I would like to know how I could join these 3 queries together as I'm wanting only one JSON output, I thought INNER JOIN would do this. But don't know how to use this. Can someone guide me onto the right path please?
$json = array();
$following_string = mysqli_real_escape_string($mysqli,$_SESSION['id']);
$call="SELECT * FROM streamdata WHERE streamitem_id < '$lastID' AND streamitem_target=".$following_string." OR streamitem_creator=".$following_string." ORDER BY streamitem_id DESC LIMIT 10";
$chant = mysqli_query($mysqli, $call) or die(mysqli_error($mysqli));
$json['streamdata'] = array();
while ($resultArr = mysqli_fetch_assoc($chant)) {
$json['streamitem_id'] = $resultArr['streamitem_id'];
$json['streamitem_content'] = $resultArr['streamitem_content'];
$json['streamitem_timestamp'] = Agotime($resultArr['streamitem_timestamp']);
$json['streamdata'] = $json;
}
/***** COMMENTS *****/
$check = "SELECT comment_id, comment_datetime, comment_streamitem, comment_poster, comment_content FROM streamdata_comments WHERE comment_poster=".$following_string." ";
$check1 = mysqli_query($mysqli,$check);
$json['streamdata_comments'] = array();
while ($resultArr = mysqli_fetch_assoc($check1)) {
$json['comment_id'] = $resultArr['comment_id'];
$json['comment_content'] = $resultArr['comment_content'];
$json['comment_poster'] = $resultArr['comment_poster'];
$json['comment_datetime'] = Agotime($resultArr['comment_datetime']);
$json['comment_streamitem'] = $resultArr['comment_streamitem'];
$json['streamdata_comments'] = $json;
}
/***** USERS *****/
$check = "SELECT * FROM users WHERE id=".$following_string."";
$check1 = mysqli_query($mysqli,$check);
$json['users'] = array();
while ($resultArr = mysqli_fetch_assoc($check1)) {
$json['username'] = $resultArr['username'];
$json['id'] = $resultArr['id'];
$json['first'] = $resultArr['first'];
$json['middle'] = $resultArr['middle'];
$json['last'] = $resultArr['last'];
$json['users'] = $json;
}
echo json_encode($json);
}
?>
You're fetching unrelated data, so you can't use a join at the SQL level.
But JSON couldn't care less WHAT you feed it, or how. Just build the appropriate PHP-level data structure, e.g.
$data = array();
$data['streamdata'] = array();
... insert data from 'streamdata' query...
$data['streamdata_comments'] = array();
... insert comment data ...
$data['users'] = array();
... insert user data ...
which will give you a 3-way array containing the data from each of your queries. You then pass that entire $data structure to json_encode, and boom - you've got your 3 unrated queries in a single data structure, without every touching an SQL join.
Some previous answers have suggested that you can't join unrelated tables, but these are clearly not unrelated tables. The streamdata and streamdata_comments tables are quite closely related, and the users table maps user ID values in the other tables to names.
At the SQL level, these can be combined easily:
SELECT d.*, c.*, u.*
FROM streamdata AS d
JOIN streamdata_comments AS c ON d.streamitem_ID = c.comment_streamitem
JOIN users AS u ON u.user_id = c.comment_poster
WHERE c.comment_poster = '$following_string'
AND d.streamitem_id < '$lastID'
AND (d.streamitem_target = '$following_string' OR
d.streamitem_creator = '$following_string');
Whether the result makes sense for wrapping into a JSON string is a different matter, on which I can't pontificate. This would give you one record from the comments information for each comment associated with each stream item.
You are fetching unrelated data. Joining data is only usefull when the data to join has a relation.
You can't join apples, cows and monkeys.
Related
I've got the following code which queries a table. Then it uses the result to make another query. That result is then used to make a third query.
But how do I grab the userid field from the 2nd query in order to grab a name from a users table and join that to the result of the 3rd query?
Please note once I figure out the code I will convert this to a prepared statement. It's just easier for me to work with legacy code when figuring out queries.
