[Last Call] Learn how to a build a cloud-first strategyRegister Now

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 413
  • Last Modified:

PHP MySQL multiple mysql_num_rows in one query

I have a MySQL table that I need to generate reports from.  One column is called reason_code.  There are over 100 possible reason codes in the table.  I need to generate a report that tells me how many records  are in the table for each reason_code value.  

I know how to find the number of records for one reason code, but don't know how to do this for over 100 reason codes, without running 100 querries.  Please help me simplify my code below.  The code below shows a separate query for each reason code.  What I want is one querry that assigns the number of records to a variable for all reason codes in the table.  The following code works, but it is a killer script if I have to write a separate query for each reason code type.
$query = sprintf("SELECT amount, card_type FROM pdf_made_cb WHERE mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 AND reason_code = 'Code 1' ");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
$rca1 = mysql_num_rows($rs_v_c);
 
$query = sprintf("SELECT amount FROM pdf_made_cb WHERE mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 AND reason_code = 'Code 2' ");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
$rca2 = mysql_num_rows($rs_v_c);
 
$query_rs_v_c = sprintf("SELECT amount FROM pdf_made_cb WHERE  mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 AND reason_code = 'Code 3' ");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
$rca3 = mysql_num_rows($rs_v_c);
 
$query = sprintf("SELECT amount FROM pdf_made_cb WHERE mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 AND reason_code = 'Code 4' ");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
$rca4 = mysql_num_rows($rs_v_c);

Open in new window

0
fastfind1
Asked:
fastfind1
  • 4
  • 3
2 Solutions
 
Guy Hengel [angelIII / a3]Billing EngineerCommented:
something this would do:
$query = sprintf("SELECT reason_code, count(*) rows, sum(amount) sum_amount FROM pdf_made_cb WHERE mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 group by reason_code ");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
$rca1 = mysql_num_rows($rs_v_c);

Open in new window

0
 
bleach77Commented:
try this.
$query_rs_v_c = sprintf("SELECT reason_code, count(amount) AS amount FROM pdf_made_cb WHERE  mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 GROUP BY reason_code");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
 
$code = array();
 
while ($row = mysql_fetch_assoc($rs_v_c)) {
   $code["$row['reason_code']"] = $row['amount'];
}  

Open in new window

0
 
fastfind1Author Commented:
Thank you for your help.  I need to assign a variable to each count of records for each reason code.  So if reason code 1 had 4 records, I need to assign $rca = 4.  If reason code 2 had 9 records, I need to assign $rca = 2.  How do I do this based on your code?
0
Configuration Guide and Best Practices

Read the guide to learn how to orchestrate Data ONTAP, create application-consistent backups and enable fast recovery from NetApp storage snapshots. Version 9.5 also contains performance and scalability enhancements to meet the needs of the largest enterprise environments.

 
fastfind1Author Commented:
i meant, if reason code 2 had 9 records i need to assign $rcb = 9.  Sorry for the confusion.
0
 
fastfind1Author Commented:
In the end, I need something like this:

$rca = the number of records with reason code 1;
$rcb = the number of records with reason code 2;
$rcc = the number of records with reason code 3;
$rcd = the number of records with reason code 4;
0
 
bleach77Commented:
The old code would be like
$code['Code 1'] = the number of records with reason code 1;
$code['Code 2'] = the number of records with reason code 2;
$code['Code 3'] = the number of records with reason code 3;

This code will produce like your first post
$rca1 =  the number of records with reason code 1;
$rca2 =  the number of records with reason code 2;
$rca3 =  the number of records with reason code 3;
$query_rs_v_c = sprintf("SELECT reason_code, count(amount) AS amount FROM pdf_made_cb WHERE  mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 GROUP BY reason_code ORDER BY reason_code");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
 
$count = 0;
 
while ($row = mysql_fetch_assoc($rs_v_c)) {
  $count++;
  ${"rca".$count} = $row['amount'];
}  

Open in new window

0
 
bleach77Commented:
This should work like your new post
$rca = the number of records with reason code 1;
$rcb = the number of records with reason code 2;
$rcc = the number of records with reason code 3;
$rcd = the number of records with reason code 4;
$query_rs_v_c = sprintf("SELECT reason_code, count(amount) AS amount FROM pdf_made_cb WHERE  mid = '$mid' AND receive_date >= '$date1' AND receive_date <= '$date2' AND amount >= $amt1 AND amount <= $amt2 GROUP BY reason_code ORDER BY reason_code");
$rs_v_c = mysql_query($query, $connect_crm) or die(mysql_error());
 
$count = 'a';
 
while ($row = mysql_fetch_assoc($rs_v_c)) {
  ${"rc".$count} = $row['amount'];
  $count++;
}  

Open in new window

0
 
bleach77Commented:
Both way you need to either declare the variables first.
$rca1 = $rca2 = $rcb = "";

OR

when you want to use it, you have to check using isset, otherwise an error will occur.

if(isset($rca1)) echo $rca1;
if(isset($rca2)) echo $rca2;
if(isset($rcb)) echo $rcb;
0

Featured Post

Hire Technology Freelancers with Gigs

Work with freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely, and get projects done right.

  • 4
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now