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PHP-访问MS SQL视图和回显结果_Php_Sql_Sql Server_View - Fatal编程技术网

PHP-访问MS SQL视图和回显结果

PHP-访问MS SQL视图和回显结果,php,sql,sql-server,view,Php,Sql,Sql Server,View,我正在尝试使用PHP从MS SQL数据库访问和读取视图。现在我只是想在屏幕上显示结果。当我测试时,我的页面根本没有显示任何内容。 这就是我所拥有的: <?php $myServer = "localhost"; $myUser = "username1"; $myPass = "password1"; $myDB = "database1"; //connect to database $dbhandle = mssql_connect($myServer, $myUser, $myPa

我正在尝试使用PHP从MS SQL数据库访问和读取视图。现在我只是想在屏幕上显示结果。当我测试时,我的页面根本没有显示任何内容。 这就是我所拥有的:

<?php
$myServer = "localhost";
$myUser = "username1";
$myPass = "password1";
$myDB = "database1";

//connect to database
$dbhandle = mssql_connect($myServer, $myUser, $myPass)
  or die("Couldn't connect to SQL Server on $myServer");

//select database
$selected = mssql_select_db($myDB, $dbhandle)
  or die("Couldn't open database $myDB");

//declare statement
$query = "SELECT ProductId";
$query .= "FROM Inventory ";
$query .= "WHERE UPC='15813658428' and ManufacturerID=465";

//execute the query
$result = mssql_query($query);

$numRows = mssql_num_rows($result);
echo "<h1>" . $numRows . " Row" . ($numRows == 1 ? "" : "s") . " Returned </h1>";

//display results
while($row = mssql_fetch_array($result))
{
  echo "<li>" . $row["id"] . $row["name"] . $row["year"] . "</li>";
}
//close
mssql_close($dbhandle);
?> 

通常,当您从PHP获得完全空白的输出时,这是因为服务器在解析或解释代码时遇到问题。在没有其他错误的情况下,处理此问题的最佳方法是去掉(或注释掉)大块代码,并使用数字echo语句验证未去掉的代码是否正在执行。然后慢慢地将越来越多的代码添加回

在这种特定情况下,我猜问题可能会归结到php文件中
?>
后面的空格。这会让服务器感到困惑。尝试完全删除
?>

如果仍然没有输出,则注释掉
$myDB=“database1”之后的所有内容,在注释之前添加一条echo语句,然后慢慢地将代码移回未注释部分,直到找到导致问题的行