Question

<?php
if (isset($_POST['submitted'])){
include('connect.php');
$name=$_POST['name'];
$addr=$_POST['addr'];
$phno=$_POST['phno'];
$tr=$_POST['tr'];
$trd=$_POST['trd'];
$sname=$_POST['sname'];
$sphno=$_POST['sphno'];
$sfield=$_POST['sfield'];
$med=$_POST['med'];
$apaid=$_POST['apaid'];
$apend=$_POST['apend'];

$sqlinsert="INSERT INTO patient (name , addr , phno , tname , tdate , sName , sphno , sSpeciality , med , amtpaid , amtrem) VALUES 
('$name','$addr','$phno','$tr','$trd','$sname','$sphno','$sfield','$med','$apaid','$apend')";

if(!mysql_query($dbcon , $sqlinsert))
{
        die('error inserting new record');
}
else
echo "new";
}
?>

<html>
<head>
    <title>Insert Data</title>
</head>
<body>
<h1>Insert Data</h1>
<form method="post" action="insert.php">
<input type="hidden" name="submitted" value="true" />
<fieldset>
<legend>New Patient</legend>
<label>Name : <input type="text" name="name" /></label>
<label>Address : <input type="text" name="addr" /></label>
<label>Phone no : <input type="text" name="phno" /></label>
<label>Treatment : <input type="text" name="tr" /></label>
<label>Treatment date : <input type="text" name="trd" /></label>
<label>Specialist Recommended : <input type="text" name="sname" /></label>
<label>Specialist phno : <input type="text" name="sphno" /></label>
<label>Specialist field : <input type="text" name="sfield" /></label>
<label>Medicine : <input type="text" name="med" /></label>
<label>Amount Paid : <input type="text" name="apaid" /></label>
<label>Amount Pending : <input type="text" name="apend" /></label>
</fieldset>
<br />
<input type="submit"value="add new record" />
</form>
</body>
</html>

即使表中有字符串变量

，插入时也会出现此错误

警告：mysql_query（）要求参数1为字符串，对象在第18行的C：\ xampp \ htdocs \ insert.php中给出插入新记录时出错

请帮忙......

Answer 1

阅读docs：

mixed mysql_query ( string $query [, resource $link_identifier = NULL ] )

侧点：

mysql_*个功能不再支持，它们为officially deprecated，不再维护，将为removed个未来。您应该使用PDO或MySQLi更新代码，以确保将来的项目功能。

Answer 2

查看本手册。 http://php.net/manual/en/function.mysql-query.php 参数1是查询，2是conn

Answer 3

您将mysql连接对象作为参数传递给mysql_query()函数。 mysql_query()只需要1个字符串参数。使用mysqli_query() 你必须改变你的代码..

<?php
if (isset($_POST['submitted'])){
include('connect.php');
$name=$_POST['name'];
$addr=$_POST['addr'];
$phno=$_POST['phno'];
$tr=$_POST['tr'];
$trd=$_POST['trd'];
$sname=$_POST['sname'];
$sphno=$_POST['sphno'];
$sfield=$_POST['sfield'];
$med=$_POST['med'];
$apaid=$_POST['apaid'];
$apend=$_POST['apend'];

$sqlinsert="INSERT INTO patient (name , addr , phno , tname , tdate , sName , sphno , sSpeciality , med , amtpaid , amtrem) VALUES 
('$name','$addr','$phno','$tr','$trd','$sname','$sphno','$sfield','$med','$apaid','$apend')";

if(!mysqli_query($dbcon , $sqlinsert))
{
        die('error inserting new record');
}
else
echo "new";
}

＆GT;

在php中插入mysql时出错

3 个答案: