I have two tables in mysql one is quote and the fields are quote_id, status, created and submit_by. in submit_by field i am saving username of the employee and in employee table the fields are firs_name, last_name, username, password. I want to show in a php table the employee first name instead of username of the employee.
full quote table stucture
full employee table sructure
this is the code i am using but not getting any result
$sql = "SELECT q.quote_id, q.client_name, q.status, e.first_name
FROM `quote` AS q
LEFT JOIN `employee` AS e ON q.submit_by = e.username";
Use this one
$sql = "SELECT * , employee.first_name FROM quote
LEFT JOIN employee ON quote.submit_by = employee.username ORDER BY quote.client_name";
$result = $dbLink->query($sql);
try this:
SELECT quote.quote_id, quote.client_name, quote.status, employee.first_name FROM quote
LEFT JOIN employee ON quote.submit_by = employee.username ORDER BY quote.client_name;
You can use below Query to get the result.
SELECT q.`quote_id`, q.`client_name`, q.`status`, e.`first_name` FROM `quote` q , `employee` e WHERE q.`submit_by` = e.`username`;
Related
i have a question. my english isn't well. so i hope i explain well...
i have two tables, tbl_home and tbl_office, the question is
how do i make a select statement from 2 tables which have identical value from column 'case_no' where it is referenced in both table..
$a=$_POST['home_id']
the code above is where i get the home_id from,
while the statement below is how i try to select both tables based on value in column 'case_no' of both table. but it is based on variable $a which i retrieved from form
<?php
$sql2 = "SELECT * FROM tbl_office WHERE case_no IN (SELECT * FROM tbl_home WHERE home_id = '$
$result2=$conn->query($sql2);
while($row = $result2->fetch_assoc()){
$a=$row['case_no'];
$bc=$row['colour'];
echo " $a <br/> ";
echo " $bc2 <br/>";
?>
is the select statement above correct??
soo, i just want anybody to take a look a this specific statement and how to make it right
$sql2 = "SELECT * FROM tbl_office WHERE case_no IN (SELECT * FROM tbl_home WHERE home_id = '$a'";
You need inner join to use:
" SELECT t_office.home_id,t_office.case_no,t_office.name FROM tbl_office
t_office INNER JOIN tbl_home t_home ON t_office.case_no = t_home.case_no;
where t_office.case_no ='$a'";
u can use "inner join" for example:
"SELECT t.home_id,t.case_no,t.name FROM tbl_office
t INNER JOIN tbl_home h ON h.case_no = h.case_no"
**select tbl_home.name,tbl_office.case_no,tbl_office.color from tbl_office
INNER JOIN tbl_home on tbl_office.case_no = tbl_home.case_no
where tbl_office.case_no ='$a';**
I hope this will be working fine until $a(case_no) value is existed in tbl_home or else it doesn't give any rows
I was using this query to connect my student table and attendance table,
My Problem is, sometimes, attendance table has no value.
It's not returning any value.
<?php
if($_SERVER['REQUEST_METHOD']=="POST"){
include('include/connection.php');
showData();
}
function showData(){
global $connect;
$teacher_id = $_POST['teacher_id'];
$subject_id = $_POST['subject_id'];
$date = $_POST['date'];
$query ="
SELECT s.student_name
, s.student_number
, s.student_section
, s.subject_id
, s.fingerprint_id
, s.teacher_id
, a.status
FROM tbl_student s
LEFT
JOIN tbl_attendance a
on s.subject_id=a.subject_id
WHERE s.subject_id = '$subject_id'
and a.date='$date'
and s.teacher_id = '$teacher_id';";
$result =mysqli_query($connect,$query);
$number_of_rows = mysqli_num_rows($result);
$temp_array=array();
if($number_of_rows>0){
while($row=mysqli_fetch_assoc($result)){
$temp_array[]=$row;
}
}
header('Content-Type: application/json');
echo json_encode(array("student"=>$temp_array));
mysqli_close($connect);
}
?>
What I want to achive is even if attendance table has no value,
I can still see the student fields.
Is it even possible with SQL query? Thanks
You have to move the fields of table attendance from where to the on condition:
$query ="SELECT student.student_name,student.student_number,student.student_section,student.subject_id,student.fingerprint_id,student.teacher_id,attendance.status
FROM tbl_student student
LEFT JOIN tbl_attendance attendance on student.subject_id=attendance.subject_id and attendance.date='$date'
WHERE student.subject_id='$subject_id' and student.teacher_id='$teacher_id';";
Because first the join Statement will be executed and then the where, if you access the table tbl_attendance in where ans all the columns are null, they will filtered out.
Hint: read about prepared Statements to provide SQL-injection
SELECT student.student_name,student.student_number,student.student_section,student.subject_id,student.fingerprint_id,student.teacher_id,attendance.status
FROM tbl_student student
LEFT JOIN tbl_attendance attendance on student.subject_id=attendance.subject_id and attendance.date='$date'
WHERE student.subject_id='$subject_id' and student.teacher_id='$teacher_id';
Try above code.Hope this will helps.