$selectaudioid = "SELECT audioid FROM subscribe WHERE userid = $userid";
$audioResult=$dblink->query($selectaudioid);
if ($audioResult->num_rows>0) {
while ($row = $audioResult->fetch_assoc()) {
$newaudio = $row[audioid];
$getallaudio = "SELECT opid, userid from audioposts WHERE audioid = $newaudio" ;
$getallresult = $dblink->query($getallaudio);
if ($getallresult->num_rows>0) {
while ($row = $getallresult->fetch_assoc()) {
$opid = $row[opid];
$opuserid = $row[userid];
$getreplies =
"SELECT * from audioposts ap WHERE opid = $opid AND opid
NOT IN (SELECT opid FROM audioposts WHERE audioposts.opid = '0' )";
$getreplyresults = $dblink->query($getreplies);
if ($getreplyresults->num_rows>0) {
while ($row = $getreplyresults->fetch_assoc()) {
$dbdata[]=$row;
}
}
}
}
}
} "SELECT * from audioposts ap WHERE opid = $opid AND opid
NOT IN (SELECT opid FROM audioposts WHERE audioposts.opid = '0' )";
$getreplyresults = $dblink->query($getreplies);
if ($getreplyresults->num_rows>0) {
while ($row = $getreplyresults->fetch_assoc()) {
$dbdata[]=$row;
}
}
}
}
}
}
echo json_encode($dbdata);
The result I need are rows of json encoded instances of $getreplyresults with the $row[userid] from the original result joined to each row.
Here's what I did in the end. Now I just have to figure out how to convert this to a prepared statement in order to avoid malicious injection.
$selectaudioid = "SELECT audioid FROM subscribe WHERE userid = $userid";
$audioResult=$dblink->query($selectaudioid);
if ($audioResult->num_rows>0) {
while ($row = $audioResult->fetch_assoc()) {
$newaudio = $row[audioid];
$getallaudio = "
SELECT ap.audioid, ap.title, us.name FROM audioposts ap
INNER JOIN audioposts a2 ON a2.audioid = ap.opid
INNER JOIN users us ON us.id = a2.userid
WHERE ap.opid = $newaudio AND ap.opid <> '0'
";
$getallresult = $dblink->query($getallaudio);
if ($getallresult->num_rows>0) {
while ($row = $getallresult->fetch_assoc()) {
$dbdata[]=$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';
I'm trying to display a list of status updates from artists that a logged in user is following.
So far I have this:
#Get the list of artists that the user has liked
$q = "SELECT * FROM artist_likes WHERE user_id = '1' ";
$r = mysqli_query($dbc,$q);
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
#Now grab the statuses for each artist
$status_query = "SELECT * FROM status_updates WHERE artist_id = '".$row['artist_id']."' ";
$status_result = mysqli_query($dbc,$status_query)
}
But i'm not sure how to loop through and display the returned status updates?
This isn't a strong point of mine, so any pointers would be greatly appreciated!
What prevented you from doing similar to what you'd already done for the first query? Something like follows:
#Get the list of artists that the user has liked
$q = "SELECT * FROM artist_likes WHERE user_id = '1' ";
$r = mysqli_query($dbc,$q);
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
#Now grab the statuses for each artist
$status_query = "SELECT * FROM status_updates WHERE artist_id = '".$row['artist_id']."' ";
$status_result = mysqli_query($dbc,$status_query)
while($status_result_row = mysqli_fetch_assoc($status_result)) {
echo $status_result_row['mycol']; // This is where you know better than us
}
}
Or if those two tables artist_likes and status_updates have artist_id in common then you could just use one query with a join. (But don't know if you are asking for that).
Just for avoiding multiple query, you can use one query like this:
SELECT l.*, s.*
from artist_likes l, status_updates s
WHERE
l.artist_id = s.artist_id and
l.user_id = '1'
or
SELECT l.*, s.*
from artist_likes l
JOIN status_updates s on (l.artist_id = s.artist_id)
WHERE
l.user_id = '1'
I currently have this query with an array that outputs the variables within using a dynamic input in my form (term), this creates a Dynamic Search with auto complete to fill in all of the details for a product.
$return_arr = array();
$param = $_GET["term"];
$fetch = mysql_query("SELECT * FROM crd_jshopping_products WHERE `name_en-GB` REGEXP '^$param'");
while ($row = mysql_fetch_array($fetch, MYSQL_ASSOC)) {
//$row_array['category_id'] = $row ['category_id'];
$row_array['product_id'] = $row['product_id'];
$row_array['product_names'] = $row['name_en-GB'];
$row_array['jshop_code_prod'] = $row['product_ean'];
$row_array['_ext_price_html'] = number_format($row['product_price'],2);
if (!empty($row['product_thumb_image']) AND isset($row['product_thumb_image'])){
$row_array['image'] = $row['product_thumb_image'];
}else {
$row_array['image'] = 'noimage.gif';
}
array_push( $return_arr, $row_array);
}
mysql_close($conn);
echo json_encode($return_arr);
Unfortunately I also need to get the category_id which is not in the same table, I have tried to modify my query as such, but to no avail:
$fetch = mysql_query("SELECT * FROM crd_jshopping_products WHERE `name_en-GB` REGEXP '^$param' AND `crd_jshopping_products_to_categories` = `product_id` ");
What step am I missing here ? The product_id's match in both tables?
try this query instead and try to understand what I have written in it:
$fetch = mysql_query("
SELECT
p.*,
c.category_id
FROM
crd_jshopping_products as p
INNER JOIN crd_jshopping_products_to_categories as c
ON p.product_id = c.product_id
WHERE
`p.name_en-GB` REGEXP '^$param'
");
This means:
SELECT:
Give me everything from p and the category_id from c.
