i have retrieved mysql data from one table in json using the following script
$table_first = 'abc';
$query = "SELECT * FROM $table_first";
$resouter = mysql_query($query, $conn);
$set = array();
$total_records = mysql_numrows($resouter);
if($total_records >= 1){
while ($link = mysql_fetch_array($resouter, MYSQL_ASSOC)){
$set[] = $link;
}
}
echo json_encode($set);
how can i retrieved data from two other tables in which there is a foreign key of this table in both of those tables. OR simply how can i retrieved data from 3 mysql tables in php.
I believe the best way to go here is using a JOIN or just something like this:
$sql = "SELECT
tabl1.*, table2.*, tabl3.* FROM table1, table2, table3
WHERE
table1.fk1 = table2.id AND
table1.fk2 = table2.id";
//Do the whole selection process...
If you make the queries separately, you'll be forcing 3 queries onto your database and will end in a performance hit that you dont need. So, the idea is load all the data from the DB using joins or similar that and then encode the results. Is faster and you'll leave the merging work to MySQL
Hope I can help
You can get all data firstly.
Then merge the data array.
Finally use json_encode to change the data format.
There is a foreign key of this table in both so you can use "join" to retrieve values from other tables.
Suppose that there are two tables as State(st_id,st_name) and City(ct_id,ct_name,state_id). Now, primary key are st_id & ct_id respectively of tables State & City.
Connection between this two table can be establish by joining State.st_id and City.state_id.
Now, coming to your problem to retrieve data from two table State & City, we can make sql query like following,
$sql="select s.*, c.* from State s, City c
where s.st_id=c.state_id ";
Using above query you can fetch data from database and convert into json format and can send it to android system. here is a good article http://blog.sptechnolab.com/2011/02/10/android/android-connecting-to-mysql-using-php/. i hope you like it.
I believe your code roughly will look like this:
$query = "SELECT
A.column1 AS First_1
A.column2 AS First_2
B.column2 AS Second
C.column3 AS Third
FROM table1 A, table2 B, table3 C
WHERE
A.fk1 = B.id AND
B.fk2 = C.id";
where a column is a relevant record you want to show. Meanwhile,
AS will act as a key name in JSON.
Related
So i fetch my data from two tables in my php and encode it in one json object. I got everything i needed except that it doubles the display. And my teamone is located in the matches tables. instead of starting from array 0, it starts after the schedules tables. Which is array 7. I dont know why this happen.
Here is my php.
$sql = "SELECT * from schedule, matches;";
$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$db_name);
$result = mysqli_query($con,$sql);
$response = array();
while($row=mysqli_fetch_array($result))
{
array_push($response, array("start"=>$row[4],"end"=>$row[5],"venue"=>$row[6], "teamone"=>$row[8], "teamtwo"=>$row[9],
"s_name"=>$row[17]));
}
echo json_encode (array("schedule_response"=>$response));
mysqli_close($con);
?>
Here is my display. As you can see there are four displays but in my database it only has 2. It doubles the display. How do i fix this? Thanks
{ "schedule_response":[
{"start":"2016-11-10 00:00:00","end":"2016-11-04 00:00:00","venue":"bbbb","teamone":"aaa","teamtwo":"bbb",
"s_name":"hehehe"},
{"start":"2016-11-23 00:00:00","end":"2016-11-24 00:00:00","venue":"bbbbbbbb","teamone":"aaa","teamtwo":"bbb",
"s_name":"hehehe"},
{"start":"2016-11-10 00:00:00","end":"2016-11-04 00:00:00","venue":"bbbb","teamone":"ehe","teamtwo":"aha",
"s_name":"aaa"},
{"start":"2016-11-23 00:00:00","end":"2016-11-24 00:00:00","venue":"bbbbbbbb","teamone":"ehe","teamtwo":"aha",
"s_name":"aaa"}]}
I need to get the teamone, teamtwo and s_name values from the matches while i need the start, end and the venue from the schedule table in one query.
