如何显示一个表中的数据，然后将该数据插入到php中的另一个表中？

Question

I need to fetch data from table whose mem_count is equal to 2 and then insert that data to another table after last unique_id.

Here is my code :

$mem_count2=mysql_query("SELECT mem_count,mem_id FROM member_status WHERE mem_count = 2 order by mem_id ASC");

if(mysql_num_rows($mem_count2 )==0){

      die(mysql_error()); 
     }
 else{

          $start_value=00;
  // $start_value dynamically comes from other table which i am not mentioning here
       // $start_value can starts from any number For Eg. 2, 5.

        while ($row=mysql_fetch_array($mem_count2)) {

       $uniq_code="PHR-" . str_pad((int) $start_value, 2, "0", STR_PAD_LEFT);


         //if mem_id already inserted then use IGNORE in INSERT i.e INSERT IGNORE

            $cql = "INSERT IGNORE INTO member_code (mem_id, temp_code,unique_code) values ('".$row['mem_id']."','".$row['mem_count']."','".$uniq_code."')";

          mysql_query($cql) or exit(mysql_error()); 

        $start_value++; 

       }
    }

It runs smoothly but sometimes i am getting this output:

+------------+-------------+
| mem_id     | unique_code |
+------------+-------------+
|  1         | PHR-01      |
+------------+-------------+
|  5         | PHR-02      |
+------------+-------------+
|  3         | PHR-04      |
+------------+-------------+

Some problem is surely in the INSERT IGNORE query!I have made mem_id as UNIQUE. Please rectify this as i am completely stuck over it !

Member_code structure

+------------+-------------+-----+
| mem_id     | unique_code | Id  |
+------------+-------------+-----+
|  1         | PHR-01      | 1   | 
+------------+-------------+-----+
|  5         | PHR-02      | 5   |
+------------+-------------+-----+
|  3         | PHR-04      | 3   |
+------------+-------------+-----+

Answer 1

It is probably because you are getting an error at some point at this line

mysql_query($cql) or exit(mysql_error());

and then your increment variable will be incremented for the next insert. You should test the insert and if doesn't get error you do the increment.

Answer 2

当结果存储在$mem_count时，您在代码中使用了一个名为$mem_count2的变量。