As you had made condition on student table using attendance.date='$date' on WHERE clause it exclude that record which are not satisfy this condition.
So instead of where i had put that condition through ON clause on LEFT JOIN.
This will achieve your goal.
I'm working with a mysql query to select data from multiple tables using LEFT OUTER JOIN. Now i get the following error when i exequte the query:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near 'wg.werkbon_global_id = wk.werkbon_klant_globalid LEFT OUTER
JOIN users AS u' at line 16
Only the problem is that i can't find out what's wrong with my query.
PHP Query:
$query = '
SELECT
wg.werkbon_global_id AS id,
wg.werkbon_global_status AS status,
wg.werkbon_global_date_lastedit AS date,
usr.user_firstname AS monteur_vn,
usr.user_insertion AS monteur_tv,
usr.user_lastname AS monteur_an,
wg.werkbon_global_type AS type,
wg.werkbon_global_layout AS layout,
wg.werkbon_global_werkzaamheden AS werkzaamheden,
wg.werkbon_global_opmerkingen AS opmerkingen,
wk.werkbon_klant_nummer AS klantnr
FROM
werkbon_klant AS wk
LEFT OUTER JOIN werkbon_global AS wg
wg.werkbon_global_id = wk.werkbon_klant_globalid
LEFT OUTER JOIN users AS usr
usr.user_id = wg.werkbon_global_monteur_finish
WHERE
wk.werkbon_klant_nummer = '.$db->Quote($klantid).'
ORDER BY id ASC;
$result = $db->loadAssoc($query);
I think my problem has something todo with left outer join but what?
You are missing the ON operator in your joins!
The correct syntax for a join is:
SELECT * FROM x LEFT JOIN y ON condition WHERE...
$query = "
SELECT
wg.werkbon_global_id AS id,
wg.werkbon_global_status AS status,
wg.werkbon_global_date_lastedit AS date,
usr.user_firstname AS monteur_vn,
usr.user_insertion AS monteur_tv,
usr.user_lastname AS monteur_an,
wg.werkbon_global_type AS type,
wg.werkbon_global_layout AS layout,
wg.werkbon_global_werkzaamheden AS werkzaamheden,
wg.werkbon_global_opmerkingen AS opmerkingen,
wk.werkbon_klant_nummer AS klantnr
FROM
werkbon_klant AS wk
LEFT OUTER JOIN werkbon_global AS wg
wg.werkbon_global_id = wk.werkbon_klant_globalid
LEFT OUTER JOIN users AS usr
usr.user_id = wg.werkbon_global_monteur_finish
WHERE
wk.werkbon_klant_nummer = '.$db->Quote($klantid).'
ORDER BY id ASC";
$result = $db->loadAssoc($query);
Make sure there isn't missing quote
Problem soved thanks to arkascha
The fixed query is now:
$query = '
SELECT
wg.werkbon_global_id AS id,
wg.werkbon_global_status AS status,
wg.werkbon_global_date_lastedit AS date,
usr.user_firstname AS monteur_vn,
usr.user_insertion AS monteur_tv,
usr.user_lastname AS monteur_an,
wg.werkbon_global_type AS type,
wg.werkbon_global_layout AS layout,
wg.werkbon_global_werkzaamheden AS werkzaamheden,
wg.werkbon_global_opmerkingen AS opmerkingen,
wk.werkbon_klant_nummer AS klantnr
FROM
werkbon_klant AS wk
LEFT OUTER JOIN werkbon_global AS wg ON
wg.werkbon_global_id = wk.werkbon_klant_globalid
LEFT OUTER JOIN users AS usr ON
usr.user_id = wg.werkbon_global_monteur_finish
WHERE
wk.werkbon_klant_nummer = '.$db->Quote($klantid).'
ORDER BY id ASC';
$result = $db->loadAssoc($query);
#fred i don't need to add quotes by column names. You only need to add quotes by string/blob values.
#johny my $db->Quote() function will add qoutes automaticly. I don't need to add them and put everything in quote's.
Thanks all for help.
i'm in the process of joining two tables together under two different conditions. For primary example, lets say I have the following nested query:
$Query = $DB->prepare("SELECT ID, Name FROM modifications
WHERE TYPE =1 & WFAbility = '0'");
$Query->execute();
$Query->bind_result($Mod_ID,$Mod_Name);
and this query:
$Query= $DB->prepare("SELECT `ModID` from `wfabilities` WHERE `WFID`=?");
$Query->bind_param();
$Query->execute();
$Query->bind_result();
while ($Query->fetch()){ }
Basically, I want to select all the elements where type is equal to one and Ability is equal to 0, this is to be selected from the modifications table.
I further need to select all the IDs from wfabilities, but transform them into the names located in modifications where WFID is equal to the results from another query.
