Question

我创建了临时表 - temporary_ticket

我想将主表中的数据插入临时表ticket

在导出数据之前，我需要将数据从主表插入临时表，然后排除不需要的列

这是我的sql代码。

$sql = "SELECT id,subject,created_by,type,status,cat_id,DATE_FORMAT(create_time, '%m/%d/%Y 00:00:00')as create_time ,DATE_FORMAT(update_time, '%m/%d/%Y 00:00:00')as update_time,priority,updated_by,assigned_to,ticket_type,agent_created,cust4 FROM  `$DB_TBLName` ORDER BY id ASC";

我试过

Insert* INTO temporary_ticket  FROM ticket /* Drop the cloumns that are not needed */
ALTER TABLE temporary_ticket  DROP COLUMN guest_name,guest_email /* Get results and drop temp table */
SELECT * FROM temporary_ticket DROP TABLE temporary_ticket

但它未来

此查询有什么问题？

这是我的完整代码

<?php
$DB_Server = "localhost"; //MySQL Server    
$DB_Username = "*****"; //MySQL Username     
$DB_Password = "******";             //MySQL Password     
$DB_DBName = "i******";         //MySQL Database Name  
$DB_TBLName = "wp3_wpsp_ticket"; //MySQL Table Name   
$filename = "Support Ticket_Export";         //File Name


$sql="insert into wp3_wpsp_temporary_ticket  
      SELECT *   FROM wp3_wpsp_ticket 
      commit;" 


$Connect = @mysql_connect($DB_Server, $DB_Username, $DB_Password) or die("Couldn't connect to MySQL:<br>" . mysql_error() . "<br>" . mysql_errno());

//select database   
$Db = @mysql_select_db($DB_DBName, $Connect) or die("Couldn't select database:<br>" . mysql_error(). "<br>" . mysql_errno());   
//execute query 
$result = @mysql_query($sql,$Connect) or die("Couldn't execute query:<br>" . mysql_error(). "<br>" . mysql_errno());    
$file_ending = "xls";
//header info for browser
header("Content-Type: application/xls");    
header("Content-Disposition: attachment; filename=$filename.xls");  
header("Pragma: no-cache"); 
header("Expires: 0");


 $sep = "\t"; //tabbed character
 //start of printing column names as names of MySQL fields
 for ($i = 0; $i < mysql_num_fields($result); $i++) {
 echo mysql_field_name($result,$i) . "\t";
 }
 print("\n");   

 //end of printing column names  
 //start while loop to get data
 while($row = mysql_fetch_row($result))
 {
  $schema_insert = "";
  for($j=0; $j<mysql_num_fields($result);$j++)
    {
   if(!isset($row[$j]))
    $schema_insert .= "NULL".$sep;
    elseif ($row[$j] != "")
    $schema_insert .= "$row[$j]".$sep;
    else
    $schema_insert .= "".$sep;
    }
    $schema_insert = str_replace($sep."$", "", $schema_insert);
    $schema_insert = preg_replace("/\r\n|\n\r|\n|\r/", " ", $schema_insert);
    $schema_insert .= "\t";
    print(trim($schema_insert));
    print "\n";
    }   
    ?>

Answer 1

在MySQL中select ... into ...不用于根据另一个表填充表，该语句用于为变量赋值或将数据导出到外部文件中。

您需要使用insert ... select ...语句：

insert into temporary_ticket
select * from ticket

但是，您在问题中描述的一系列陈述可以简单地用

代替

select <list of fieldnames excluding guest_name,guest_email) from ticket

语句。无需临时表创建和所有其他复杂的东西。

Answer 2

您应该使用插入选择

insert into temporary_ticket  
SELECT *   FROM ticket ;

并确保两个表之间的列匹配类型和编号（也是相同的序列）

在导出之前，我需要将数据从一个表插入另一个表-php

2 个答案: