使用 PHP 将数据插入 MSSQL DB
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Insert Data into MSSQL DB using PHP
提问by user3315848
Hello there am trying to insert data into MSSQL using PHP. I have tried many times to figure out what the problem might be but i seem not to find it. Is there something am not getting right or missing?
您好，我正在尝试使用 PHP 将数据插入到 MSSQL 中。我已经多次尝试找出问题所在，但我似乎没有找到它。有什么地方没有得到正确或遗漏吗？
<?php
//pull form fields into php variables
$no = $_POST['no'];
$name= $_POST['name'];
$date = $_POST['date'];
$leave = $_POST['leave'];
$days= $_POST['days'];
$empno= $_POST['empno'];
//connect to sql
$dbc = mssql_connect('Server-PC','user','password','database')or die('Error connecting to
the SQL Server database.');
// Input into staff database
$query = "INSERT INTO dbo.[CAGD$Leave Plan] ([No_],[Employee No_],[Employee Name],
[Leave Name], [Start Date],[Leave Days],Satus) VALUES
('$no','$name','$leave','$date','days','empno')";
$r esult = mssql_query($query,$dbc)or die('Error querying MSSQL database');
//close to sql
mssql_close($dbc);
echo $name . 'Your submission has been received<br />';
echo 'If you need change this request please contact your HR Manager<br />';
echo 'Thank you <br />';
echo 'HR Manager';
?>
I get this error message:
Warning: mssql_query() [function.mssql-query]: message: Invalid object name 'dbo.CAGD Plan'.
(severity 16) in C:\xampp\htdocs\CAGD\leave_request.php on line 110
我收到此错误消息：警告：mssql_query() [function.mssql-query]：消息：对象名称“dbo.CAGD 计划”无效。
（严重性 16）在 C:\xampp\htdocs\CAGD\leave_request.php 第 110 行
Warning: mssql_query() [function.mssql-query]: Query failed in C:\xampp\htdocs
\CAGD\leave_request.php on line 110
Error querying MSSQL database
回答by Jayakarthik Appasamy
First Specify your database Connection...
首先指定您的数据库连接...
mssql_connect('Server-PC','user','password','database')
like -> "localhost","root","XXXX", "DBNAME"
then query like
然后像这样查询
$query = "INSERT INTO TABLENAME (id,name) VALUES
('$id','$name')";
$result = mssql_query($query,$dbc)
回答by Alireza
You can use SQLSRV Driver instead of MSSQL Driver and then try this
您可以使用 SQLSRV 驱动程序代替 MSSQL 驱动程序，然后试试这个
<?php
$serverName = "serverName";
$options = array( "UID" => "sa", "PWD" => "Password", "Database" => "DBname");
$conn = sqlsrv_connect($serverName, $options);
if( $conn === false )
{
echo "Could not connect.\n";
die( print_r( sqlsrv_errors(), true));
}
$no = $_POST['no'];
$name= $_POST['name'];
$query = "INSERT INTO dbo.Test
(No_,FirstName)
VALUES(?, ?)";
$params1 = array($no,$name);
$result = sqlsrv_query($conn,$query,$params1);
sqlsrv_close($conn);
?>
This is more useful, and you can learn more here:
这更有用，您可以在此处了解更多信息：
https://msdn.microsoft.com/en-us/library/cc626305(v=sql.105).aspx
https://msdn.microsoft.com/en-us/library/cc626305(v=sql.105).aspx
回答by Cosmin Hutanu
Hmm, it seems to me that you have 7 fields in the table but only 6 values submitted - you are missing the value for the first column, [No_].
Besides, the last column satus(i suppose it should be 'status') does not have de []delimiters.
The error returned tells you that the name of the table is wrong.
And yes variable names are case sensitive in PHP, it should be $leave - best to exit the string and concatenate - something like "bla bla".$leave."anything here with spaces or not".
嗯，在我看来，您在表中有 7 个字段，但只提交了 6 个值 - 您缺少第一列的值[No_]. 此外，最后一列satus（我想它应该是“状态”）没有[]分隔符。返回的错误告诉您表的名称是错误的。是的，变量名在 PHP 中区分大小写，它应该是 $leave - 最好退出字符串并连接 - 类似于"bla bla".$leave."anything here with spaces or not".
回答by Marc B
Is this supposed to be a variable?
这应该是一个变量吗？
$query = "INSERT INTO dbo.[CAGD$Leave Plan] ([No_],[Employee No
^^^^^^
If so, then it's apparently undefined in your code, and the generated query string will contain dbo.[CAGD Plan], and not whatever value was supposed to be in that variable. If the $is literally in your table name, then it should be CAGD\$Leave, so that $Leaveisn't treated as a variable.
如果是这样，那么它显然在您的代码中未定义，并且生成的查询字符串将包含dbo.[CAGD Plan]，而不是该变量中应该包含的任何值。如果$您的表名中字面上是 ，则它应该是CAGD\$Leave，因此$Leave不会将其视为变量。
