public function addemail(){
$email_name=$_post['name'];
$email_email=$_post['email'];
$con=mysqli_connect(localhost,root,root,bottlelover)
or die('connect error');
mysqli_set_charset($con,utf8);
$sql=select * from email_list where email_email = $email_email;
$result=mysqli_query($con,$sql)
or die('query1 error');
if($row=mysqli_fetch_array($result)){
$this->assign(res,'邮箱已被使用,用户添加失败');
mysqli_close($con);
$this->display(bottleloveradd);
}
else {
$sql=insert into email_list(email_name,email_email) values ('$email_name','$email_email');
$result=mysqli_query($con,$sql)
or die('query2 error');
$this->assign(res,'新用户添加成功');
mysqli_close($con);
$this->display(bottleloveradd);
}
}
--------------------------------------------以上是代码--------------------
我的本意是添加新用户之前要检查邮箱是不是已经被使用,
$sql=select * from email_list where email_email = $email_email;
$result=mysqli_query($con,$sql)
or die('query1 error');
这段代码提示query1 error错误,其他运行正常;为什么,明显没有错啊,表名没错,列名没错,变量名都没错.....
------解决方案--------------------
$sql=select * from email_list where email_email = '$email_email';
------解决方案--------------------
字符类型的量要用引号括起
正因为太简单了，所以容易忽视
------解决方案--------------------
楼上正解,少了单引号,因为这是字符串