duancai7002
duancai7002
2015-10-14 12:46

PHP + SQL数据库Azure错误

已采纳

I am a beginner of web developing and have a question about the PHP + SQL Database connection and displaying the result.

<?php

    ini_set('display_errors',1);
    ini_set('display_startup_errors',1);
    error_reporting(-1);

    function OpenConnection()
    {
        try
        {
            $serverName = "tcp:***,1433";
            $connectionOptions = array("Database"=>"flan",
                "Uid"=>"***", "PWD"=>"***");
            $conn = sqlsrv_connect($serverName, $connectionOptions);
            if($conn == false)
                die(FormatErrors(sqlsrv_errors()));
        }
        catch(Exception $e)
        {
            echo("Error!");
        }
    }

    function ReadData()
    {
        try
        {
            $conn = OpenConnection();
            $tsql = "SELECT * FROM tour_id";
            $getProducts = sqlsrv_query($conn, $tsql);
            if ($getProducts == FALSE)
                die(FormatErrors(sqlsrv_errors()));
            $productCount = 0;
            while($row = sqlsrv_fetch_array($getProducts, SQLSRV_FETCH_ASSOC))
            {
                echo($row['tour_title']);
                echo("<br/>");
                $productCount++;
            }
            sqlsrv_free_stmt($getProducts);
            sqlsrv_close($conn);
        }
        catch(Exception $e)
        {
            echo("Error!");
        }
    }

    echo ReadData();
?>

Result :

Warning: sqlsrv_query() expects parameter 1 to be resource, null given in D:\home\site\wwwroot\test.php on line 29 Fatal error: Call to undefined function FormatErrors() in D:\home\site\wwwroot\test.php on line 31

enter image description here

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2条回答

  • duanke1286 duanke1286 6年前

    Your Openconnection() function is not returning anything, so $conn will always be null.

    Add a return line in your function like so to return the connection:

    function OpenConnection()
    {
        try
        {
            $serverName = "***";
            $connectionOptions = array("Database"=>"***", "Uid"=>"***", "PWD"=>"***");
            $conn = sqlsrv_connect($serverName, $connectionOptions);
            if($conn == false)
                die(FormatErrors(sqlsrv_errors()));
    
            return $conn; // <--- Here
        }
        catch(Exception $e)
        {
            echo("Error!");
        }
    }
    
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  • duanpendan8067 duanpendan8067 6年前

    You are not returning a connection from your OpenConnection function.

    //...
    $conn = sqlsrv_connect($serverName, $connectionOptions);
    if($conn == false)
        die(FormatErrors(sqlsrv_errors()));
    return $conn;
    //...
    

    Furthermore: you should't post your credentials online.

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