<?php
$db_host="localhost";
$db_user="root";
$db_pass="";
$db_name="test";
$table_name="student";
$list_num=10;
$link=mysqli_connect($db_host,$db_user,$db_pass,$db_name)or die("不能连接到服务器".mysql_error());
$sql="create table $table_name(
id int(5) not null auto_increment primary key,
name varchar(12) not null,
sex varchar(4) not null,
birthday varchar(16) not null,
s_id varchar(10) not null,
parent varchar(12) not null,
phone varchar(14) not null,
address varchar(30) not null
)";
if (mysqli_query($link,"CREATE TABLE $table_name"))
echo "表".$table_name."创建成功";
else echo "wrong";
?>
为什么会输出wrong,哪里写错了吗?
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<?php$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Perform queries
mysqli_query($con,"SELECT *
FROM Persons");
mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age)
VALUES ('Glenn','Quagmire',33)");
mysqli_close($con);
?>
加上临时变量也不行
追答那数据库里面有这信息吗?sql代码能不能执行成功呀?
追问新手照着书打的,结果新版本不支持就版本代码,就照着手册改啊改,所以根本不知道怎么算成功
追答你直接看看mysql里面有没有你这段代码创建的表呀。mysqli_query只是执行一段sql语句,本身是不具备判断条件的。
本回答被网友采纳if (mysqli_query($link,$sql))追问
依旧不行的- -
追答那是你mysqli的环境没配置好吧?你输出 phpinfo() 看看有没有mysqli选项!~
如果没有那就开启php.ini 中的 extension=php_mysqli.dll 并把 php/ext/php_mysqli.dll 复制到 c:\windows\system32 目录下 重启apache
if (mysqli_query($link,"CREATE TABLE $table_name")) 改成
if (mysqli_query($link,$sql)) 肯定是可以执行的!~