No response from database query as JSON
I want to get an array of articles that are liked by the current user that's logged in but I don't get any response from my server. I am doing a http post with userId set that will be used in my query that I'm using to get liked articles by that specific user. My query is as following:
SELECT *, (CASE likes.userId WHEN $userId THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'
I do get the userId from the http post in my PHP file but somehow I get the feeling that the query isn't getting runned by my PHP file.
The code of my PHP file is as following:
require "dbconnect.php";
$data = file_get_contents("php://input");
if(isset($data))
{
$request = json_decode($data);
$userId = $request->userId;
}
$sql = "SELECT *, (CASE likes.userId WHEN '$userId' THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$likedata = $row;
}
$count = mysqli_num_rows($result);
if($count > 0)
{
echo json_encode($likedata);
}
Thanks,
php http ionic-framework
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
|
show 5 more comments
I want to get an array of articles that are liked by the current user that's logged in but I don't get any response from my server. I am doing a http post with userId set that will be used in my query that I'm using to get liked articles by that specific user. My query is as following:
SELECT *, (CASE likes.userId WHEN $userId THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'
I do get the userId from the http post in my PHP file but somehow I get the feeling that the query isn't getting runned by my PHP file.
The code of my PHP file is as following:
require "dbconnect.php";
$data = file_get_contents("php://input");
if(isset($data))
{
$request = json_decode($data);
$userId = $request->userId;
}
$sql = "SELECT *, (CASE likes.userId WHEN '$userId' THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$likedata = $row;
}
$count = mysqli_num_rows($result);
if($count > 0)
{
echo json_encode($likedata);
}
Thanks,
php http ionic-framework
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57
|
show 5 more comments
I want to get an array of articles that are liked by the current user that's logged in but I don't get any response from my server. I am doing a http post with userId set that will be used in my query that I'm using to get liked articles by that specific user. My query is as following:
SELECT *, (CASE likes.userId WHEN $userId THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'
I do get the userId from the http post in my PHP file but somehow I get the feeling that the query isn't getting runned by my PHP file.
The code of my PHP file is as following:
require "dbconnect.php";
$data = file_get_contents("php://input");
if(isset($data))
{
$request = json_decode($data);
$userId = $request->userId;
}
$sql = "SELECT *, (CASE likes.userId WHEN '$userId' THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$likedata = $row;
}
$count = mysqli_num_rows($result);
if($count > 0)
{
echo json_encode($likedata);
}
Thanks,
php http ionic-framework
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
I want to get an array of articles that are liked by the current user that's logged in but I don't get any response from my server. I am doing a http post with userId set that will be used in my query that I'm using to get liked articles by that specific user. My query is as following:
SELECT *, (CASE likes.userId WHEN $userId THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'
I do get the userId from the http post in my PHP file but somehow I get the feeling that the query isn't getting runned by my PHP file.
The code of my PHP file is as following:
require "dbconnect.php";
$data = file_get_contents("php://input");
if(isset($data))
{
$request = json_decode($data);
$userId = $request->userId;
}
$sql = "SELECT *, (CASE likes.userId WHEN '$userId' THEN 1 END) AS liked FROM article INNER JOIN likes on likes.articleId = article.id WHERE likes.userId = '$userId'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$likedata = $row;
}
$count = mysqli_num_rows($result);
if($count > 0)
{
echo json_encode($likedata);
}
Thanks,
php http ionic-framework
php http ionic-framework
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
asked Nov 21 '18 at 11:58
Batuhan BeyogullariBatuhan Beyogullari
33
33
echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57
|
show 5 more comments
echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57
echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57
|
show 5 more comments
1 Answer
1
active
oldest
votes
Try this json_encode($likedata,JSON_UNESCAPED_UNICODE); It may due to non-utf8 encoded characters
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
Try this json_encode($likedata,JSON_UNESCAPED_UNICODE); It may due to non-utf8 encoded characters
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
add a comment |
Try this json_encode($likedata,JSON_UNESCAPED_UNICODE); It may due to non-utf8 encoded characters
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
add a comment |
Try this json_encode($likedata,JSON_UNESCAPED_UNICODE); It may due to non-utf8 encoded characters
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
Try this json_encode($likedata,JSON_UNESCAPED_UNICODE); It may due to non-utf8 encoded characters
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
answered Nov 26 '18 at 12:32
Chris shiChris shi
1531111
1531111
add a comment |
add a comment |
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echo the query $sql and then exit; and check whether you are getting specific query value
– Chris shi
Nov 21 '18 at 12:10
@Chrisshi I ran the query multiple times in the database itself aswell with my own userId and it shows my liked articles and this is the output I get picture
– Batuhan Beyogullari
Nov 21 '18 at 12:13
print the likedata array and check whether the query is returning any value
– Chris shi
Nov 21 '18 at 12:33
@Chrisshi I printed the array and all it returns is this imgur.com/a/EfAWPdX
– Batuhan Beyogullari
Nov 26 '18 at 10:18
Can you remove the if($count > 0) { } condition and just echo json_encode($likedata);
– Chris shi
Nov 26 '18 at 10:57