如何从PHP中的mssql DB获取值

时间:2018-07-03 13:19:59

标签: php json sql-server

首先我从BIRTHDAYS得到了工人的名字,然后又想从e-mail address得到了USERS。从Table1取得工人的名字是没有问题的,但是当我尝试时要获取数据库的电子邮件地址,数据库将返回NULL。我的数据库是mssql。

<?php
include_once("connect.php");
$today = '05.07';
$today1 = $today . "%";
$sql = "SELECT NAME FROM BIRTHDAYS WHERE BIRTH LIKE '$today1' ";
$stmt = sqlsrv_query($conn,$sql);
if($stmt == false){
    echo "failed";
}else{
   $dizi = array();
   while($rows = sqlsrv_fetch_array($stmt,SQLSRV_FETCH_ASSOC))
   {
    $dizi[] = array('NAME' =>$rows['NAME']);
    $newarray = json_encode($dizi,JSON_UNESCAPED_UNICODE);
   }
 }

    foreach(json_decode($newarray) as $nameObj) 
{
    $nameArr = (array) $nameObj;
    $names = reset($nameArr);
    mb_convert_case($names, MB_CASE_UPPER, 'UTF-8');
    echo $sql2 = "SELECT EMAIL FROM USERS WHERE NAME = '$names' ";
    echo "<br>";
    $stmt2 = sqlsrv_query($conn,$sql2); 
        if($stmt2 == false)
        {
        echo "failed";
        }
        else
        {
        $dizi2 = array();   
        while($rows1 = sqlsrv_fetch_array($stmt2,SQLSRV_FETCH_ASSOC))
        {   
            $dizi1[] = array('EMAIL' =>$rows['EMAIL']);
            echo $newarray1 = json_encode($dizi1,JSON_UNESCAPED_UNICODE);
        }   
    }
}   
?>

1 个答案:

答案 0 :(得分:1)

 while($rows1 = sqlsrv_fetch_array($stmt2,SQLSRV_FETCH_ASSOC))
        {   
            $dizi1[] = array('EMAIL' =>$rows['EMAIL']);
            echo $newarray1 = json_encode($dizi1,JSON_UNESCAPED_UNICODE);
        } 

您放入$ rows1并从$ rows中取出NULL是正确答案:)

使用$rows1['EMAIL'],它将起作用

以及为什么foreach =?

您可以将语句放入while循环中,如下所示:

  while ($rows = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {

    $names = $rows['NAME'];
    $sql2 = "SELECT EMAIL FROM USERS WHERE NAME = '$names' ";
    echo "<br>";
    $stmt2 = sqlsrv_query($conn, $sql2);
    if ($stmt2 == false) {
        echo "failed";
    } else {
        $dizi2 = array();
        while ($rows1 = sqlsrv_fetch_array($stmt2, SQLSRV_FETCH_ASSOC)) {
            $dizi1[]        = array('EMAIL' => $rows1['EMAIL']);
            echo $newarray1 = json_encode($dizi1, JSON_UNESCAPED_UNICODE);
        }
    }

}