Here I want to access two table field but I cant get success. Here is my little code. please check that. I want to access Analysis.Right and tabl3.right.
I am printing its with foreach loop. like $res['Right'] of Analysis and $res['right'] of tabl3. when I try to print this it's show me error
Undefind index Right.
any one can help me
$qry = "select Analysis.Q_Id, tabl3.Q_Id, Analysis.Right, tabl3.right from tabl3 INNER JOIN Analysis ON Analysis.Q_Id = tabl3.Q_Id where Analysis.Q_Id = 3";
please help..
you have tow column with right name related to different table so there is not a column name right but 'Analysis.Right ' or 'tabl3.right'
or you can assign an alias for set the column name equalt to Right where you need .. eg:
$qry = "select
Analysis.Q_Id
, tabl3.Q_Id
, Analysis.Right as Right
, tabl3.right as Right_t3
from tabl3
INNER JOIN Analysis ON Analysis.Q_Id = tabl3.Q_Id where Analysis.Q_Id = 3";
Your result set has columns with the same name. Give them different names:
select t3.Q_Id, a.Right as a_right, t3.right as t3_right
from tabl3 t3 inner join
Analysis a
on a.Q_Id = t3.Q_Id
where a.Q_Id = 3;
When you look for the names in your code, look for a_right and t3_right.
Note that you don't need to return Q_Id twice. The ON clause guarantees that the values are the same.
Related
So I'm trying to implement a JOIN in my PHP but I don't know how to include the WHERE clause with JOIN in my query.
I'm trying to do:
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts
INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpeciesID="G. cuvier"
Basically I'm trying to make sure the join happens on the tables where the SpeciesID is G. cuvier but everything I have tried so far doesn't work and the error it is giving me now is "Column SpeciesID in where clause is ambiguous".
Here is my relevant PHP code:
<?php
include 'connect.php';
$result = mysqli_query($connect,"SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located FROM SpecialFacts INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID WHERE SpeciesID='G. cuvier'") or die("fail" . mysqli_error($connect));
$i = 0;
while($row = mysqli_fetch_array($result))
{
echo $row[$i];
$i = $i + 1;
}
Column SpeciesID exists in both tables, so it doesn't know which need to be compared to value given as G. cuvier. As you are writing every column name as table name with dot(.), the same should be in where condition column SpecialFacts.SpeciesID = "G. cuvier".
So query should be like:
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts
INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpecialFacts.SpeciesID="G. cuvier"
It's not giving you that error because you're using WHERE incorrectly. It's giving you that error because there are multiple tables in your query that have a column with that name, hence the "ambiguous". You just need to disambiguate it by adding the table name to the identifier.
WHERE SpecialFacts.SpeciesID="G. cuvier"
or
WHERE Habitat.SpeciesID="G. cuvier"
Since you're inner joining the tables on that column, either table should work for the WHERE clause. I would suggest using the smaller table for performance reasons, but honestly I'm not 100% certain if it will matter or not. You can do EXPLAIN on each one to see how they compare.
Use below query instead
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts INNER JOIN Habitat
ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpecialFacts.SpeciesID='G. cuvier'
I have used single quote and used column name with table name.
I am creating a query and I used the LEFT JOIN to join two tables. But I'm having trouble in getting the fb_id value from the table, it gives me an empty result. Here is my code:
$sql = "SELECT * FROM tblfeedback a LEFT JOIN tblreply b ON a.fb_id = b.fb_id WHERE a.fb_status = 1 ORDER BY a.fb_id DESC";
$res = $con->query($sql);
....
....
The query above would give me a result like this :
I think that's why I don't get the fb_id value is because the last column is null. How do I get the value of fb_id from the first column? Thanks. I am really having trouble with this. I hope someone can enlightened my mind.
You should give an alias to the column in the parent table, because the column names are the same in both tables. When fetch_assoc() fills in $row['fb_id'], it gets the last one in the result row, which can be NULL because it comes from the second table.
SELECT a.fb_id AS a_id, a.*, b.*
FROM tblfeedback a
LEFT JOIN tblreply b ON a.fb_id = b.fb_id
WHERE a.fb_status = 1
ORDER BY a_id DESC
Then you can access $row['a_id'] to get this column.
More generally, I recommend against using SELECT *. Just select the columns you actually need. So you can select a.fb_id without selecting b.fb_id, and it will always be filled in.
Because you are using a left join, the 2 rows in your result set image are the rows from tblfeedback whose fb_id were not found in tblreply. We know this is true because all the tblreply columns in the result set are null.
With that said, its not real clear what you are asking for. If you are asking how you access the tblfeedback.fd_id column from your query via php, you can use the fetch_array method and use the 0 index.