Here is my current semi-working code.
$Get_ID = $DB->prepare("SELECT ID FROM warframes WHERE Name=?");
$Get_ID->bind_param('s',$_GET['Frame']);
$Get_ID->execute();
$Get_ID->bind_result($FrameID);
$Get_ID->fetch();
$Get_ID->close();
echo $FrameID;
$WF_Abilties = $DB->prepare("SELECT ModID FROM `wfabilities` WHERE WFID=?");
$WF_Abilties->bind_param('i',$FrameID);
$WF_Abilties->execute();
$WF_Abilties->bind_result($ModID);
$Mod_IDArr = array();
while ($WF_Abilties->fetch()){
$Mod_IDArr[] = $ModID;
}
print_r($Mod_IDArr);
$Ability_Name = array();
foreach ($Mod_IDArr AS $AbilityMods){
$WF_AbName = $DB->prepare("SELECT `Name` FROM `modifications` WHERE ID=?");
$WF_AbName->bind_param('i',$AbilityMods);
$WF_AbName->execute();
$WF_AbName->bind_result($Mod_Name);
$WF_AbName->fetch();
$Ability_Name[] = $Mod_Name;
}
print_r($Ability_Name);
See below:
SELECT ModID,
ID,
Name
FROM modifications M
LEFT JOIN wfabilities WF
ON WF.ModID = M.ID
WHERE TYPE =1 & WFAbility = '0'
To do this, you need to join your tables, I'm not quite sure what you are trying to do so you might have to give me more info, but here is my guess.
SELECT ID, Name, ModID
FROM modifications
JOIN wfabilities
ON WFID = ID
WHERE TYPE = '1'
AND WFAbility = '0'
In this version I am connecting the tables when WFID is equal if ID. You will have to tell me exactly what is supposed to be hooking to what in your requirements.
To learn more about joins and what they do, check this page out: MySQL Join
Edit:
After looking at your larger structure, I can see that you can do this:
SELECT modifications.Name FROM modifications
JOIN wfabilities on wfabilities.ModID = modifications.ID
JOIN warframes on warframes.ID = wfabilities.WFID
WHERE warframes.Name = 'the name you want'
This query will get you an array of the ability_names from the warframes name.
This is the query:
"SELECT A.ID, A.Name,B.ModID,C.Name
FROM modifications as A
LEFT JOIN wfabilities as B ON A.ID = B.WFID
LEFT JOIN warframes as C ON C.ID = B.WFID
WHERE A.TYPE =1 AND A.WFAbility = '0' AND C.Name = ?"
Hey guys need some more help
I have 3 tables USERS, PROFILEINTERESTS and INTERESTS
profile interests has the two foreign keys which link users and interests, they are just done by ID.
I have this so far
$statement = "SELECT
InterestID
FROM
`ProfileInterests`
WHERE
userID = '$profile'";
Now I want it so that it selects from Interests where what it gets from that query is the result.
So say that gives out 3 numbers
1
3
4
I want it to search the Interests table where ID is = to those...I just don't know how to physically write it in PHP...
Please help.
Using a JOIN:
Best option if you need values from the PROFILEINTERESTS table.
SELECT DISTINCT i.*
FROM INTERESTS i
JOIN PROFILEINTERESTS pi ON pi.interests_id = i.interests_id
WHERE pi.userid = $profileid
Using EXISTS:
SELECT i.*
FROM INTERESTS i
WHERE EXISTS (SELECT NULL
FROM PROFILEINTERESTS pi
WHERE pi.interests_id = i.interests_id
AND pi.userid = $profileid)
Using IN:
SELECT i.*
FROM INTERESTS i
WHERE i.interests_id IN (SELECT pi.interests_id
FROM PROFILEINTERESTS pi
WHERE pi.userid = $profileid)
You are on the right track, lets say you execute the query above using this PHP code:
$statement = mysql_query("SELECT InterestID FROM `ProfileInterests`
WHERE userID = '$profile'");
Then you can use a PHP loop to dynamically generate an SQL statement that will pull the desired IDs from a second table. So, for example, continuing the code above:
$SQL = "";
while ($statementLoop = mysql_fetch_assoc($statement)) {
//Note the extra space on the end of the query
$SQL .= "`id` = '{$statementLoop['InterestID']}' OR ";
}
//Trim the " OR " off the end of the query
$SQL = rtrim($SQL, " OR ");
//Now run the dynamic SQL, using the query generated above
$query = mysql_query("SELECT * FROM `table2` WHERE {$SQL}")
I haven't tested the code, but it should work. So, this code will generate SQL like this:
SELECT * FROM `table2` WHERE `id` = '1' OR `id` = '3' OR `id` = '4'
Hope that helps,
spryno724
Most likely you want to join the tables
select
i.Name
from
ProfileInterests p
inner join
interests i
on
p.interestid = i.interestid
where
p.userid = 1