Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 7 years ago.
Improve this question
I'm having a strange error with my PHP code. it says there is a unexpected T_CONSTANT_ENCAPSED_STRING error, but I simply can't see it.
$dateResult = mysqli_query($connection, 'select county, cuisine,
count(*) from inspection, restaurant where inspection.rid =
restaurant.rid and inspection.passfail = ''PASS'' and idate like '$date%'
group by county, cuisine');
I'm assuming its an issue with either my like pattern '$date%' or with 'PASS'.
Thank you!
Try this your are conflicting ' and "
$dateResult = mysqli_query($connection, "select county, cuisine,
count(*) from inspection, restaurant where inspection.rid =
restaurant.rid and inspection.passfail = 'PASS' and idate like '$date%'
group by county, cuisine");
Related
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 2 years ago.
Improve this question
I want to Select one record from a table Named 'sms_students' by using Student Id or Student Telephone
Here is My Code
$student_rand_id=$_GET['std_rand'];
$std_tell=$_GET['std_tell'];
$view_student = $config->prepare("SELECT * FROM sms_students WHERE st_rand = :random_id || st_tel =: st_tel");
$view_student->execute(['random_id' => $student_rand_id]);
$view_student->execute(['st_tel' => $std_tell]);
$row = $view_student->fetch();
Since you call execute twice, this executes twice, both times with an incomplete set of arguments. It's an easy fix though:
$view_student = $config->prepare("SELECT * FROM sms_students WHERE st_rand = :random_id OR st_tel = :st_tel");
$view_student->execute(['random_id' => $_GET['std_rand'], 'st_tel' => $_GET['std_tell'] ]);
$row = $view_student->fetch();
Try and get rid of single-use variables, they're almost always unnecessary, and do try and steer towards having names that match precisely. Seeing st_tel and std_tell together is a sign something's not quite right. Get your code to agree on names and stick with them.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 2 years ago.
Improve this question
I have to find the total number or records but count is based on distinct of multiple columns.
I have following lines in my query.
$query = $this->createQueryBuilder("j");
$query->select(
"COUNT(DISTINCT
'j.mark',
'j.model'
) as total");
But this is giving me error:
[Syntax Error] line 0, col 76: Error: Expected Doctrine\ORM\Query\Lexer::T_CLOSE_PARENTHESIS, got ','
Can anybody please help me solve this issue?
Thank You.
I used concat instead which worked for me. But I am not sure if its the correct solution or not.
$query->select(
"COUNT(DISTINCT(concat(j.mark, j.model))
) as total");
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 4 years ago.
Improve this question
i getting data sql from controller and here is my code
$qr = $this->db->query("select journalDetail.JOURNAL_ID, JOURNAL_TYPE_CODE, JOURNAL_NUMBER, JOURNAL_MEMO, COA_CODE, JOURNAL_DETAIL_DESC, CURRECY_ID, JOURNAL_DETAIL_ORIG, JOURNAL_DETAIL_SUM, JOURNAL_DETAIL_TYPE, date_format(JOURNAL_DATE,'%d %M %Y') AS DATE, from t_journal_detail journalDetail left join t_journal journal on journalDetail.journal_id=journal.journal_id where JOURNAL_DETAIL_ID = '".$journalDetailId."'");
$gen = $qr->result();
But my code was Error Number: 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 'from t_journal where JOURNAL_ID = '83'' at line 1
select JOURNAL_NUMBER, JOURNAL_MEMO, date_format(JOURNAL_DATE,'%d %M %Y') AS DATE, from t_journal where JOURNAL_ID = '83'
Please i need help i don't know to fix this syntax
You have a comma (,) before the from keyword: and after 'AS DATE'-
date_format(JOURNAL_DATE,'%d %M %Y') AS DATE, from t_journal_detail
Remove the comma and I think your problem will be solved.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
I am trying to run following query:
$result = mysql_query(SELECT SecurityQues FROM reg_indi WHERE UserName='$usrnm') UNION (SELECT SecurityQues FROM reg_ac WHERE UserName='$usrnm');
But i am getting error in this syntax.
What is the error?
You need to put the query in quotes because for PHP it's a string, your parenthesis are wrong too:
$result = mysql_query(
"SELECT SecurityQues
FROM reg_indi
WHERE UserName='" . $usrnm . "'
UNION
SELECT SecurityQues
FROM reg_ac
WHERE UserName='" . $usrnm . "'"
);
Btw mysql_ extension is deprecated and removed from more recent PHP versions. Use PDO or MySQLi instead.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
When I run:
$query = "UPDATE subjects SET
menu_name = '{$menu_name}',
position = {$position},
visible = {$visible},
WHERE ID = {$ID}";
$result = mysql_query($query, $connection);
I get back:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE ID = 1' at line 5
Remove this comma before the WHERE clause. Since there are no more values to update, the comma is not needed and hence causes a syntax error.
visible = {$visible},
^