store value return by mysql in php variable

Question

For the code below i am trying to get the gender value in gen then store it in $gender but i am getting wrong output. I am getting Resource id #8 as my output instead of male or female. What shall i do to get the gender value and store it in $gender with use of minimal coding.

$gender=mysql_query("SELECT `gender` as gen FROM `register` WHERE `reg_id` = 'reg252'");
echo $gender;

One way i can do is by calling something like the way below -

$gender=mysql_query("SELECT `gender` FROM `register` WHERE `reg_id` = 'reg252'");
while($erow = mysql_fetch_array($gender))
{
    echo $erow['gender'];
}

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — PeeHaa
– PeeHaa, Commented Aug 22, 2012 at 8:43
Well, yes, mysql_query returns a result set resource, and you have figured out yourself how to get results from that result set. That's the way it works. — deceze
– deceze ♦, Commented Aug 22, 2012 at 8:45
it copy paste problem i corrected the code below as gen is removed — Kundan Raj
– Kundan Raj, Commented Aug 22, 2012 at 10:02

Peter · Accepted Answer · 2012-08-22 08:44:19Z

1

list($gender) = mysql_fetch_row(mysql_query("SELECT `gender` as gen FROM `register` WHERE `reg_id` = 'reg252'"));

answered Aug 22, 2012 at 8:44

Peter

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4 Comments

deceze Over a year ago

I'd advice against such one-liners, you should properly error handle the result of any and all mysql_query calls, not blindly feed them into the next function.

Peter Over a year ago

I'd advice to use PDO instead of mysql_ functions but obviously this guy is still learning

Sergey Telshevsky Over a year ago

In any case, PDO or mysql_, if you want to use one-liners, you should make helper functions with proper returns and exception handling.

Peter Over a year ago

OK so if you put next line if(is_null($gender)) you can handle your error.

Mina Abadir · Accepted Answer · 2012-08-22 08:52:27Z

0

You are returning the column as gen, so you have to check gen echo $erow['gen'];

In general check the comment from PeeHaa

edited Aug 22, 2012 at 8:52

answered Aug 22, 2012 at 8:46

Mina Abadir

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3 Comments

Mina Abadir Over a year ago

Sorry to see other answers mentioning the same. When I submitted my answer, no answers were here. Probably we submitted them in the same second.

Kundan Raj Over a year ago

coorected the second one as gen was mistakenly there

Mina Abadir Over a year ago

did you try to run this Query direct on the DB, and ensure that it returns a value? "SELECT gender FROM register WHERE reg_id = 'reg252'"

Sergey Telshevsky · Accepted Answer · 2012-08-22 08:53:11Z

0

You have in your SQL

SELECT `gender` as gen ...

While you output echo $erow['gender'];

So either output echo $erow['gen'];, either SELECT gender FROM ...

edited Aug 22, 2012 at 8:53

answered Aug 22, 2012 at 8:44

Sergey Telshevsky

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2 Comments

Waygood Over a year ago

+1'd this as it is part of the solution (which was re-done by Mina)

Mina Abadir Over a year ago

probably it's downvoted because you made another logical error. So either output echo $erow['gender'];, either SELECT gender FROM...It should be So either output echo $erow['gen'];, either SELECT gender FROM

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