Schedule table
Matches Table
Because of your SQL query, you should not forget to perform some form of a grouping (the way you select results from both table defines that):
$sql = "SELECT * from schedule s, matches m GROUP BY s.id"; //I assume that your table schedule has an `id`
Or, you can rework the query to be more readable:
$sql = "SELECT s.*, m.* FROM schedule s
INNER JOIN matches m ON m.schedule_id = s.id
GROUP BY s.id"; //I assume that you have such database design that you have defined foreign key on table `matches`.
Of course, the INNER JOIN above could be LEFT OUTER JOIN - it all depends on your database design.
I think it is the problem with mysqli_fetch_array().
Can you please change mysqli_fetch_array() to mysqli_fetch_assoc()
I have to select data from multiple tables based on single key value. I have one table called maintable where I will get all the ids from, and i have another 10 tables in the same database which have maintable.id as a foreign key. Now I have to retrieve data from the 10 tables where maintable.id matches in one single table.
The code I have tried is:
$sql = select id from maintable;
$runsql = mysql_query($sql);
while($sqlRow = mysql_fetch_array($runsql ,MYSQL_ASSOC)) {
for($i=1;$i<=10(This is another table count);$i++) {
$servSql = "select * from table.$i where ref_id = ".$sqlRow['id'];
$runServerSql = mysql_query($servSql);
while($serverRow = mysql_fetch_array($runServSql,MYSQL_ASSOC)) {
}
}
}
Try something like this in a join:
SELECT * FROM maintable m
INNER JOIN othertable o
ON m.id = o.id
That will select from both tables using an inner join ON the id column. You may want to look up a basic SQL tutorial to learn the basic types of joins you can use. Good luck!
Hi there I have 2 tables
|id|musicname|url|image|type
and the second table
|id|user|songslist|
inside songsids theres an array like this
1,3,5,6,8 etc ...
What Im aiming to do is select * from table1 and echo out the table1 as in an array but instead of tables two array , the actual row of table1.
So basically To take out each row that contains the id in songslist and put them all into a php array.
I have learned a lot about PHP arrays , but I'm not that good with mysql , Any Idea of how can I do that ?
EDIT
$selectmusiclist = mysql_query("SELECT * FROM music");
$songslist = array();
while ($songs = mysql_fetch_assoc($selectmusiclist)){
$songslist[] = $songs;
}
and then table 2 select:
$username="user1";
$selectuser = mysql_query("SELECT * FROM usersmusic where user=$username");
$user = mysql_fetch_assoc($selectuser);
$songslist = $user['songslist'];
NOW I need to tell the array $songslist[] to output only the songs with id $songslist contained ids
I think running a join like this will give you the results you are after.
SELECT * FROM usersmusic as um
join music as m
on um.songslist = m.id
where user = '$username'
If $username is not a static value make sure you escape it; don't want to get SQL injected in the future.
Also note the mysql_ driver is now deprecated you should consider updating to mysqli or PDO.
This should be a basic question, but I haven't used Mysql for a very long time and forgot all the basic stuff. So SO programmers please bear with me.
I have 2 tables like this:
Table 1 (events): here
Table 2 (users): here
I would like to select all rows in the events table where event_invitees contains a username. I was able to do this using:
SELECT * FROM meetmeup_events WHERE event_invitees LIKE '%$username%'
Now I'd like to also select the event_invitees's photo from the users table (column called user_userphoto). My attempt to this was this:
$result = mysql_query("SELECT meetmeup_events.*, meetmeup_user.user_photo
FROM meetmeup_events
WHERE event_invitees LIKE '%$username%'
INNER JOIN meetmeup_user
ON meetmeup_user.user_username = meetmeup_events.event_inviter");
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows['meetmeup_user'][] = $r;
}
echo json_encode($rows);
This gave me an error: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
How can I do this? What am I missing? Can you give me some examples?
Thanks in advance! I'll be sure to accept the working answer!
