how to solve this php mysql warning

Question

My code is as following:

$br_code=$_SESSION["br_code"];
echo $sqlstr="select DISTINCT branch_assets.s_id,s_name
              from branch_assets,assets
              where assets.s_id=branch_assets.s_id and
              br_code='$br_code'";
echo $result=mysql_query($sqlstr);
while($row1=mysql_fetch_array($result))
{
}

But on the line of while loop it show me warning as

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\System_management\send_in_maintanance1.php

where is my mistake please help to find this.

add mysql_error to your query and use error reporting, what does that throw you? — Funk Forty Niner, Oct 05 '16 at 12:53
Why `echo $result=mysql_query($sqlstr)` that is going to be a result object.. — chris85, Oct 05 '16 at 12:55
use like this `$result = mysql_query($sqlstr) or trigger_error(mysql_error()." ".$sqlstr);` — Karthi, Oct 05 '16 at 12:56
make sure you started the session also; did you? does that session array have a value to start with? — Funk Forty Niner, Oct 05 '16 at 12:57
Smit: you need to get in the convo here. What does `var_dump($br_code);` produce? Your question may get closed based on the possible duplicate I posted. — Funk Forty Niner, Oct 05 '16 at 13:01
I have a few thoughts why it's failing but you're not telling us the whole story here. Again; did you start the session? Does the session array have a value? are you connecting with mysql_? mysqli_? PDO? Other? the question is too unclear. — Funk Forty Niner, Oct 05 '16 at 13:04
its better to run your query manuly in PHP my admin, `echo $sqlstr="select DISTINCT branch_assets.s_id,s_name from branch_assets,assets where assets.s_id=branch_assets.s_id and br_code=$br_code";` — devpro, Oct 05 '16 at 13:04
@devpro in phpmyadmin when i run this query it shows me output which i wants — Smit Saraiya, Oct 05 '16 at 13:06
ok, I'm out of this one. Ask the guy with the answer below. I've done what I could here. — Funk Forty Niner, Oct 05 '16 at 13:07
@Fred-ii-: u r right, and OP dont want to read your comment :p — devpro, Oct 05 '16 at 13:08
@devpro Pffft... try and help and ask questions; I get midnight when it's really 9am here. — Funk Forty Niner, Oct 05 '16 at 13:09
alright, u tried `'$br_code'` not work, u tried echo YOUR QUERY and run into php myadmin, still not work, than you must need to check your session and mysql_error() — devpro, Oct 05 '16 at 13:10

score 3 · Accepted Answer · edited Oct 05 '16 at 13:13

3

$sqlstr="select DISTINCT branch_assets.s_id,s_name from branch_assets,assets where assets.s_id = branch_assets.s_id and br_code='".$br_code."'";

It's best to switch to either mysqli or PDO with a prepared statement instead of mysql_, since it is deprecated and deleted from PHP 7.0.

Reference:

https://en.wikipedia.org/wiki/Prepared_statement

edited Oct 05 '16 at 13:13

Funk Forty Niner

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answered Oct 05 '16 at 12:55

Mike Wight

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did u mean this `where assets.s_id = branch_assets.s_id = '".$br_code."'`? – devpro Oct 05 '16 at 12:57
yes. sorry I missed the first equal – Mike Wight Oct 05 '16 at 12:58
I edited it for you. – Mike Wight Oct 05 '16 at 12:59
but seriously, i didn't down vote this .. – devpro Oct 05 '16 at 13:00
@devpro actually that didn't solve it and I have a feeling why, yet the OP isn't yet giving us the right time of day yet. OP said: *"@PurushotamThakur still showing me same worning – Smit Saraiya 28 secs ago"* – Funk Forty Niner Oct 05 '16 at 13:03
@Fred-ii-: true – devpro Oct 05 '16 at 13:04

how to solve this php mysql warning

1 Answers1