需要在表“ doctors”中插入（id是autoincrement）。不知道查询出了什么问题。我收到错误消息-“网页不可用”。 帮助解决它

Question

<?php

$host = 'localhost';
$username = 'uai';
$password = '';
$database = 'Pharma';
$table = 'doctors';
mysql_connect( $host, $username, $password )
or  die(" could not connect to database..." .mysql_error ());
mysql_select_db($database)
or die ("Could not connect to database..." .mysql_error ());
$firstname =$_POST['firstname'];
$lastname =$_POST['lastname'];
$speciality =$_POST['speciality'];
$hospital=$_POST['hospital'];
$city=$_POST['city'];
$state=$_POST['state'];
$country=$_POST['country'];
$sql = "INSERT INTO $table (firstname,lastname,speciality,hospital,city,state,country)VALUES ('$firstname','$lastname','$speciality','$hospital','$city','$state','$country')";
$result =mysql_query($sql);
if ($result) {
echo "New record created successfully";
}
else {
echo "Insertion Error: ";
}
?>
<?php
mysql_close();
?>

<form action="doctors.php" method = "post">

<table align="center">
<tr>
<td>First Name:</td>
<td><input type="text" name="firstname" /></td>
</tr>
<tr>
<td>Last Name:</td>
<td><input type="text" name="lastname" /></td>
</tr>
<tr>
<td>Speciality:</td>
<td><input type="text" name="speciality" /></td>
</tr>
<tr>
<td>Hospital:</td>
<td><input type="text" name="hospital" /></td>
</tr>
<tr>
<td>City:</td>
<td><input type="text" name="city" /></td>
</tr>
<tr>
<td>State:</td>
<td><input type="text" name="state" /></td>
<tr>
<td>Country:</td>
<td><input type="text" name="country" /></td>
</tr>
</table>
<table align="center">
<tr>
<td><center> <input type="submit" name="submit" >
<input type="reset" name="clear">
</td>
</form>

Answer 1

$country=$_POST[country'];缺少' $country=$_POST[country'];

更改此$country=$_POST[country']; 到$country=$_POST['country'];

$country=$_POST['country'];
$sql = "INSERT INTO $table (firstname,lastname,speciality,hospital,city,state,country)VALUES ('$firstname','$lastname','$speciality','$hospital','$city','$state','$country')";
$result =mysql_query($sql);
if ($result) {

更新的代码：

改变这个

mysql_connect( $host, $username, $password ) or  die(" could not connect to database..." .mysql_error ());
mysql_select_db($database) or die ("Could not connect to database..." .mysql_error ());

至

$con = mysql_connect( $host, $username, $password ) or  die(" could not connect to database..." .mysql_error ());
$db = mysql_select_db($database,$con) or die ("Could not connect to database..." .mysql_error ());

句法

mysql_select_db（connection，dbname）;

connection =>   Required. Specifies the MySQL connection to use
dbname =>       Required. Specifies the default database to be used

因此，连接和数据库名称是必填字段。 将连接值传递给mysql_select_db。

[ 注意 ： mysql functions已弃用。 使用PDO或mysqli.* functions ]

<?php

$host = 'localhost';
$username = 'uai';
$password = '';
$database = 'Pharma';
$table = 'doctors';

$con = mysql_connect( $host, $username, $password ) or  die(" could not connect to database..." .mysql_error ());
$db = mysql_select_db($database,$con) or die ("Could not connect to database..." .mysql_error ());

$firstname =$_POST['firstname'];
$lastname =$_POST['lastname'];
$speciality =$_POST['speciality'];
$hospital=$_POST['hospital'];
$city=$_POST['city'];
$state=$_POST['state'];
$country=$_POST['country'];
$sql = "INSERT INTO $table (firstname,lastname,speciality,hospital,city,state,country)VALUES ('$firstname','$lastname','$speciality','$hospital','$city','$state','$country')";
$result =mysql_query($sql);
if ($result) {
echo "New record created successfully";
}
else {
echo "Insertion Error: ";
}
?>
<?php
mysql_close();
?>

---

使用mysqli。*函数

自2012年末PHP 5.5.0起，MySQL扩展正式被弃用，将来将被删除。 从PHP 5和更高版本开始的替代版本是MySQLi（“ i”代表“改进”）和PDO（PHP数据对象）。

<?php
$host = 'localhost';
$username = 'uai';
$password = '';
$database = 'Pharma';
$table = 'doctors';

$con = mysqli_connect($host, $username, $password,$database) or die(" could not connect to database..." .mysqli_connect_error());

$firstname =$_POST['firstname'];
$lastname =$_POST['lastname'];
$speciality =$_POST['speciality'];
$hospital=$_POST['hospital'];
$city=$_POST['city'];
$state=$_POST['state'];
$country=$_POST['country'];

$result = mysqli_query($con,"INSERT INTO $table (firstname,lastname,speciality,hospital,city,state,country)VALUES ('$firstname','$lastname','$speciality','$hospital','$city','$state','$country')");

if ($result) {
    echo "New record created successfully";
}
else {
    echo "Insertion Error: ";
}

mysqli_close($con);
?>