You should change your mysql functions to either mysqli / PDO, although the problem seems to be the query itsef. Should be:
SELECT meetmeup_events.*, meetmeup_user.user_photo
FROM meetmeup_events
INNER JOIN meetmeup_user
ON meetmeup_user.user_username = meetmeup_events.event_inviter
WHERE event_invitees LIKE '%$username%'
(the WHERE clause at the end)
Sql fiddle demo: http://sqlfiddle.com/#!2/852a2/1
Its just a matter of getting the query coded in the correct order, and you might like to make it a little more managable by using alias's for the table names
Try this :-
SELECT me.*,
mu.user_photo
FROM meetmeup_events me
INNER JOIN meetmeup_user mu ON mu.user_username = me.event_inviter
WHERE me.event_invitees LIKE '%$username%'
This of course assumes that all the column names are correct and the mu.user_username = me.event_inviter does in fact make sence because those fields are in fact equal
Additional Suggestion
You are not actually issuing the query for execution by mysql.
You have to do this :-
$sql = "SELECT me.*,
mu.user_photo
FROM meetmeup_events me
INNER JOIN meetmeup_user mu ON mu.user_username = me.event_inviter
WHERE me.event_invitees LIKE '%$username%'";
$result = mysql_query($sql);
$rows = array('mysql_count' => mysql_num_rows($result) );
while($r = mysql_fetch_assoc($result)) {
$rows['meetmeup_user'][] = $r;
}
echo json_encode($rows);
Now in your browser using the javascript debugger look at the data that is returned. There should at least be a mysql_count field in it even if there is no 'meetmeup_user' array, and if it is zero you know it found nothing using your criteria.
I'm trying to generate a list of events that a user is attending. All I'm trying to do is search through columns and comparing the userid to the names stored in each column using LIKE.
Right now I have two different events stored in my database for testing, each with a unique eventID. The userid i'm signed in with is attending both of these events, however it's only displaying the eventID1 twice instead of eventID1 and eventID2.
The usernames are stored in a column called acceptedInvites separated by "~". So right now it shows "1~2" for the userid's attending. Can I just use %like% to pull these events?
$userid = $_SESSION['userid'];
echo "<h2>My Events</h2>";
$myEvents = mysql_query("select eventID from events where acceptedInvites LIKE '%$userid%' ");
$fetch = mysql_fetch_array($myEvents);
foreach($fetch as $eventsAttending){
echo $eventsAttending['eventID'];
}
My output is just 11 when it should be 12
Change your table setup, into a many-to-many setup (many users can attend one event, and one user can attend many events):
users
- id (pk, ai)
- name
- embarrassing_personal_habits
events
- id (pk, ai)
- location
- start_time
users_to_events
- user_id ]-|
|- Joint pk
- event id ]-|
Now you just use joins:
SELECT u.*
FROM users u
JOIN users_to_events u2e
ON u.id = u2e.id
JOIN events e
ON u2e.event_id = e.id
WHERE u.id = 11
I'm a bit confused by your description, but I think the issue is that mysql_fetch_array just returns one row at a time and your code is currently set up in a way that seems to assume $fetch is filled with an array of all the results. You need to continuously be calling mysql_fetch_array for that to happen.
Instead of
$fetch = mysql_fetch_array($myEvents);
foreach($fetch as $eventsAttending){
echo $eventsAttending['eventID'];
}
You could have
while ($row = mysql_fetch_array($myEvents)) {
echo $row['eventID'];
}
This would cycle through the various rows of events in the table.
Instead of using foreach(), use while() like this:
$myEvents = mysql_query("SELECT `eventID` FROM `events` WHERE `acceptedInvites` LIKE '".$userid."'");
while ($fetch = mysql_fetch_array($myEvents))
{
echo $fetch['eventID'];
}
It will create a loop like foreach() but simpler...
P.S. When you make a MySQL Query, use backticks [ ` ] to ensure that the string is not confused with MySQL functions (LIKE,SELECT, etc.).