sqlsrv_get_field()

I'm needing a little help here. Would anyone know why my sqlsrv_get_field($myData,0) would be complaining with "Warning: sqlsrv_get_field() expects parameter 1 to be resource, boolean given in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\TMCTranscriptionViewer\downloadfile.php on line 43  "

Here's the output for $myData,
$myData = bool(true)
<?php
error_reporting(E_ALL | E_STRICT);
ini_set('display_errors', 1);
date_default_timezone_set('America/Tegucigalpa') ;
$REPORTID = $_GET['REPORTID'];

echo 'ReportID = '.$REPORTID.'<br>';
if(!is_numeric($REPORTID))
    die("Invalid REPORTID specified. REPORTID = " . $REPORTID);

$serverName = "TEXPARRPT";
$connectionInfo = array( "Database"=>"MEDQUIST","UID"=>"xx","PWD"=>"xxxxx");

/* Connect */
$conn2 = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn2 ){
//    echo "<br>Connection ready to go!<br>";
}
else
{
     echo "<br>Could not connect to ".$serverName ."." . $connectionInfo["Database"]."<br><br>";
     die( print_r( sqlsrv_errors(), true));
}

$dbQuery = "SELECT MIMETYPE, CONTENT
            FROM REPORTCONTENTT
            INNER JOIN REPORT_DESCT ON REPORTCONTENTT.REPORTID = REPORT_DESCT.REPORTID
            WHERE REPORTCONTENTT.REPORTID = ".$REPORTID;
$result = sqlsrv_query($conn2, $dbQuery);
// Note: I do get results from sqlsrv_query() above.
if($result){
//    echo '<br>Got results from sqlsrv_query()!<br><br>';
    $myData = sqlsrv_fetch($result);
// Note: I do get results from sqlsrv_fetch() above.
    if($myData){
        echo '<br>sqlsrv_fetch() suceeded! sqlsrv_fetch($result) = '.$myData.'<br>';
//        var_dump($myData);
        }
    else{
        echo '<br>Error with sqlsrv_fetch($result). $myData = <br>';
        var_dump($result);
        }
    $fileType = sqlsrv_get_field($myData,0);
// Warning: sqlsrv_get_field() expects parameter 1 to be resource, boolean given...
    if($fileType){
        echo '<br>sqlsrv_get_field() suceeded!<br>FileType = '.$fileType;
        }
    else{
        echo '<br>sqlsrv_get_field($myData,0) failed. $myData = ';
        var_dump($myData);
        echo '<br><br>';
        print_r( sqlsrv_errors(), true);
        }


    }
else{
    echo '<br>Something happened with sqlsrv_query()<br>';
    die( print_r( sqlsrv_errors(), true));
    }


////    echo $fileContent;
//    }
//else
//    {
//    echo "Record doesn't exist.";
//    }
sqlsrv_free_stmt($result);
sqlsrv_close($conn2);
?>

Open in new window

ajfjrAsked:
Who is Participating?
 
wmadrid1Connect With a Mentor Commented:
Try this:

$f = 0;
$type = sqlsrv_field_metadata($result);
$fileType = ($type[$f]['Type'] == SQLSRV_SQLTYPE_FLOAT) ? sqlsrv_get_field($result, $f) : sqlsrv_get_field($result, $f, SQLSRV_PHPTYPE_STRING(SQLSRV_ENC_CHAR)); 

Open in new window

//$fileType = sqlsrv_get_field($myData,0);

I am supposing $f is the first SELECT field
0
 
Marco GasiFreelancerCommented:
It seems sqlsrv_get_field() expects to find something like a string identifying a MIME file type and it find a boolean value. If this is the case, then I would look at sqlsrv_fetch($result) function which return a boolean value and not a string. Can you post the code for sqlsrv_fetch()?
0
 
ajfjrAuthor Commented:
This was interesting and it works! Could you explain what's going on so a PHP newbie like me can understand what's going on?

Thank You!
0
 
wmadrid1Commented:
Basically your error was the first paramater of function
$fileType = sqlsrv_get_field($myData,0);
the correct line is a valid resource
$fileType = sqlsrv_get_field($result,0);

The rest  is a good validating returning data types
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.