douxiangbiao1899 2014-06-29 14:04
浏览 65
已采纳

Foreach和while循环打印相同的值

I'm trying to get the info from each user, this is what I am trying:

$db_conexion = mysqli_connect("localhost", "foo", "bar", "users");
if(mysqli_connect_errno()){
    printf("Ocurrio un error en la conexion de la base de datos: %s
", mysqli_connect_error()); exit();
}

// I got $evUsers from previous MySQL query
// It looks like: 1,2,4,6,10,25;
// Each number = userID
$cleanUsers = preg_replace('/\;$/', '', $evUsers);
$members = explode(',', $cleanUsers);

foreach($members as $member){
    $result = mysqli_query($db_conexion, "SELECT * FROM users WHERE userID=$member");
    while($row = mysqli_fetch_array($result)){
        $memberID = $row['userID'];
        $memberName = $row['userName'];
        $memberAddress = $row['userAddress'];

        echo "ID: ".$memberID." - Name: ".$memberName." - Address: ".$memberAddress.PHP_EOL;
    }
}

mysqli_free_result($result);
mysqli_close($db_conexion);

This code prints the info from the first user over and over again.

What am i doing wrong?

  • 写回答

1条回答 默认 最新

  • douzhan8652 2014-06-29 14:10
    关注

    Try this out

    $members = preg_replace('/\;$/', '', $evUsers);
    //$members = explode(',', $cleanUsers); <--- take this off
    
    //foreach($members as $member){ <--- take this off as well
        $result = mysqli_query("SELECT * FROM users WHERE userID IN ($members)");
        while($row = mysqli_fetch_array($result)){
            $memberID = $row['userID'];
            $memberName = $row['userName'];
            $memberAddress = $row['userAddress'];
    
            echo "ID: ".$memberID." - Name: ".$memberName." - Address: ".$memberAddress.PHP_EOL;
        }
    //}
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥50 silvaco GaN HEMT有栅极场板的击穿电压仿真问题
  • ¥15 谁会P4语言啊,我想请教一下
  • ¥20 win11无法启动 持续蓝屏且系统还原失败,无法开启系统保护
  • ¥15 哪个tomcat中startup一直一闪而过 找不出问题
  • ¥15 这个怎么改成直流激励源给加热电阻提供5a电流呀
  • ¥50 求解vmware的网络模式问题 别拿AI回答
  • ¥24 EFS加密后,在同一台电脑解密出错,证书界面找不到对应指纹的证书,未备份证书,求在原电脑解密的方法,可行即采纳
  • ¥15 springboot 3.0 实现Security 6.x版本集成
  • ¥15 PHP-8.1 镜像无法用dockerfile里的CMD命令启动 只能进入容器启动,如何解决?(操作系统-ubuntu)
  • ¥30 请帮我解决一下下面六个代码