Php JSON输出包含的列数超出了其应包含的列数

我正在用PHP从MSSQL数据库检索数据，我想将其格式化为JSON，因为这是我开发图表的唯一方法（借助于am charts）

这是我的代码：

<?php
    header('Content-Type: applicaton/json');
    $serverName = "server\sqlexpress";
    $connectionInfo = array( "Database"=>"myDB", "UID"=>"myUID", "PWD"=>"myPWD");

    $conn = sqlsrv_connect( $serverName, $connectionInfo);
    if( $conn==true ) {
 //echo "Connection established.<br />";
    }
    else
    {
    echo "Connection could not be established.<br />";
    die( print_r( sqlsrv_errors(), true));}

//QUERY

$sql = "SELECT FIRSTNAME, LASTNAME, JOBTYPE from directory ORDER BY JOBTYPE";

$stmt = sqlsrv_query( $conn, $sql);//, $params);

if( $stmt) {
    // echo "Statement executed.<br>\n";
}
else
{
    echo "error in statement execution.\n";
    die (print_r (sqlsrv_errors(), true));
}

//PRINT ROWS
$data = array();
while ($row=sqlsrv_fetch_array($stmt, SQLSRV_FETCH_BOTH))
{
$data[] = $row;
//echo $row['LASTNAME']. ' | ' .$row['FIRSTNAME']. ' | ' .$row['JOBTYPE']. "<br>" ;
}
echo json_encode($data);
?>

但它是这样输出的：

[{"LASTNAME": "MEINE",
   "FIRSTNAME": "KLAUS",
   "JOBTYPE": "LEADVOICE"
},
{"LASTNAME": "SCHENKER",
   "FIRSTNAME": "RUDOLF",
   "JOBTYPE": "RYTHMGUITAR"
}]

[{"0":"KLAUS",
"FIRSTNAME":"KLAUS",
"1":"MEINE",
"LASTNAME":"MEINE",
"2":" LEADVOICE",
"JOBTYPE":" LEADVOICE"
},
{"0":"RUDOLF",
"FIRSTNAME":"RUDOLF",
"1":"SCHENKER",
"LASTNAME":"SCHENKER",
"2":" RYTHMGUITAR",
"JOBTYPE":" RYTHMGUITAR"
}]

发生了什么以及为什么

---

提前谢谢

只需在第行下方替换即可

$data[] = $row;

与

---

非常感谢你，真是妙极了！所以，如果我检索另一个带有其他字段的表，我必须像您的示例那样执行，对吗？是的，当您希望存储数据库表中的一些值时，请遵循相同的操作。

$row_data['LASTNAME'] = $row['LASTNAME'];
$row_data['FIRSTNAME] = $row['FIRSTNAME'];
$row_data['JOBTYPE'] = $row['JOBTYPE'];

array_push($data,$row_data);