I want two execute to query at the same time but I get mysql_error .
Query
select L.PLesName,
L.ELesName,
L.LesTotalUnit,
L.TheoryUnit,
L.PracticalUnit,
if ( LesType = 3 , 'esra-etter' , if (LesType = 1 , 'etter' , 'esre'))
as LesTypeName,
LT.PLesTypName,
ES.PEduSecName,
L.LesMinMark
from lessons L
left join LessonTypes LT on (L.LesTypCode=LT.LesTypCode)
left join EducationalSections ES on (L.EduSecCode=ES.EduSecCode)
where L.LesCode=2133004; drop table SpamLog;--
Error
Error description: 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 'drop table SpamLog;--' at line 7
I test diffrence cases but I get same error. What am I going to do ?
Thanks.
Is there a reason you can't simply call mysql_query() twice?
e.g.
mysql_query("select L.PLesName, L.ELesName, L.LesTotalUnit, L.TheoryUnit, L.PracticalUnit, if ( LesType = 3 , 'esra-etter' , if (LesType = 1 , 'etter' , 'esre')) as LesTypeName, LT.PLesTypName,ES.PEduSecName,L.LesMinMark from lessons L left join LessonTypes LT on (L.LesTypCode=LT.LesTypCode) left join EducationalSections ES on (L.EduSecCode=ES.EduSecCode) where L.LesCode=2133004");
mysql_query("drop table SpamLog");
Since your SELECT statement isn't modifying anything, there's no real need to enforce grouping.
Related
Okay so the code works without WHERE but i need to add WHERE cause i want to display only the records where price is >400. When I execute this code
$results=mysqli_query($db,"SELECT aktfil.id, filmat.titulli, filmat.cmimi,
GROUP_CONCAT( Concat( aktoret.emri,' ',aktoret.mbiemri ) SEPARATOR ',' ) AS Aktori
FROM aktfil JOIN filmat ON aktfil.id=filmat.id LEFT JOIN aktoret ON aktfil.idakt=aktoret.idakt
GROUP BY filmat.id ORDER BY aktoret.emri WHERE filmat.cmimi>'400'") or die(mysqli_error($db));
It says
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'WHERE filmat.cmimi>'400'' at line 4
$results=mysqli_query($db,"SELECT aktfil.id, filmat.titulli, filmat.cmimi,
GROUP_CONCAT( Concat( aktoret.emri,' ',aktoret.mbiemri ) SEPARATOR ',' ) AS Aktori
FROM aktfil JOIN filmat ON aktfil.id=filmat.id LEFT JOIN aktoret ON aktfil.idakt=aktoret.idakt
WHERE filmat.cmimi>'400' GROUP BY filmat.id ORDER BY aktoret.emri") or die(mysqli_error($db));
The syntax of your query was not correct, WHERE clause should be placed before GROUP BY and ORDER BY clause should be at the end. Also please use quotation '' only if the data type of the column filmat.cmimi is some sort of text, otherwise use it like WHERE filmat.cmimi > 400
I need to get specific columns from three tables through joins.every time it goes wrong.my code is
$saf=mysqli_query($db , "select pod.mobile, tpaidamount.Aptdate, follow.cent, pdate, time from pod,tpaidamount, follow where tpaidamount.pid = follow.pid and pod.Customer Id = tpaidamount.pid and pod.Customer Id =follow.pid ");
$i=1;
while($sfg=mysqli_fetch_assoc($saf) or die(mysqli_error($db)))
;
?>
pod,tpaidamount,follow are tables and other coloumns
Getting error
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 'Id = tpaidamount.pid and pod.Customer Id =follow.pid' at line 1
is it a typo? Is setfollowup.pid or follow.pid?
select pod.mobile, tpaidamount.Aptdate, follow.cent, <table>.pdate, <table>.time
from pod, tpaidamount, follow
where tpaidamount.pid = follow.pid
and pod.`Customer Id` = tpaidamount.pid
and pod.`Customer Id` = follow.pid
You shouldn't Ever create columns names with spaces. The name: "pod.Customer Id" is a Bad attribute name, and you need to avoid use names like datatypes (or any SGBD reserved word) like: 'date', 'time', 'char', 'table', 'column'....
However, if you need to do it, try this SQL:
SELECT p.mobile
, t.Aptdate
, f.cent
, ???.pdate
, ???.`time`
FROM pod AS p
JOIN tpaidamount AS t ON o.`Customer Id` = t.pid
JOIN follow AS f ON t.pid = f.pid ON p.`Customer Id` = f.pid
Use alias for easy coding SQL queryes. Ex: table_name AS tn, table_a_name AS tan.
!!! I sugest you to watch some basic SQL Lessons.
Good Luck.