$sql = "SELECT * FROM tblfeedback a LEFT JOIN tblreply b ON a.fb_id = b.fb_id WHERE a.fb_status = 1 ORDER BY a.fb_id DESC";
$res = $con->query($sql);
while($row = $res->fetch_array()) {
echo "fb_id: " . $row[0] . "<br>";
}
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()
Ok, here's a database.
http://i.stack.imgur.com/j05AB.png
Say I've inserted values into the database for each of these tables, although the IDs would be auto incrementing. There are many BVALUES from each AVALUE, thus the AB table. I have all the AVALUEs from TABLE A in a drop-down list. A user selects an AVALUE which I put into a variable using
$AVALUE = $_POST['AVALUE']
Then I do an sql statement to get the AVALUEs from TABLE A that equal $AVALUE.
$sql = "SELECT AVALUE FROM TABLEA WHERE" . $AVALUE . " = AVALUE";
How do I then get the NAMEID from TABLEA for that AVALUE, then reference to AB where TABLEANAMEID = NAMEID from TABLEA? Then I want to get all the BVALUES by getting all the TABLEBNAMEIDs that correspond to the TABLEANAMEIDs.
I then want those BVALUES in a drop-down list on a seperate HTML page. After a bit of Googling the solution I think would be to do some sort of a loop putting the BVALUES into a variable as all the NAMEIDs from TABLE B increment where the variable would be in an $BVALUE loop and the list values would show with all the BVALUES.
I hope I explained that right. I think I know what I'm trying to do but I have no idea how to actually implement it. Please help guys.
You need to join those table together. What you are describing is an m:n relation. In this case you have do use 2 joins like this:
SELECT * FROM TableA AS a WHERE a.avalue = $AVALUE
JOIN TableAB AS a2b ON a.namevalue = a2b.id_a
JOIN TableB AS b ON a2b.id_b = b.id
Maybe u means, u want to get all BVALUE which has relation with selected AVALUE in table AB
$sql = "SELECT B.NAMEID as id, BVALUE as value FROM TABLEA A
LEFT JOIN AB ON TABLEANAMEID=A.NAMEID
LEFT JOIN TABLEB B ON TABLEBNAMEID=B.NAMEID
WHERE AVALUE = $AVALUE";`
get mysql result
$result = mysql_query($sql);
iterate
echo "<select>";
foreach ($r = mysql_fetch_object($result)) {
echo '<option value="'.$r->id.'">'.$r->value.'</option>';
}
echo "</select>";
I have an issue getting data from three tables, which I want to return using one query.
I've done this before using a query something like this one:
$query = mysql_query("SELECT
maintable.`id`,
maintable.`somedata`,
maintable.`subtable1_id`,
subtable1.`somedata`,
subtable1.`subtable2_id`,
subtable2.`somedata`
FROM
`maintable` maintable,
`subtable1` subtable1,
`subtable2` subtable2
WHERE
maintable.`somedata` = 'some_search' AND
subtable1.`id` = maintable.`subtable1_id` AND
subtable2.`id` = subtable1.`subtable2_id`
")or die(mysql_error());
The problem this time is that the extra details might not actually apply. I need to return all results that match some_search in maintable, even if there is no subtable1_id specified.
I need something that will go along the lines of
WHERE
maintable.`somedata` = 'some_search'
AND IF maintable.`subtable1_id` IS NOT NULL (
WHERE subtable1.`id` = maintable.`subtable1_id` AND
subtable2.`id` = subtable1.`subtable2_id`
)
As you will probably guess, I am not an advanced mysql user! Try as I might, I cannot get the syntax right, and I have had no luck searching for solutions on the web.
Any help much appreciated!
It seems like the basic distinction you're looking for is between an INNER JOIN and a LEFT JOIN in MySQL.
An INNER JOIN will require a reference between the two tables. There must be a match on both sides for the row to be returned.
A LEFT JOIN will return matches in both rows, like an INNER, but it will also returns rows from your primary table even if no rows match in your secondary tables -- their fields will be NULL.
You can find example syntax in the docs.
If I got this right, you need to use MySQL LEFT JOIN. Try this:
SELECT
m.id,
m.somedata,
m.subtable1_id,
s1.somedata,
s1.subtable2_id,
s2.somedata
FROM
maintable m
LEFT JOIN
subtable1 s1
ON
m.subtable1_id = s1.id
LEFT JOIN
subtable2 s2
ON
s1.subtable2_id = s2.id
WHERE
m.somedata = 'some search'
This will return the data of maintable even if there's no equivalent data in subtable1 or 2 OR data of maintable and subtable1 if there's no equivalent data in subtable2.
How about this:
WHERE
maintable.`somedata` = 'some_search'
AND (maintable.`subtable1_id` IS NULL OR (
subtable1.`id` = maintable.`subtable1_id` AND
subtable2.`id` = subtable1.`subtable2_id` )
)
Keep in mind that this will result in a cross product of subtable1 and subtable2 with maintable when subtable1_id is NULL.