PHP和mySQL不工作
PHP和mySQL不工作
提问于 2014-03-25 01:53:06
PHP代码：
<?php
// Create connection
$con=mysqli_connect("localhost","root","root","demo1");
echo "Connection was successful";
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysql_select_db("demo1",$con);
$sqli="INSERT INTO employee (Employee ID,NAME,Date Hired,Position,Salary,Department Code,Can HIRE,BOSSID)
VALUES('$_POST[EMPID]','$_POST[NAME]','$_POST[DATEHIRED]','$_POST[POSITION]','$_POST[SALARY]','$_POST[D EPTCODE]','$_POST[CANHIRE]','$_POST[BOSSID]')";
if (!mysqli_query($con,$sqli))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>HTML代码
<!DOCTYPE html>
<html>
<body>
<h1> EMPLOYEE </h1><br>
<form action="LAB5.php" method="post">
Employee ID: <input type="text" name="EMPID" ><br>
NAME: <input type="text" name="NAME" ><br>
Date Hired <input type="text" name="DATEHIRED" ><br>
Position: <input type="text" name="POSITION" ><br>
Salary: <input type="text" name="SALARY" ><br>
Department Code: <input type="text" name="DEPTCODE" ><br>
Can HIRE <input type="text" name="CANHIRE" ><br>
BOSSID: <input type="text" name="BOSSID" ><br>
<input type="image" src="Submit.gif" alt="Submit" width="100" height="50"><br>
</body>
</html> 错误消息：
连接是successfulError:您的SQL语法有错误；请检查与MySQL服务器版本相对应的手册，以获得在第1行使用“ID、名称、雇用日期、职位、薪资、部门代码、可以雇用、BOSSID)值”的正确语法。
回答 5
Stack Overflow用户
回答已采纳
发布于 2014-03-25 02:00:08
您已经将代码与MySQL和MySQLi混合使用。我只将您的代码写入MySQLi中。请避免在列名中使用(空格)。并且可以使用mysqli_real_escape_string防止某些SQL注入。
LAB5.php:
<?php
/* CHECK CONNECTION */
$connection=mysqli_connect("localhost","root","root","demo1");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$empID=mysqli_real_escape_string($connection,$_POST['EMPID']);
$name=mysqli_real_escape_string($connection,$_POST['NAME']);
$datehired=mysqli_real_escape_string($connection,$_POST['DATEHIRED']);
$position=mysqli_real_escape_string($connection,$_POST['POSITION']);
$salary=mysqli_real_escape_string($connection,$_POST['SALARY']);
$deptcode=mysqli_real_escape_string($connection,$_POST['DEPTCODE']);
$canhire=mysqli_real_escape_string($connection,$_POST['CANHIRE']);
$bossID=mysqli_real_escape_string($connection,$_POST['BOSSID']);
/* MYSQLI REAL ESCAPE STRING WOULD PREVENT A BIT OF SQL INJECTION */
mysqli_query($connection,"INSERT INTO employee (EMPID, NAME, DATEHIRED, POSITION, SALARY, DEPTCODE, CANHIRE, BOSSID) /* DOUBLE CHECK YOUR COLUMN NAME */
VALUES('$empID','$name','$datehired','$position','$salary','$deptcode','$canhire','$bossid')";
mysqli_close($con);
?>您的HTML代码：
<html>
<body>
<h1> EMPLOYEE </h1><br>
<form action="LAB5.php" method="post">
Employee ID: <input type="text" name="EMPID" ><br>
Name: <input type="text" name="NAME" ><br>
Date Hired <input type="date" name="DATEHIRED" ><br>
Position: <input type="text" name="POSITION" ><br>
Salary: <input type="number" name="SALARY" ><br>
Department Code: <input type="text" name="DEPTCODE" ><br>
Can HIRE <input type="text" name="CANHIRE" ><br>
BOSSID: <input type="text" name="BOSSID" ><br>
<input type='submit'>
</form>
</body>
</html> Stack Overflow用户
发布于 2014-03-25 01:58:20
字段不能有像Employee ID这样的空格，请重复检查db表列名。
此外，您的代码容易受到SQL注入的影响。您应该转义要在SQL查询中使用的任何用户输入。尝试将mysql_escape_string($_POST['value'])封装在所有$_POST、$_GET和$_REQUEST输入周围。
Stack Overflow用户
发布于 2014-03-25 01:58:25
使用mysqli为什么是"mysql_select_db"？mysqli将是：
bool mysqli_select_db ( mysqli $link , string $dbname )页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/22624052
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