Websql Database将数据插入一个表而不是另一个表?

时间:2015-11-17 06:12:32

标签: php mysql sqlite web-sql

你好朋友我面临一个奇怪的问题。我正在开发PHP桌面应用程序,

我正在使用websql Database。当我将数据提交到用户表

数据成功提交。但是当我提交到另一个表时数据不是

Submitted.table结构是一样的。我不知道该怎么办?

请帮忙。这是我的代码......

<?php

session_start();
//$WindowUser = getenv('username');
//$myPDO = new PDO("sqlite:C:/Users/".$WindowUser."/Desktop/barber_shop/webcache/databases/http_127.0.0.1_51915/2");
include("../sqlitePhpDatabase/sqlitePhpDatabase.php");   
$store_name = $_POST['store_name'];
$street = $_POST['street'];
$status = $_POST['status'];
$location_id = $_POST['location_id'];

 $sth = $myPDO->exec("INSERT INTO bs_stores(store_name,street, status,location_id)VALUES('store_name','$street','$status','$location_id')");
//$sth = $myPDO->exec("INSERT INTO bs_locations(city)VALUES('rere')");
if(isset($sth) && !empty($sth))
{
    echo "done";    
}
else
{
    echo "data is not submiting";   
}   ?>

1 个答案:

答案 0 :(得分:0)

尝试使用此代码复制所有数据: -

INSERT INTO tab1 SELECT * FROM tab2;

或复制所选数据: -

INSERT INTO tab1
SELECT col1,
       col2,
       col3,
       NULL
  FROM tab2