我正在尝试将我拥有的数据保存到mysql数据库中,但我很难做到这一点。起初,我在练习数据中取得了成功。问题是,当我向查询添加其他所需信息时,我的查询不再有效。
代码是这样的。假设所有变量都有正确的对应值。
<body>
<?php
$username = "formtemplate";
$password = "admin123";
$hostname = "localhost";
$db = "practice";
$dbhandle = mysqli_connect($hostname, $username, $password, $db);
$sql = "INSERT INTO transactions (id_num, emp_name, branch, work_sched, dayoff, transaction, date_from, date_to, time_from, time_to, total, replacement, reason, pending, approve, decline, consume, paid) VALUES ( '$employeeNum', '$name', '$branch', '$workSched', '$dayoff', '$temptable', '$tempdatefrom', '$tempdateto', '$tempTimeFrom', '$tempTimeTo', '$tempTotal', '$tempReplacement', '$tempReason', 'yes', 'no', 'no', 'no', 'no')";
mysqli_query($dbhandle, $sql)
or die ( mysql_error() );
echo "success";
?>
</body>
这是数据库信息
trans_num int(11)
id_num varchar(10)
emp_name varchar(30)
branch varchar(20)
work_sched varchar(25)
dayoff varchar(15)
transaction varchar(10)
date_from date
date_to date
time_from varchar(25)
time_to varchar(25)
总varchar(25)
替换varchar(30)
原因varchar(100)
待定varchar(5)
批准的varchar(5)
拒绝varchar(5)
消费varchar(5)
付费varchar(5)
答案 0 :(得分:0)
<?php
mysql_connect("username","password","") or die("databse connected");
mysql_select_db("dbname") or die("databse not found");
$username=$_post['username'];
$password=$_post['password'];
mysqli`enter code here`_query("INSERT INTO `tablename` (`username`,`password`) VALUES ('$username','$password')");
echo "successfully insert record";
?>