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();
<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";
}
$mysqli->close();
解决方案 »
- failed to open stream: Permission denied 这个错误怎么解决?
- php获取坐标值
- 不用表单怎样取得<input name="textfield" type="text" value="123456789">的value字符串
- html中textarea没有value值的问题
- 求PHP 函数引用返回例子
- 要建一个社团(以后可能是公司)的论坛,是使用php技术的论坛社区,要求实用,不用太花哨,稳定,无版权问题,请推荐一个,谢谢
- 我在一网页向另一网页POST数据name,为什么输出$name时却得不到任何值?
- bbs上发的大家图片怎么管理的?
- require() 和 require_once()
- 两个问题~~
- 插入的记录为什么不能获取最新一条记
- 同样的删除SQL语句,在phpmyadmin中可以删除数据,为何在网站中就不能删除呢?(但是那条删除的SQL语句是执行的)
<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
welcomewangkem1997-04-03china
以上是链接数据库之前的echo 页面就输出这些了
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