Question

我有一个html文件，它创建一个链接到php文件的表单，并允许对数据库进行多参数搜索。无论出于何种原因，文件都没有读取，我收到错误，因此很难测试我的sql查询并验证它是否有效。提前谢谢。

以下是代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

<html>
<head>

<title>Bicycle Store Employees</title>
<link rel="stylesheet" type="text/css" href="web.css" />

</head>
<body>

<h1>Bicycle Store Manager</h1>

<h2>Customized Business Management for Bicycle Stores</h2>

<h3>EMPLOYEE INFORMATION</h3>

<?php

ini_set('display_errors', 'On');

$mysqli = new mysqli("$database", "$username", "$password", "$username");

if ($mysqli->connect_errno) {

print "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;

}
if( $_POST["submit"] ) {

    if (!($stmt =$mysqli->prepare("SELECT * into tempTable FROM 
(SELECT Fname, Minit, Lname, Address, Hourly, Sname, Saddress, Dno
FROM EMPLOYEE
join on Employee.Dno = DEPARTMENT.Dnumber
join on LOCATION.Store_num = DEPARTMENT.Store_num
where Fname = $_POST['Fname'] OR Lname = $_POST['Lname'] 
OR Stor_num = $_POST['Store_num'] OR Dno = $_POST['Dno']
GROUP BY Store_num);"))) {

print "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
if (!$stmt->bind_param("ssii",$_POST['Fname'], $_POST['Lname'], $_POST['Dno'], $_POST['Store_num'])) {
print "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}
if (!$stmt->execute()) {
print "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
 $stmt->store_result();

if (!$stmt->bind_result($Fname,$Minit,$Lname,$Phone,$Address,$Slocation,$Saddress,$Dno,$Hourly)) {
    print "Binding output parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}
if ($stmt->num_rows == 0){
print "No results were found for the following search <p>"
  .$_POST['Fname'].$_POST['Lname'].$_POST ['Store_num'].$_POST['Dno']."</p>";
    }
else {
print "<table border=2 cellpadding=4>

        <tr bgcolor=white>

        <th>First Name</th>

        <th>Middle Initial</th>

        <th>Last Name</th>

        <th>Phone</th>

        <th>Address</th>

        <th>Store</th>

        <th>Store Location</th>

        <th>Dept #</th>

        <th>Hourly Rate</th>

        </tr>";

    while ($stmt->fetch()){

        print "<tr><td>".$Fname."</td><td>".$Minit."</td><td>".$Lname.

        "</td><td>".$Phone."</td><td>".$Address."</td><td>".$Slocation."</td>

        <td>".$Saddress."</td><td>".$Dno."</td><td>".$Hourly."</td></tr>";

    }

    print "</table>";
}

$stmt->free_result();
}

$mysqli->close();
?>


<p><a href="employee_info.html">SEARCH AGAIN</a></p>

<p><a href="index.html">HOME</a></p>

</body>
</html>

Answer 1

看看。

mysqli_error,mysqli::$error

Answer 2

$mysqli = new mysqli("$database", "$username", "$password", "$username");

首先，您不必将变量括在双引号中。

所以，就像这样：

$mysqli = new mysqli($database, $username, $password, $username);

此外，变量$database，$username，$password和$username的值定义在哪里？您应该在使用之前设置值！

PHP文件错误

2 个答案: