一直显示failed 求大神指点
<html>
<head>
        <title>php is running</title>
</head>
<body>
        <h1>step</h1>
        <?php
                $name = $_POST['name'];
                $sex = $_POST['sex'];
                $birth = $_POST['birth'];
                $birthaddr = $_POST['birthaddr'];
                echo "welcome",$name,$sex,$birth,$birthaddr  ;
                $mysqli = new mysqli("localhost","root@localhost","gdragon","first_log");
                $mysqli->query("set names 'utf8'");
                $sql = "INSERT INTO mytable(name,sex,birth,birthaddr)
                VALUES('$name','$sex','$birth','$birthaddr')";
                if($mysqli->query($sql) == TRUE){
                        echo "success";
}else{ echo "failed";
}
$mysqli->close();

解决方案 »

  1.   

    }else{ echo "failed";  echo $mysqli->error;
      

  2.   

    是把最后一行改了吗
    <html>
    <head>
            <title>php is running</title>
    </head>
    <body>
            <h1>step</h1>
            <?php
                    $name = $_POST['name'];
                    $sex = $_POST['sex'];
                    $birth = $_POST['birth'];
                    $birthaddr = $_POST['birthaddr'];
                    echo "welcome",$name,$sex,$birth,$birthaddr  ;
                    $con = mysql_connect("localhost","root@localhost","gdragon");
                    mysql_query("set names 'utf8'");
                    mysql_select_db("first_log");
                    $sql = "INSERT INTO mytable(name,sex,birth,birthaddr)VALUES('$name','$sex','$birth','$birthaddr')";
                    if(!mysql_query($sql,$con)){
                    echo"failed";   die('error'.mysql_error());
    }
    echo " add a number";
    mysql_close($con)
    这是我最新的 还是没添加到数据库
      

  3.   

    step
    welcomewangkem1997-04-03china
    以上是链接数据库之前的echo  页面就输出这些了
      

  4.   

    add a number 没有输出?
      

  5.   

    没有
    而且很奇怪 我在第一次能输出failed的代码里 在   $mysqli = new mysqli("127.0.0.1","root@localhost","gdragon","first_log");
    之后echo $mysqli的值  这个值不输出 failed也不输出