我写了一个就是在html中写连接数据库并且创建一个数据库和表,可是我始终不知道怎样将$table作为变量让他成为表明实验了好多总写法
<?php
$file="date/conn.php";
if(!is_writable($file)){
echo"<font color=red>不能够进行读写</font>";
}
else{
echo "<font color=green>能够进行读写</font>";
if(isset($_POST['tj'])){
$conn.="<?php";
$conn.="\n";
$conn.='$host="'.$_POST['sever'].'";';
$conn.="\n";
$conn.='$user="'.$_POST['user'].'";';
$conn.="\n";
$conn.='$pass="'.$_POST['pass'].'";';
$conn.="\n";
$conn.='$db="'.$_POST['db'].'";';
$conn.="\n";
$conn.='$table="'.$_POST['table'].'";';
$conn.="\n";
$conn.="?>";
$str=fopen("date/conn.php","w");
fwrite($str,$conn);
}
// ==========================================>
require("date/conn.php");
if( mysql_connect($host,$user,$pass)){echo "成功";} else{ echo "失败";}
if( mysql_query("CREATE DATABASE $db")) {echo "成功";} else{ echo "失败";}
mysql_select_db($db);
$sql_query='CREATE TABLE `$table` (' //问题所在处怎样让变脸$table让数据库识别,
//想了好几种方法"'.$table.'" ,'".$table."'都无法成为让mysql识别这个变量从而见表
. ' `id` INT NULL AUTO_INCREMENT PRIMARY KEY, '
. ' `companyName` VARCHAR(255) NULL, '
. ' `name` VARCHAR(255) NOT NULL, '
. ' `email` VARCHAR(255) NULL, '
. ' `tel` VARCHAR(15) NOT NULL, '
. ' `fax` VARCHAR(15) NULL, '
. ' `add` VARCHAR(255) NOT NULL, '
. ' `business` VARCHAR(25) NOT NULL, '
. ' `time` VARCHAR(255) NULL, '
. ' `requirement` VARCHAR(255) NOT NULL, '
. ' `when` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP'
. ' )'
. ' TYPE = innodb;';
if( mysql_query($sql_query)){echo "成功";} else{ echo "失败";}
}
?>
<div id="head">
<div id="body">
<fieldset><legend>欢迎进入Mysql连接</legend>
<form method="post" action="for.php" name="myform">
<div id="sever"> 服务器名:<input type="text" name="sever" /></div>
<div id="user"> 输入账号:<input type="text" name="user" /> </div>
<div id="pass"> 输入密码:<input type="text" name="pass" /> </div>
<div id="db"> 数据库名:<input type="text" name="db" /> </div>
<div id="button"> <button name="tj" type="submit" size="30" >连接 </button> </div>
</form>
</fieldset>
</div>
</div>
