array problem

raffael3d

I fetch a value form the column "catg" from a table:

$r = mysql_query("select catg from jobs WHERE approve=0");

then I have this:

$catg_array = array (1 => 'Available Jobs', 2 => 'Looking for Job');

now I want to print the result depening on if the value in the databse was 1 or 2

I have this: print $catg_array;

but this prints "array" instead of the value, what is wrong? please correct my code, thanks.

philipolson

$catg = mysql_result($r, 0);

print $catg_array[$catg];

See also mysql_fetch_assoc() for details on retrieving information from a mysql result. And, the section on arrays.

raffael3d

thanks a lot!

raffael3d

I discovered a problem:
if the value of "catg" is 1 it displays correctly. however if the value is 2 it also displays the same...
what can I do?

philipolson

Please be more specific, not sure what you mean.

raffael3d

the array is not working correctly:
it should fetch the data from the column "catg" and depending if the value there is 1 it should print "available job", if the value is 2 it should print "looking for job"
now I experienced that if the value is 1 or 2, in both cases it prins "available job".

philipolson

That's impossible, I can't think of any reason that would happen so it must be something else. Print $catg and make sure it's actually printing out 2.

raffael3d

it doesnt' print anything in this case. ok here is the code, mabye you see something

$r = mysql_query("select catg from jobs WHERE approve=0"); 

$catg_array = array (1 => 'Available Jobs', 2 => 'Looking for Job');
$catg = mysql_result($r, 0);

print $catg_array[$catg];

raffael3d

I guess the problem is in the "while".

result = mysql_query("select id,title,company,city,country FROM jobs WHERE approve=0");
$r = mysql_query("select catg from jobs"); 


while ( $row = mysql_fetch_array($result) ) {
$catg = mysql_result($r, 0);
$catg_array = array (1 => '1', 2 => '2', 3 => '3');




print $catg_array[$catg];

}

philipolson

That code has issues, how about:

$result = mysql_query("select id, title, company, city, country, catg FROM jobs WHERE approve=0");

$info = array(1 => 'Available', 2 => 'Not Available');

while ($row = mysql_fetch_assoc($result)) {

print 'Title: ' . $row['title'];
print 'Job Status: ' . $info[$row['catg']];

}

So, if catg is 1 it will print Available, if it is 2 it will print Not Available. Also notice how we just need one query, and define the $info array just once.