I have to fetch values from two tables, I have wrote separate query for fetching from both tables. and both queries worked properly. but I need to get this in one result object. so that I have joined queries using UNION operator. but it throws some error. the query is given below
$query1 = "SELECT dev_members.name,dev_members.id,dev_members.age,dev_members.family_id,dev_family.house_name,dev_ib_account_registration.account_id FROM (dev_members)
JOIN dev_family ON dev_family.id=dev_members.family_id
JOIN dev_ib_account_registration ON dev_ib_account_registration.member_id=dev_members.id
UNION
SELECT dev_members.name,dev_members.id,dev_members.age,dev_members.family_id, dev_family.
house_name,dev_ib_sub_member_registration.account_id FROM (dev_members)
JOIN dev_family ON dev_family.id=dev_members.family_id
JOIN dev_ib_sub_member_registration ON dev_ib_sub_member_registration.member_id=dev_members.id";
$result = $this->db->query($query);
return $result->result();
and the error is:
1064 - 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 'JOIN dev_family ON dev_family.id=dev_members.family_id JOIN dev_ib_s' at line 8
There is something wired with the usage of brackets in your second query
(dev_members) <---
when i use your query without them in Fiddle query works ,but using with brackets it produces syntax error ,so try them without ()
updated query
SELECT
dev_members.name,
dev_members.id,
dev_members.age,
dev_members.family_id,
dev_family.house_name,
dev_ib_account_registration.account_id
FROM
dev_members
JOIN dev_family
ON dev_family.id = dev_members.family_id
JOIN dev_ib_account_registration
ON dev_ib_account_registration.member_id = dev_members.id
UNION
SELECT
dev_members.name,
dev_members.id,
dev_members.age,
dev_members.family_id,
dev_family.house_name,
dev_ib_sub_member_registration.account_id
FROM
dev_members
JOIN dev_family
ON dev_family.id = dev_members.family_id
JOIN dev_ib_sub_member_registration
ON dev_ib_sub_member_registration.member_id = dev_members.id
I am trying to do following.
Select a Hospital with id = hid from table named as hospital.
Add a value "overall_rating" to it and get all the ratings and make avg of it from another table named as hrating
here is my query
$statement = $conn->prepare('SELECT hospital.*,(SELECT AVG(hrating.rating_h) FROM hrating WHERE hid = hospital.hid) as overall_rating WHERE hid=:hid LIMIT 1');
Getting this error
{"error":"SQLSTATE[42000]: Syntax error or access violation: 1064 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 'WHERE hid='44' LIMIT 1' at line 1"}
Where am i being wrong.
It appears you don't have a " FROM hospital " bit in your query?
SELECT hospital.*,(SELECT AVG(hrating.rating_h)
FROM hrating
WHERE hid = hospital.hid) as overall_rating
FROM hospital -- this line seems to be missing ??
WHERE hid=:hid LIMIT 1
Try this:
SELECT hospital.*, temp.overall_rating
FROM hospital
LEFT JOIN (SELECT hid AVG(rating_h) as overall_rating
FROM hrating group by hid
) temp
on hid = hospital.hid
WHERE hospital.hid=:hid LIMIT 1
hey guys im trying to query something in php but the WHERE is a index from array this is what i did
$data=array();
while ($row = mysql_fetch_array($result))
{
$item = array(
'sec_id'=>$row['section_id'],
'sec_name'=>$row['section_name'],
'sec_dept'=>$row['section_department'],
'sec_lvl'=>$row['section_level'],
'advisory_id'=>$row['advisory_id'],
'first_name'=>$row['f_firstname'],
'last_name'=>$row['f_lastname'],
'middle_name'=>$row['f_middlename'],
'advisor_id'=>$row['faculty_id'],
);
$get_subjects = "SELECT subject_name
FROM subjects
WHERE level = '".$row['section_level']."' ";
$result_get_subjects =mysql_query($get_subjects)or die(mysql_error());
$subjects_count = mysql_num_rows($result_get_subjects);
$check_archive_subjects = " SELECT b.subject_name
FROM registrar_grade_archive a
LEFT JOIN subjects b ON(a.subject_id=b.subject_id)
LEFT JOIN section c ON(a.section_id = c.section_id)
WHERE a.advisor_faculty_id ='".$row['faculty_id']."'
WHERE a.section_id ='".$row['section_id']."'
GROUP BY b.subject_name ASC
" ;
$query_checking =mysql_query($check_archive_subjects)or die(mysql_error());
$subjects_count_sent = mysql_num_rows($query_checking);
but unfortunately i got an error in $check_archive_subjects that says:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE a.section_id ='24'
is there any other way to put an array index in where clause in mysql. I know mysql is deprecated and ill be switching to mysqli after i finished this project so pardon me guys. thanks in advance
Multiple WHERE conditions should be joined using boolean keywords AND or OR. You don't issue multiple WHERE clauses.
Also, please read this regarding the MySQL extension - https://stackoverflow.com/a/12860046/283366
Try the below query
SELECT b.subject_name
FROM registrar_grade_archive a
LEFT JOIN subjects b ON(a.subject_id=b.subject_id)
LEFT JOIN section c ON(a.section_id = c.section_id)
WHERE a.advisor_faculty_id ='".$row['faculty_id']."' AND a.section_id ='".$row['section_id']."'
GROUP BY b.subject_name ASC
Explanation
Using multiple WHERE clause like above is not accepted in MySQL. Try replacing the 2nd WHERE using an AND or using an OR depending on your requirement. I have used AND