Question

我把以下脚本放在一起，它将搜索mysql数据库以查找在表单文本框中提交的某些成员ID，并为找到的每条记录写一个mail（）字符串，但它似乎不起作用，我知道我正在连接，因为如果在“成功连接”后我删除了所有内容，我将获得“成功连接”。如果我按原样运行脚本，我会得到一个空白屏幕。任何有关逻辑的帮助都会得到批准。

op

FORM PAGE HTML

<?php

$servername = "";
$username = "";
$password = "";
$db = "";

// Create connection
$conn = new mysqli($servername, $username, $password, $db);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 
echo "Connected Successfully";

if(isset($_POST['Submit'])) {
    $MemberIDs = $_POST['MemberIDs']; // contains the values "100,107"
    $SearchIDs = array($MemberIDs);
    $SearchStr = implode(',', $SearchIDs); // returns 100,107
    $sql = "SELECT * FROM members where id in ({$SearchStr})";
    $sql->execute();
    $raw_results = $sql->fetchAll();
    $num_rows = count($raw_results);

    if($num_rows > 0 ){
    while($results = $sql->fetch($raw_results)) {
    echo "mail(".$results['name']." <".$results['requestemail'].", $Subject2, $message2, $headers2);";
    }
    else
    {
    echo "No results";
    }
    }
}

$conn->close();
?>

Answer 1

您希望对$ conn对象进行操作，这是您与数据库的mysqli连接。当然我无法测试这个或检查语法，但这应该让你接近。

if(isset($_POST['Submit'])) {
   // Create connection
   $conn = new mysqli($servername, $username, $password, $db);

   // Check connection
   if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
   } 
   echo "Connected Successfully";

    $MemberIDs = $_POST['MemberIDs']; // contains the values "100,107"
    $SearchIDs = array($MemberIDs);
    $SearchStr = implode(',', $SearchIDs); // returns 100,107
    $sql = "SELECT * FROM members where id in ({$SearchStr})";
    $results = $conn->query($sql);
    if( $results->num_rows > 0 ) {
        while($row = $results->fetch_assoc()) {
            echo 'mail('.$row['name'].');';
        }
    }
    else {
        echo "No results";
    }

    $results->free();
    $conn->close();
}

使用文本框中的多个ID来回显来自mysql db的电子邮件

1 个答案: