I am trying to send this function in php but it keeps coming back with this 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 ''tolu)' at line 2
function getDuplicate($case, $select,$from,$where,$equals,$and="",$equals2=""){
global $database_conndb;
global $conndb;
switch($case){
case 1:
$sql= "SELECT {$select} FROM {$from} WHERE {$where}='{$equals}'";
break;
case 2:
$sql= "SELECT {$select} FROM {$from} WHERE {$where}='{$equals}' AND {$and} != '{$equals2}'";
break;
}
looks like you did not escape the parameters correctly. look at this function: http://php.net/manual/en/function.mysql-real-escape-string.php
If you are sure that there is an error in SQL statement then echo the SQL then copy it and execute that query manually into database. U will got the actual error occurence point.
Hope this why u will fix your bug
Related
I tried to use the following code to insert,
$op=$_POST["ans"];
$username=$_GET["username"];
mysql_query("insert into $username values('Q3','$op')")
or die(mysql_error());
But I got the following 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 'values('Q1','Wrong')' at line 1
Why am I getting this error? How can I fix it?
Your query structure is not making any sense. You're inserting into $username? That's not the name of the table, is it?
mysql_query("INSERT INTO `tablename` values('Q3','" . mysql_real_escape_string($op) . "')") or die(mysql_error());
Always be very careful to escape any and all user data being put into your queries, and please, please stop using mysql_query in new code.
Need your advice about my PHP MySQL syntax. I work hard with that, but still facing problem with the query and the error is:
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 '$bagianWhere = ""' at line 1: $bagianWhere = ""
You can see my demo here
If "bagianWhere" is empty, the SQL query becomes invalid with an orphan "WHERE" at the end. You can do
$bagianWhere = "1 = 1";
at the top to counter for that.
Also get rid of "$bagianWhere .= " and make it "$bagianWhere = "
Also the error you are mentioning here is from SQL Fiddle, as you have put in PHP code in there. Can you run the PHP on your web server and show us the error you get.
I'm getting these weird errors, and I've been up and down the code, commenting and rewriting, and googling all the things.
Perhaps you guys will see what I'm not seeing:
$mysqli = new mysqli('host','login','passwd','db');
if($mysqli->connect_errno > 0){ die('Cannot connect: '. $mysqli->connect_error); }
// See if there is one term or multiple terms
if (count($search) == 1) {
// If one term, search for that
$like = $search[0];
$stmt = "SELECT
gsa_committees.id,
gsa_committees.committee,
gsa_committees.appointer,
gsa_committees.representatives,
gsa_committees.contact,
gsa_committees.category,
gsa_committees.attachments,
gsa_committees.labels,
gsa_committee_reports.committee,
gsa_committee_reports.title,
gsa_committee_reports.author,
gsa_committee_reports.link,
gsa_funds.id,
gsa_funds.fund,
gsa_funds.attachments,
gsa_funds.labels,
gsa_meeting_minutes.title,
gsa_meeting_minutes.link,
gsa_officers.office,
gsa_officers.dept,
gsa_officers.name,
gsa_representatives.program_dept,
gsa_representatives.representatives,
gsa_representatives.alternate
FROM
gsa_committees,
gsa_committee_reports,
gsa_funds,
gsa_meeting_minutes,
gsa_officers,
gsa_representatives
WHERE
(gsa_committees.committee LIKE $like) AND
gsa_committees.committee IS NOT NULL";
}
if(!$result = $mysqli->query($stmt)){ die('Bad query: '. $mysqli->error); }
This gives me this error message:
Bad 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 '%ARCHAC%) AND gsa_committees.committee IS NOT NULL' at line 34
Which I know isn't true. If I change that las part to just this:
WHERE gsa_committees.committee LIKE $like";
I get this error message:
Bad query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%ARCHAC%' at line 34
Everywhere I've looked, the string "%".search."%" seems to be the correct method, but my server doesn't seem to like it here.
Interesting side note: I have a different LIKE statement working on another page on the same server, this just won't work for some reason.
Thanks!
Try putting single quotes around your search term ($like variable).
for example: (gsa_committees.committee LIKE '$like')
You need to wrap the variable in quotes for like to work:
WHERE gsa_committees.committee LIKE '$like';
See reference documentation on String Comparison Function.
it looks like missing quotes:
"WHERE gsa_committees.committee LIKE '$like' ";
Ok, I got it. The answer on this post solved my issue:
MYSQLI SQL query over multiple tables fail
As soon as I assigned the tables t1,t2,etc and did INNER JOIN, the results came in as expected, with %$search% or $search.
Thanks all!
Full error message:
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 '' at line 1
So it hasn't really told me much... Is there a way to find out more?
It has returned this message from two PHP files. Here are the first MySQL queries that I made in each file:
$query = mysql_query("SELECT * FROM `questions` WHERE `id`=".$currentId.";") or die( mysql_error() );
$query = mysql_query("SELECT * FROM `questions` WHERE `id`=".$theNextId.";") or die( mysql_error() );
There is PHP code before this though which opens the database etc.
Here is a similar problem: Link
Perhaps there an error in my concatenation?
Thanks.
$currentId is null or empty.
And don't forget about SQL-injection!
Remove semicolons. The docs say "the query string should not end with a semicolon".
It seems that your final ` (back-tick) character is missing.
The exact error message is:
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 rfflag='0'' at line 1
Hi,
I'm trying to get some php scripts working and it dies with the above error message. There are two locations where rfflag is used in the SQL query:
$_SESSION['lang']=$objTerm->my_get_one("select min(id) from "
.$objTerm->TABLE['languages']." where status='1' and rfflag='0'");
$rs_lang=$objTerm->execute_query("select id,language from "
.$objTerm->TABLE['languages']." where `status`='1' and `rfflag`='0'");
How do I determine which one is causing the problem? Or is the problem something else altogether?
Echo this:
"select id,language from ".$objTerm->TABLE['languages']." where status='1' and rfflag='0'"
and this:
"select min(id) from ".$objTerm->TABLE['languages']." where status='1' and rfflag='0'"
Then run execute each output in your favorite sql developer tool.
Errors will be displayed there.
How do I determine which one is causing the problem?
Remove one of the queries. See if it still happens.
On a secondary thought, I would suggest that you change your MySQL query code so, that it doesn't use die() to print out the error message. Use trigger_error or exceptions instead, this way you will automatically get a trace of which line caused it.
How do I determine which one is causing the problem?
use trigger_error() to output an error message.
I guess (I have to guess because you supply no code) that you are using die() to output an error.
if you change this bad practice function to trigger_error(), you will be able to see the line number, where error occurred.
If you add non only mysql_error() to it's output, but also query itself, you will be able to see the problem code too.
I guess $objTerm->TABLE['languages'] is undefined or does not have the value you’re expecting.
As sheeks06 has already suggested, just echo the query to see if everything is as expected:
$query = "select min(id) from "
.$objTerm->TABLE['languages']." where status='1' and rfflag='0'";
echo $query;
$_SESSION['lang']=$objTerm->my_get_one($query);
$query = "select id,language from "
.$objTerm->TABLE['languages']." where `status`='1' and `rfflag`='0'";
echo $query;
$rs_lang=$objTerm->execute_query($query);