Printing a variable from a query

[provision]

Hey all,

 

Im new to all this php stuff so bear with me.

 

Ill just post my code rather then trying to explain myself i will just get confused.

 

$query = 'SELECT COUNT(*) FROM performer WHERE performer_type = "Band"';
$result = run_sql( $query, true );
$print= mysql_fetch_assoc( $result );
print $print

 

When I view the page it prints out "Array".

 

Thanks

Josh

[mmarif4u]

So what u want to get from ur query.

[provision]

Ok, sorry I didnt explain myself enough, what I am doing is my site is a cultural events site and I want to be able to print out the amount of types of cultural events from the database examle. Band (2), Artist (5).

 

The query works but when I go to print it it just puts "array" instead of the number.

 

 

[teng84]

$print= mysql_fetch_assoc( $result );

 

$print this is array so when you prin you will have the word array

 

use print_r($print );

 

hope you know hoew to get array content

[provision]

Ok,

 

Im a real beginner at this im sorry.

 

I dont know how to get array content im just kinda winging this site till i get the hang of it.

 

When i use print_r($print );

 

it prints Array ( [COUNT(*)] => 2 )

 

I know it must be something easy to fix but...... I just dont know.

 

 

[mmarif4u]

Ok, you want to print a specific row from db or want to print all the records.

If that is the issue than u have to use while loop to loop ur records.

please explain ur need clearly.

[provision]

Ok sorry.

 

I just want to print the number of a certain event/artist type. Say if in my database there is 5 bands and 13 artists and 4 anual events. I want to print in the nav bar something like Band (5), Artist (13), Anual Event (4).

 

I only want to print the number of rows in the db with a certain value in the column. My query works the only problem I have is printing that number out.

 

 

[mmarif4u]

try this bcoz u mention band in ur query so it will give out put for band.

 

$query = 'SELECT COUNT(*) FROM performer WHERE performer_type = "Band"';
$result = run_sql( $query, true );
$print= mysql_num_rows( $result );
print $print

[provision]

Well that prints a number but its only 1, It wont change to any other number.

[mmarif4u]

This is bcoz of your query:

WHERE performer_type = "Band"

Still i am confused about ur q?.

ok let me clear,

tell me when u insert a record to that table so u just update the performer_type field or increment it.

Like if u have 10 records for band when u insert new it goes to 11,

OR it make a new row in db.

 

Try this code for single row output.

instead of this

$print= mysql_num_rows( $result );
print $print

try this:

$print= mysql_fetch_object( $result );
$band=$print->performer_type;
echo $band;

The above  code depends on your db table field.

 

 

[provision]

When a band member creates a new account to the site a new row in the db is created with all their details including a colum that says they are a band.

 

That code that you gave me then did nothing.

 

 

[mmarif4u]

ok try this, yeh how many rows u have in db are u check it.

$query = 'SELECT * FROM performer WHERE performer_type = "Band"';
$result = mysql_query($query);
$print= mysql_num_rows($result);
print $print

 

[suma237]

hey,

query is not clear..are you trying to do this

 

 

$query = 'SELECT * FROM performer WHERE performer_type = "Band"';

$result=mysql_query($query);

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

 

echo $row['event'];

}

 

 

 

[drummer101]

 

$query = 'SELECT * FROM performer WHERE performer_type = "Band"';

$result=mysql_query($query);

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

 

echo $row[####'];

}

 

This is what you want, I just stopped by to see if this was resolved and I was beaten by one post

 

#### = a column header in your database.

 

Edit 2: while(); executes until it evaluates to FALSE, in your case, it's executing mysql_query($query); until there are no more rows that meet the criteria.

[mmarif4u]

I want to print in the nav bar something like Band (5), Artist (13), Anual Event (4).

 

I think he want to show the number of rows for band or something not  showing results for it..

[drummer101]

I want to print in the nav bar something like Band (5), Artist (13), Anual Event (4).

 

I think he want to show the number of rows for band or something not  showing results for it..

 

$artist = "SELECT * FROM `performer` WHERE performer_type = 'Artist'";
$artist_result = mysql_query($artist) or die ("Query Failed" . mysql_error());
$artist_numrows = mysql_num_rows($artist_result);

$band = "SELECT * FROM `performer` WHERE performer_type = 'Band'";
$band_result = mysql_query($band) or die ("Query Failed" . mysql_error());
$band_numrows = mysql_num_rows($band_result);

$event = "SELECT * FROM `performer` WHERE performer_type = 'Event'";
$event_result = mysql_query($event) or die ("Query Failed" . mysql_error());
$event_numrows = mysql_num_rows($event_result);


echo "Artist(" . $artist_numrows . ")";
echo "Band(" . $band_numrows . ")";
echo "Annual Event(" . $event_numrows . ")";

 

Hope that gets you started in the right direction.

 

or die ("Query Failed" . mysql_error())

This will let you know if your query is incorrect. If it comes back with any errors just post and let me know and I'll try and help you some more.

 

[drummer101]

$band = "SELECT * FROM `performer` WHERE performer_type = 'Band'";

$band_result = mysql_query($band) or die ("Query Failed" . mysql_error());

$band_numrows = mysql_num_rows($band_result);

 

echo "Bands(" . $band_result . ")";

 

I'm not exactly sure as to your database or form structure, so I'm going to take back my above code and let you use the band query as a template.

 

Best of luck!  And happy coding.

[trq]

<?php

  $query = 'SELECT COUNT(*) AS total FROM performer WHERE performer_type = "Band"';
  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) {
      $row = mysql_fetch_assoc($result);
      print $row['total'];
    }
  }

?>