory_attribute_options AS c JOIN #__jbusinessdirectory_company_attributes AS cp
on c.id = cp.option_id';
$db->setQuery($query);
if( $rows = $db->loadObjectList() ) {
foreach( $rows as $row ){
echo $row->name;
}
}
}
echo GetEmploees($company->id);
?>
I want select $row->name as $company->id
how can I insert code WHERE company_id = '.$company->id.' in query?
company_id is in table #__jbusinessdirectory_company_attributes
Try
$sql = "SELECT * FROM #__jbusinessdirectory_attribute_options AS c INNER JOIN #__jbusinessdirectory_company_attributes AS cp (on c.id = cp.option_id) WHERE c.company_id = '.$company->id.'";
Should do the trick
I was working on a post system..
So, I have to show posts by friends of the user and the groups in which user has participated..
Here is my code to show posts..
<?php
$sql = "SELECT * FROM posts WHERE uploader_id=:friend_id ORDER BY id DESC";
$query = $db->prepare($sql);
$query->execute(array(
":friend_id" => $friend_id
));
$rows = $query->fetchAll(PDO::FETCH_ASSOC);
foreach ($rows as $row) {
$name = $row['name'];
echo "POST BY $name";
}
$sql = "SELECT * FROM group_posts WHERE id=:member_group ORDER BY id DESC";
$query = $db->prepare($sql);
$query->execute(array(
":member_group" => $group_id
));
$rows = $query->fetchAll(PDO::FETCH_ASSOC);
foreach ($rows as $row) {
$name = $row['name'];
echo "POST BY $name";
}
?>
Now, I want all these posts to be shuffled in a way that all the posts of the post table and group_posts table are shown in the descending order.
UPDATE
I edited my code to this..
I figured out that first I'll have to code this before coding my post system..
<?php
$sql = "SELECT * FROM friends WHERE user_one=:me OR user_two=:me2 UNION SELECT * FROM group_members WHERE member_id=:me3";
$query = $db->prepare($sql);
$query->execute(array(
":me" => $my_id,
":me2" => $my_id,
":me3" => $my_id
));
$rows = $query->fetchAll(PDO::FETCH_ASSOC);
foreach ($rows as $row) {
$user_one = $row['user_one'];
$user_two = $row['user_two'];
$group_id = $row['group_id'];
if ($user_one == $my_id) {
$friend_id = $user_two;
} else {
$friend_id = $user_one;
}
echo $friend_id . "<BR>" . $group_id;
}
?>
Now, here's the problem..
This is successfully printing the $friend_id but, it shows an undefined index 'group_id' while printing $group_id.
I have checked all the fields are correct.
Try using just one query with UNION
SELECT *
FROM (
SELECT name, id FROM posts WHERE uploader_id=:friend_id
UNION
SELECT name, id FROM group_posts WHERE id=:member_group
) p
ORDER BY p.id DESC
Note, your inner queries must return the same number of columns in the same order (and I think with the same name/alias, too).
I am using the following tables:
artists
related_artists
The idea is when I enter an artist's page with an artist_id I can use that id to load related artists. The following code works, but how do I put it in a single query? I can't figure it out.
To connect related_artists with artists I created the following code:
$sql = "SELECT related_artist_id FROM related_artists WHERE artist_id = 1";
$res = mysqli_query($db, $sql);
if (!$res) {
echo "Er is een fout opgetreden.";
exit;
} else {
while ($row = mysqli_fetch_array($res)) {
$query = 'SELECT * FROM artists WHERE artist_id = '.$row["related_artist_id"];
print_r($query."<br />\n");
$result = mysqli_query($db, $query);
if ($result) {
while ($test = mysqli_fetch_array($result)) {
echo $test["lastName"]."<br />\n";
}
} else {
echo "It doesn't work";
exit;
}
}
}
You can just try :
select *
from artists
where artist_id in (
select related_artist_id
from related_artists
WHERE artist_id = 1
);
You can use a LEFT JOIN, try this:
SELECT b.*
FROM related_artist a
LEFT JOIN artists b
USING(artist_id)
WHERE a.artist_id = 1
Should return * from artists, where I aliased artists as b and related_artist as a.
Didn't test, does it work for you / return the expected result?
SELECT * FROM artists where artists.arist_id = 1
INNER JOIN related_artist ON related_artist.artist_id = artists.artist_id
This provides a join on the artist_id columns of both tables, having artist_id = 1. I'm not sure if you need an Inner or a Left join