尝试使用PHP连接到我的数据库时出错。如果有人不介意为我看一下,我将不胜感激! (需要一组不同的眼睛哈哈)。
以下是我遇到的错误:
Warning: mssql_query() [function.mssql-query]: Unable to connect to server: (null) in H:\root\home\... on line 24
Warning: mssql_query() [function.mssql-query]: A link to the server could not be established in H:\root\home\... on line 24
Warning: mssql_fetch_array(): supplied argument is not a valid MS SQL-result resource in H:\root\home\... on line 25
Warning: mssql_query() [function.mssql-query]: message: Incorrect syntax near ','. (severity 15) in H:\root\home\... on line 35
Warning: mssql_query() [function.mssql-query]: Query failed in H:\root\home\... on line 35
这是PHP文件:
<?php
$hostname_localhost ="********";
$database_localhost ="*****";
$username_localhost ="******";
$password_localhost ="*****";
$table_name = "VehicleInformation";
//$phone = $_REQUEST['Phone'];
//$rpm = $_REQUEST['RPM'];
//$fuelLevel = $_REQUEST['FuelLevel'];
//$engineRunTime = $_REQUEST['EngineRunTime'];
//$speed = $_REQUEST['Speed'];
//$troubleCode = $_REQUEST['TroubleCodes'];
//$oilPressure = "0";
$phone = "2125559852";
$rpm = "1000";
$fuelLevel = "20";
$engineRunTime = "200";
$speed = "25";
$oilPressure = "420";
$ownedByQuery = "Select o.Id from OwnedBy o JOIN tblUser u ON o.UserId = u.ID where u.Phone = '$phone'";
$sql2 = mssql_query($ownedByQuery);
$row = mssql_fetch_array($sql2);
$query = "INSERT INTO $table_name (OwnedById,RPM,FuelLevel,OilPressure,EngineRunTime,Speed) VALUES ($row[0],$rpm,$fuelLevel,$oilPressure,$engineRunTime,$speed)";
$server = mssql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
die('MSSQL Error: ' . mssql_get_last_message());
mssql_select_db("*******", $server);
//$sql2 = mssql_query($ownedByQuery);
$sql=mssql_query($query);
//$row = mssql_fetch_array($sql2);
//$row = mssql_fetch_array($sql);
//echo $row[0];
?>
答案 0 :(得分:2)
<?php
$myServer = "localhost";
$myUser = "your_name";
$myPass = "your_password";
$myDB = "examples";
//connection to the database
$dbhandle = mssql_connect($myServer, $myUser, $myPass)
or die("Couldn't connect to SQL Server on $myServer");
//select a database to work with
$selected = mssql_select_db($myDB, $dbhandle)
or die("Couldn't open database $myDB");
//declare the SQL statement that will query the database
$query = "Select o.Id from OwnedBy o JOIN tblUser u ON o.UserId = u.ID where u.Phone = '$phone'";
//execute the SQL query and return records
$result = mssql_query($query);
$numRows = mssql_num_rows($result);
echo "<h1>" . $numRows . " Row" . ($numRows == 1 ? "" : "s") . " Returned </h1>";
//display the results
while($row = mssql_fetch_array($result))
{
// do something
}
//close the connection
mssql_close($dbhandle);
?>
尝试使用此代码..
答案 1 :(得分:1)
此错误似乎数据库连接失败。因此,在获取sql quires后首先连接数据库。您可以在下面看到示例编码。
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
}
else{
echo "0 results";
}
mysqli_close($conn);