TRYING TO UNDERSTAND MYSQL QUERY for ZIP CODE COMPARISON

[robman2100]

I am trying to figure out how to grab a

$number=mysql_num_rows($query);

and only get the sum from Zip codes in a group that i have defined such as

$query = mysql_query("SELECT * FROM table WHERE zip ='22303,22304,22305,22306");
echo $number;

 

This is nesscecary since only certain Zip codes are in Certain Counties. I will then Echo the results to each one of the Counties to show the amount of records i have in each County.

[Mchl]

SELECT * FROM table WHERE zip IN (22303,22304,22305,22306)

 

??

[robman2100]

SELECT * FROM table WHERE zip IN (22303,22304,22305,22306)

 

??

 

What is the purpose of the IN?

[Mchl]

To select records where zip is equal to any of those values

[robman2100]

To select records where zip is equal to any of those values

 

So This will allow me to get the sum of the amount of records that equal to the zips i have placed in the database

 

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = mysql_query("SELECT * FROM locator_store WHERE zip IN='22303,22304,22305,22306');
$number=mysql_num_rows($query);


echo $number;

[Mchl]

To count how many records with these sips are use this

 

$query = mysql_query("SELECT COUNT(*) AS zipCount FROM locator_store WHERE zip IN('22303','22304','22305','22306')";
$row = mysql_fetch_assoc($query);
$number = $row['zipCount'];

echo $number;

[robman2100]

To count how many records with these sips are use this

 

$query = mysql_query("SELECT COUNT(*) AS zipCount FROM locator_store WHERE zip IN('22303','22304','22305','22306')";
$row = mysql_fetch_assoc($query);
$number = $row['zipCount'];

echo $number;

 

I get a Syntax error using the code you provided on line 8 which is the $query line. Im not sure what the syntax error means because it says the error is ; . But that dosent make sense since you have to end each command with ;

[Mchl]

Check if you did not forget to put ; at the end of previous line

 

[robman2100]

Check if you did not forget to put ; at the end of previous line

 

I found what was originaly wrong we didnt close the TAG with a ) lol

But now i am getting a error

called

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

 

I dont understand this is so frustrating im so close but so far away.

[ignace]

I get a Syntax error

 

You also mind telling what that is? I also want to point out you take a look at:

 

$query = mysql_query("SELECT COUNT(*) AS zipCount FROM locator_store WHERE zip IN('22303','22304','22305','22306')";
$row = mysql_fetch_assoc($query);
$number = $row['zipCount'];

echo $number;

 

Now take a look at:

 

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = mysql_query("SELECT * FROM locator_store WHERE zip IN='22303,22304,22305,22306');
$number=mysql_num_rows($query);


echo $number;

 

You notice something?

 

Hint #1:

 

I found what was originaly wrong we didnt close the TAG with a ) lol

 

But now i am getting a error

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

 

Hint #2:

 

You sure it was )??

[robman2100]

I get a Syntax error

 

You also mind telling what that is? I also want to point out you take a look at:

 

$query = mysql_query("SELECT COUNT(*) AS zipCount FROM locator_store WHERE zip IN('22303','22304','22305','22306')";
$row = mysql_fetch_assoc($query);
$number = $row['zipCount'];

echo $number;

 

Now take a look at:

 

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = mysql_query("SELECT * FROM locator_store WHERE zip IN='22303,22304,22305,22306');
$number=mysql_num_rows($query);


echo $number;

 

You notice something?

 

Hint #1:

 

I found what was originaly wrong we didnt close the TAG with a ) lol

 

But now i am getting a error

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

 

Hint #2:

 

You sure it was )??

well if you take a look this is what i have so far

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = mysql_query("SELECT (*) AS zip FROM locator_store WHERE zip IN('22303','22304','22305','22306')");
$row = mysql_num_rows($query);
$number = $row['zip'];
echo $number;

and now i am getting that error i posted before

[ignace]

SELECT (*) AS zip

 

should be

 

SELECT count(*) AS zip

[Mchl]

And

$row = mysql_num_rows($query);

should be

$row = mysql_fetch_assoc($query);

[robman2100]

And

$row = mysql_num_rows($query);

should be

$row = mysql_fetch_assoc($query);

 

Now i get this error?

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

 

I dont know this is really frustrating me. Im sure its something simple. Its a small Script too. So idk?

[ignace]

Did you change??

 

SELECT (*) AS zip

 

should be

 

SELECT count(*) AS zip