我想从Firebird数据库中选择一些列并将它们插入MSSQL数据库表中。
当我从Firebird读取表格时,它代表我在localhost中需要的内容,然后当我想将数据插入MSSQL时,它只插入最后一条记录。我不知道是什么原因引起的,这是我第一次尝试这个想法。
有人会告诉我出了什么问题吗?
这是我的PHP代码:
<?php
$host = 'localhost:c:\firebird.fdb';
$username='';
$password='';
$dbh = ibase_connect($host, $username, $password);
$stmt = "SELECT * FROM TWEEKDAY";
$sth = ibase_query($dbh, $stmt);
while ($row = ibase_fetch_object($sth)) {
$R1 = $row->CODE;
$R2 = $row->DAYNAME;
echo $R1 ." ". $R2 . "\n";
echo "</br>";
}
ibase_close($dbh);
/**********************************************/
$host = "servername\instancename";
$connectionInfo = array( "Database"=>"MSSQLdatabase", "UID"=>"", "PWD"=>"");
$conn = sqlsrv_connect( $host, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
echo "</br>";
$tsql = "INSERT INTO [AccessCard].[dbo].[WEEKDAY] (DAYID,DAYN) VALUES ('$R1','$R2')";
$stmt = sqlsrv_query( $conn, $tsql);
if ( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
while( $obj = sqlsrv_fetch_object( $stmt , SQLSRV_FETCH_ASSOC) ) {
echo $obj->DAYID ."  ". $obj->DAYN . "<br />";
}
sqlsrv_free_stmt( $stmt);
sqlsrv_close( $conn);
?>
当我运行我的代码时,它在本地主机中显示为: PIC1 当我打开mssql数据库时,该表有什么: PIC2 它似乎插入$ R1 $ R2的最后一个值。
当我使用此代码时:
<?php
$host = 'localhost:c:\FIREBIRD.fdb';
$username='';
$password='';
$dbh = ibase_connect($host, $username, $password);
$stmt = "SELECT * FROM TWEEKDAY";
$sth = ibase_query($dbh, $stmt);
while ($row = ibase_fetch_object($sth)) {
$R1 = $row->CODE;
$R2 = $row->DAYNAME;
echo $R1." ". $R2 . "\n";
echo "</br>";
}
ibase_close($dbh);
/**********************************************/
$host = "SERVERNAME\INSTANCENAME";
$connectionInfo = array( "Database"=>"AccessCard", "UID"=>"", "PWD"=>"");
$conn = sqlsrv_connect( $host, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
echo "</br>";
for ($RIdx = 0; $RIdx < count($R1); $RIdx++) { // each $R array value
$tsql = "INSERT INTO [AccessCard].[dbo].[WEEKDAY] (DAYID,DAYN) VALUES ('$R1[$RIdx]','$R2[$RIdx]')";
$stmt = sqlsrv_query( $conn, $tsql);
if ( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
} // end of insert loop
sqlsrv_free_stmt( $stmt);
sqlsrv_close( $conn);
?>
这是数据库中显示的内容: PIC3
答案 0 :(得分:0)
$ R1应该是一个数组,如$ R1 [] = $ row-&gt; CODE;显然也是$ R2'。相应地更改sqlserver插入。我这里没有安装sql server所以我无法提供经过测试的代码。 这是sql server代码......
$host = "servername\instancename";
$connectionInfo = array( "Database"=>"MSSQLdatabase", "UID"=>"", "PWD"=>"");
$conn = sqlsrv_connect( $host, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
echo "</br>";
for ($RIdx = 0; $RIdx < count($R1); $RIdx++) { // each $R array value
$tsql = "INSERT INTO [AccessCard].[dbo].[WEEKDAY] (DAYID,DAYN) VALUES ('$R1[$RIdx]','$R2[$RIdx]')";
$stmt = sqlsrv_query( $conn, $tsql);
if ( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
} // end of insert loop
sqlsrv_free_stmt( $stmt);
sqlsrv_close( $conn);
?>
答案 1 :(得分:0)
最终问题解决了: 这是解决方案代码
<?php
$host = 'localhost:c:\firebird.fdb';
$username='';
$password='';
$dbh = ibase_connect($host, $username, $password);
$stmt = "SELECT * FROM TWEEKDAY";
/**********************************************/
$host = "servername\instancename";
$connectionInfo = array( "Database"=>"AccessCard", "UID"=>"", "PWD"=>"");
$conn = sqlsrv_connect( $host, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
echo "</br>";
/**********************************************/
$sth = ibase_query($dbh, $stmt);
while ($row = ibase_fetch_object($sth)) {
$R1 = $row->CODE;
$R2 = $row->DAYNAME;
echo $R1." ". $R2 . "\n";
$tsql = "INSERT INTO [AccessCard].[dbo].[WEEKDAY] (DAYID,DAYN) VALUES ('$R1','$R2')";
$stmt = sqlsrv_query( $conn, $tsql);
if ( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
while( $obj = sqlsrv_fetch_OBJECT( $stmt) ) {
echo $obj->$R1."  ". $obj->$R2. "<br />";
$obj++;
}
echo "</br>";
}
ibase_close($dbh);
sqlsrv_free_stmt( $stmt);
sqlsrv_close( $conn);
?>
谢谢Ryan的帮助,在我编辑我的问题时似乎想到了解决方案。
再次感谢。