dsnw2651 2018-07-03 13:19
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如何从PHP中的mssql DB获取价值

Firstly I got the workers name from BIRTHDAYS and then want to get e-mail address from USERS.There is no problem to take workers name's from Table1 but when I try to get the e-mail addresses the db returns me NULL.My DB is 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);
        }   
    }
}   
?>
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1条回答 默认 最新

  • duandushang5148 2018-07-03 13:46
    关注
     while($rows1 = sqlsrv_fetch_array($stmt2,SQLSRV_FETCH_ASSOC))
            {   
                $dizi1[] = array('EMAIL' =>$rows['EMAIL']);
                echo $newarray1 = json_encode($dizi1,JSON_UNESCAPED_UNICODE);
            } 
    

    you put in $rows1 and would take it from $rows NULL is correct answer :)

    take $rows1['EMAIL'] and it would work

    and why foreach =?

    you can put the statement in while-loop like this:

      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);
            }
        }
    
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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