How to use a PHP variable as a column name in select query in MYSQL?

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PHP

<?php
mysql_connect("localhost","root","password");
mysql_select_db("db_name");
$towerValue_4bhk = $_GET['towerValue_4bhk']; // from AJAX call
$floor_names_query = MYSQL_QUERY("select $total_4bhk from project_master");
mysql_close();
?>

I want $towerValue_4bhk to be used as a column name in SELECT Query.

When user selects a 4BHK in the front end then I should display the no of 4BHK's available for 15 floors.

Posted 23-Dec-14 1:09am

Janardhanam Julapalli

Updated 1-May-21 4:31am

Praveen Kumar Upadhyay

v2

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3 solutions

Solution 2

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<html>
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   <meta charset="utf-8" />
    <title></title>
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<body>
<?php
print '12544';
?>
</body>
</html>

Posted 23-Dec-14 9:22am

Member 10788507

Solution 3

<?php
mysql_connect("localhost","root","password");
mysql_select_db("db_name");
$towerValue_4bhk = $_GET['towerValue_4bhk']; // from AJAX call
$floor_names_query = MYSQL_QUERY("select ($total_4bhk) from project_master");
mysql_close();
?>

Posted 1-May-21 4:31am

Member 15179218

Updated 1-May-21 4:32am

v2

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Praveen Kumar Upadhyay · Accepted Answer · 2014-12-23T01:28:00

Solution 1

Change your code to below.

PHP

<?php
mysql_connect("localhost","root","password");
mysql_select_db("rohan_iksha");
$towerValue_4bhk = $_GET['towerValue_4bhk']; // from AJAX call
$floor_names_query = MYSQL_QUERY("select ".$total_4bhk." from project_master");
mysql_close();
?>

Run below code to check if your query is returning result.

PHP

<?php

$link = mysql_connect("localhost", "root", "password");
mysql_select_db("rohan_iksha", $link);

$result = mysql_query("select total_4bhk from project_master", $link);
$num_rows = mysql_num_rows($result);

echo "$num_rows Rows\n";

?>

See what it is returning.

Posted 23-Dec-14 1:28am

Praveen Kumar Upadhyay

Updated 23-Dec-14 20:57pm

v4

Comments

Chubby Ninja 23-Dec-14 15:04pm

where is $total_4bhk defined above? FYI: you can include a variable inside "double quotes" just like the OP has

Janardhanam Julapalli 23-Dec-14 22:03pm

sorry! MYSQL_QUERY("select $towerValue_4bhk from project_master") is my requirement

Praveen Kumar Upadhyay 23-Dec-14 22:10pm

Yes. So just change variable name there.

Janardhanam Julapalli 23-Dec-14 22:13pm

I tried it.But no success.

Praveen Kumar Upadhyay 23-Dec-14 22:20pm

Try doing echo of that variable before the select statement. I guess that variable has some invalid data.

Janardhanam Julapalli 23-Dec-14 22:17pm

It is giving Resource id #4.I am not sure what it will be.

Janardhanam Julapalli 23-Dec-14 22:18pm

Can anyone please explain what exactly the query will do!

Janardhanam Julapalli 23-Dec-14 22:24pm

$towerValue_4bhk = "total_4bhk"; Even I tried with hard coded value.

Arun-23 23-Dec-14 23:12pm

select ".$total_4bhk." from project_master; this is the query did u get value in $total_4bhk variable??

Janardhanam Julapalli 23-Dec-14 23:16pm

$towerValue_4bhk = "total_4bhk";
$floor_names_query = MYSQL_QUERY("select ".$towerValue_4bhk." from project_master");
This is what I am doing.Not getting anything.

Praveen Kumar Upadhyay 24-Dec-14 1:18am

So your select query is becoming like "select total_4bhk from project_master"
Can you please execute the query in Mysql and see what output your are getting.

Janardhanam Julapalli 24-Dec-14 2:03am

Showing rows 0 - 24 (53 total, Query took 0.0010 seconds.)

Praveen Kumar Upadhyay 24-Dec-14 2:05am

How many rows it is returning?

Janardhanam Julapalli 24-Dec-14 2:11am

53 rows

Praveen Kumar Upadhyay 24-Dec-14 2:13am

Check how many rows your query is returning using below code.

echo mysql_num_rows($conn);

$conn is your connection object.

Praveen Kumar Upadhyay 24-Dec-14 2:18am

Check my updated question

Janardhanam Julapalli 24-Dec-14 2:22am

It is not returning anything for this.
$result = mysql_num_rows($conn);
echo $result;

Praveen Kumar Upadhyay 24-Dec-14 2:26am

What is your database name in which project_master table is there.

mysql_select_db("db_name");

Is "db_name" your database name??

Janardhanam Julapalli 24-Dec-14 2:34am

$conn = mysql_connect("localhost", "root", "myfourwalls");
mysql_select_db("rohan_iksha", $conn);
$num_rows = mysql_num_rows($conn);
echo $num_rows;

Praveen Kumar Upadhyay 24-Dec-14 2:36am

If your db name is "rohan_iksha" then why you are using "db_name". Just change the db name and your code will work.

Janardhanam Julapalli 24-Dec-14 2:42am

No sir.I am using the same db name before itself.

Praveen Kumar Upadhyay 24-Dec-14 2:51am

Try my second code again. I have updated the db_name there.

Praveen Kumar Upadhyay 24-Dec-14 2:51am

People who is devoting me for this solution. Please hold and see.

Janardhanam Julapalli 24-Dec-14 2:55am

It is returning "53 Rows" as result

Praveen Kumar Upadhyay 24-Dec-14 2:57am

Means it is now working fine, right?

Janardhanam Julapalli 24-Dec-14 2:59am

s

Janardhanam Julapalli 24-Dec-14 3:00am

$link = mysql_connect("localhost", "root", "");
mysql_select_db("rohan_iksha", $link);

$test = "total_4bhk";
$result = mysql_query("select $test from project_master", $link);
$num_rows = mysql_num_rows($result);

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$row_array['total_4bhk'] = $row['total_4bhk'];

echo $row['total_4bhk'];
}
It is displaying the required values for me.

Praveen Kumar Upadhyay 24-Dec-14 3:02am

Now please accept the solution buddy. I have spend a lot time on this, just for a silly mistake by you which I didn't notice.

Janardhanam Julapalli 24-Dec-14 3:02am

Thanks for your time and patience.I am grateful to u!

Janardhanam Julapalli 24-Dec-14 3:03am

Alright.What mistake exactly I have done here?

Praveen Kumar Upadhyay 24-Dec-14 3:05am

Instead of
mysql_select_db("rohan_iksha");
you were trying
mysql_select_db("db_name");

So it was searching for the database name "db_name". Rest everything is ok.

Janardhanam Julapalli 24-Dec-14 3:09am

Done Sir!

Praveen Kumar Upadhyay 24-Dec-14 3:09am

Thanks.