FROM:
Do this from rows in the tables crd_jshopping_products (referred to as p) and crd_jshopping_products_to_categories (referred to as c), where the rows match on the count of p.product_id is the same as c.product_id.
WHERE:
Only return the rows where p.name_en-GB REGEXP '^$param'.
I'm trying to mesh the below mysql query results into a single json object, but not quite sure how to do it properly.
$id = $_POST['id'];
$sql = "SELECT contracts.po_number, contracts.start_date, contracts.end_date, contracts.description, contracts.taa_required, contracts.account_overdue, jobs.id AS jobs_id, jobs.job_number, companies.id AS companies_id, companies.name AS companies_name
FROM contracts
LEFT JOIN jobs ON contracts.job_id = jobs.id
LEFT JOIN companies ON contracts.company_id = companies.id
WHERE contracts.id = '$id'
ORDER BY contracts.end_date";
$sql2 = "SELECT types_id
FROM contracts_types
WHERE contracts_id = '$id'";
//return data
$sql_result = mysql_query($sql,$connection) or die ("Fail.");
$arr = array();
while($obj = mysql_fetch_object($sql_result)) { $arr[] = $obj; }
echo json_encode($arr); //return json
//plus the selected options
$sql_result2 = mysql_query($sql2,$connection) or die ("Fail.");
$arr2 = array();
while($obj2 = mysql_fetch_object($sql_result2)) { $arr2[] = $obj2; }
echo json_encode($arr2); //return json
Here's the current result:
[{"po_number":"test","start_date":"1261116000","end_date":"1262239200","description":"test","taa_required":"0","account_overdue":"1","jobs_id":null,"job_number":null,"companies_id":"4","companies_name":"Primacore Inc."}][{"types_id":"37"},{"types_id":"4"}]
Notice how the last section [{"types_id":"37"},{"types_id":"4"}] is placed into a separate chunk under root. I'm wanting it to be nested inside the first branch under a name like, "types".
I think my question has more to do with Php array manipulation, but I'm not the best with that.
Thank you for any guidance.
Combine the results into another structure before outputting as JSON. Use array_values to convert the type IDs into an array of type IDs. Also, fix that SQL injection vulnerability. Using PDO, and assuming the error mode is set to PDO::ERRMODE_EXCEPTION:
$id = $_POST['id'];
try {
$contractQuery = $db->prepare("SELECT contracts.po_number, contracts.start_date, contracts.end_date, contracts.description, contracts.taa_required, contracts.account_overdue, jobs.id AS jobs_id, jobs.job_number, companies.id AS companies_id, companies.name AS companies_name
FROM contracts
LEFT JOIN jobs ON contracts.job_id = jobs.id
LEFT JOIN companies ON contracts.company_id = companies.id
WHERE contracts.id = ?
ORDER BY contracts.end_date");
$typesQuery = $db->prepare("SELECT types_id
FROM contracts_types
WHERE contracts_id = ?");
$contractQuery->execute(array($id));
$typesQuery->execute(array($id));
$result = array();
$result['contracts'] = $contractQuery->fetchAll(PDO::FETCH_ASSOC);
$result['types'] = array_values($typesQuery->fetchAll(PDO::FETCH_NUM));
echo json_encode($result); //return json
} catch (PDOException $exc) {
...
}
If $contractQuery returns at most one row, change the fetch lines to:
$result = $contractQuery->fetch(PDO::FETCH_ASSOC);
$result['types'] = array_values($typesQuery->fetchAll(PDO::FETCH_NUM));
It would seem like you'd be better served by consolidating the two queries with a JOIN at the SQL level. However, assuming the two arrays have equal length:
for ($x = 0, $c = count($arr); $x < $c; $x++) {
if (isset($arr2[$x])) {
$arr[$x] += $arr2[$x];
}
}
echo json_encode($arr);
Edit: you would need to change from mysql_fetch_object to mysql_fetch_assoc for this to work properly.
Why are you using 2 distinct arrays ? I would simply add the rows of the 2nd query in $arr instead of $arr2. This way, you end up with a single array containing all rows from the 2 queries.