来自 mysqli_Query 的回显结果-php问题

Echo results from mysqli_Query(来自 mysqli_Query 的回显结果)

本文介绍了来自 mysqli_Query 的回显结果的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在制作一个供我自己使用的个人脚本，我需要知道如何回显来自 mysqli_query 的结果.我的代码如下:

I'm making a personal script for my own use, and I need to know how to echo the results from a mysqli_query. My code is as follows:

$conn = mysqli_connect($servername, $username, $password, $dbname);

if(isset($_POST['commercial'])){
if (isset($_POST['0'])){
    $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 1 AND sent='a'";
    $resultsd1 = mysqli_query($conn, $sql);
    echo $resultsd1;
}   
if (isset ($_POST['1'])){
    $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 2 AND sent='a'";
    $resultsd2 = mysqli_query($conn, $sql);
    echo $resultsd2;
}   
if (isset($_POST['2'])){
    $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 3 AND sent='a'";
    $resultsd3 = mysqli_query($conn, $sql);
    echo $resultsd3;
}
if (isset ($_POST['3'])){
    $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 4 AND sent='a'";
    $resultsd4 = mysqli_query($conn, $sql);
    echo $resultsd4;
}
if (isset ($_POST['4'])){
    $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 5 AND sent='a'";
    $resultsd5 = mysqli_query($conn, $sql);
    echo $resultsd5;
}

}
?>

推荐答案

如果要输出多行

if (isset($_POST['0'])) {
 $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 1 AND sent='a'";
 $resultsd1 = mysqli_query($conn, $sql);

 while ($row = mysqli_fetch_assoc($resultsd1))
 {
    echo $row['email'];
 }
}

如果只有 1 行

if (isset($_POST['0'])){
 $sql = "SELECT email FROM CommercialEmails WHERE articleid = '$_POST[article]' AND dripid = 1 AND sent='a' LIMIT 1";
 $resultsd1 = mysqli_query($conn, $sql);

 $row = mysqli_fetch_assoc($resultsd1);

 echo $row['email'];
}

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