mysql_query() expects parameter 2 to be resource, string given in

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mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

what's wrong with this line?

23. $result = mysql_query("INSERT INTO $tbl_name('city_id', 'city', 'state_id') VALUES('NULL, '.$city.","', '4421'') or die(mysql_error())");

I get this error message each time :

Warning: mysql_query() expects parameter 2 to be resource, string given in line 23

This my full code

<?php
// Get values from form 
$city = explode(',', $_POST['city']);

//create a loop
if(isset($_POST['city'])) {


    $city = explode(',', $_POST['city']);
    $n = count($city);
    for($i=0; $i<$n; $i++) {
    $result = mysql_query("INSERT INTO $tbl_name(city_id, city, state_id) VALUES(NULL, '.$city.","', '4421')") or die(mysql_error());
    }
} if($result) {
    header('Location: index.html'); 
}


?>

You forgot to give closing `"` after `'4421'')`. It should be `'4421'')") or die(mysql_error());` — Praveen Kumar Purushothaman, Jul 25 '12 at 09:07

000 · Answer 1 · 2012-07-25T09:34:23.840

0

try mysql_query("INSERT INTO ".$tbl_name."(city_id, city, state_id) VALUES(NULL,'".$city."','4421')") or die(mysql_error());

i guess it should work this way..

edited Jul 25 '12 at 09:34

answered Jul 25 '12 at 09:03

000

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I did all the suggestions above, but still get the same error message – user1550195 Jul 25 '12 at 09:29
echo your query and then run it directly – 000 Jul 25 '12 at 09:31
what do you mean by echo my query?, sorry I kinda noob with php – user1550195 Jul 25 '12 at 09:42
use the updated query in my answer – 000 Jul 25 '12 at 09:47

Muhammad Raheel · Answer 2 · 2012-07-25T09:12:27.673

0

Your query shoul be like this

mysql_query("INSERT INTO $tbl_name(city_id, city, state_id) VALUES('NULL, '.$city.","', '4421'") or die(mysql_error()));

Columns without commas

edited Jul 25 '12 at 09:12

answered Jul 25 '12 at 09:05

Muhammad Raheel

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How can a PHP Function be inside the query string!!! :O – Praveen Kumar Purushothaman Jul 25 '12 at 09:08

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

0

You forgot to give closing `"` after `after '4421'')` and the syntax is incorrect!

Replace:

'NULL, '.$city.","', '4421'') or die(mysql_error())");

With

'NULL, '.$city.","', '4421'')") or die(mysql_error());

edited Jun 20 '20 at 09:12

Community

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answered Jul 25 '12 at 09:06

Praveen Kumar Purushothaman

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I did your suggestion but still get the same error message – user1550195 Jul 25 '12 at 09:46
Yes I just edited my first post, just learned about it, thanks – user1550195 Jul 25 '12 at 12:15

score 0 · Answer 4 · edited Jul 25 '12 at 09:10

0

mysql_query("INSERT INTO $tbl_name(city_id, city, state_id) VALUES(NULL, ".$city.",'4421')") or die(mysql_error());

Try this.

or die() is a php construct - you misplaced a bracket which meant it went inside the sql query.

This is why your error thinks that mysql_query() has a second parameter.

edited Jul 25 '12 at 09:10

Praveen Kumar Purushothaman

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answered Jul 25 '12 at 09:07

Joe Green

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I did that, but still get the same error message – user1550195 Jul 25 '12 at 09:41
That's odd. You should ~at least~ be getting a different error. Can you post a fuller code sample? – Joe Green Jul 25 '12 at 10:03
Yes I just edited the first post thanks – user1550195 Jul 25 '12 at 12:16

Omesh · Accepted Answer · 2012-07-25T09:36:04.263

0

This is because of incorrect syntax and you column names in INSERT query should not be enclosed in quotes: try

$result = mysql_query("INSERT INTO ".$tbl_name."(city_id, city, state_id) VALUES(NULL, '".$city."','4421') or die(mysql_error()";

edited Jul 25 '12 at 09:36

answered Jul 25 '12 at 09:07

Omesh

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How can a PHP Function be inside the query string!!! :O – Praveen Kumar Purushothaman Jul 25 '12 at 09:10

mysql_query() expects parameter 2 to be resource, string given in

5 Answers5

You forgot to give closing `"` after `after '4421'')` and the syntax is incorrect!

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mysql_query() expects parameter 2 to be resource, string given in

5 Answers5

You forgot to give closing " after after '4421'') and the syntax is incorrect!

Linked

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You forgot to give closing `"` after `after '4421'')` and the syntax is incorrect!