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sqlsrv_get_field()

Posted on 2010-11-22
4
2,135 Views
Last Modified: 2013-12-12
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);

?>

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0
Comment
Question by:ajfjr
  • 2
4 Comments
 
LVL 31

Expert Comment

by:Marco Gasi
ID: 34191731
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
 
LVL 5

Accepted Solution

by:
wmadrid1 earned 500 total points
ID: 34192271
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)); 

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

I am supposing $f is the first SELECT field
0
 

Author Comment

by:ajfjr
ID: 34192438
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
 
LVL 5

Expert Comment

by:wmadrid1
ID: 34192716
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

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