PHP SQLSRV 下拉列表错误

Question

I am trying to populate dropdown list from SQL Query through sqlsrv but it is not working. What am i doing wrong?

<?php

$serverName1 = "kk12334";
$connectionInfo1 = array( "Database"=>"Fruits");
$conn1 = sqlsrv_connect( $serverName1, $connectionInfo1);
$sql1="SELECT [Name] as CName,[BName] as BName from Fruits";
$stmt1 = sqlsrv_query( $conn1, $sql1 );


while ($data=sqlsrv_fetch_array($stmt1, SQLSRV_FETCH_ASSOC))

{


  echo "<option value=";
    echo $data['CName'].", ".$data['BName'].;
    echo "<br />";
        echo $data['CName'].", ".$data['BName'];
    echo "</option>";

}

?>

Answer 1

您不能将 <BR> 放在 <option> 标签内。

试试这样的东西：

while ($data=sqlsrv_fetch_array($stmt1, SQLSRV_FETCH_ASSOC))
{ $opt="$data[CName], $data[BName]";
   echo "<option value='$opt'>$opt</option>\n";
}

并且由于您希望在两个变量值之间有一个空白，因此您需要将所有内容包含在 ' 中以确保将两个部分都分配给 value=.

PHP SQLSRV 下拉列表错误

PHP SQLSRV Dropdown list error

php

sqlsrv

dropdown