【发布时间】:2015-05-04 18:25:17
【问题描述】:
我遇到的问题是我需要将用户名文本框的值发送到 php 变量,以便可以将其添加到 SQL 服务器。
我只是想出如何为价值转移编写代码。
我尝试了多种方法,我发现 $username = $_POST['username'];但这似乎不起作用。任何想法将不胜感激。
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>New Employee</title>
<style type="text/css">
body {
background-image: url();
}
</style>
</head>
<body>
<header><img src="Border.jpg" width="100%" height="55" alt=""/></header>
<p><img src="MLHC.png" width="620" height="205" alt=""/>
</p>
<p>Create New Employee</p>
<p>Enter Employee Name
<input type="text"name="name" size="35">
</p>
<p>Select Company:
<select name="company" id="Company">
<option value="MLHC">MLHC</option>
<option value="Placer Title Company">Placer Title Company</option>
<option value="RMD">RMD</option>
<option value="IRM">IRM</option>
<option value="TTLAX">TTLAX</option>
<option value="TTYL">TTYL</option>
<option value="BRB">BRB</option>
<option value="G2G">G2G</option>
<option value="IKR">IKR</option>
</select>
<br>
<br>
Select Supervisor:
<select name="supervisor" id="Supervisor">
<option value="John">John</option>
<option value="Kyle">Kyle</option>
<option value="Luke">Luke</option>
<option value="Paul">Paul</option>
<option value="Carrot man">Carrot man</option>
<option value="Employee #833">Employee #833</option>
<option value="HK-47">HK-47</option>
<option value="Whoooooo">Whoooooo</option>
<option value="Filler Name">Filler Name</option>
</select>
<br>
<br>
Enter Username <input type="text" id="username" name="username" size="35">
<br>
<br>
Enter Password <input type="text"name="password" size="35">
<br>
<br>
Enter Phone Number <input type="text"name="phone" size="15">
<br>
<br>
<label>
<input type="checkbox" name="admin" value="checkbox" id="admin">
Administration Status</label>
<br>
<label>
<input type="checkbox" name="exempt" value="checkbox" id="CheckboxGroup1_1">
Exempt Status</label>
<br>
<p> </p>
<p> <input type="button" name="Submit" id="button" value="Submit" onClick="document.write('<?php test() ?>');">
<input type="button" name="Cancel" id="button" value="Cancel" onClick="location.href='Admin.php'"></p>
<?php
function test(){
echo $username;
}
function submit(){
$username = $_POST['username'];
$password = $_POST['password'];
$name = $_POST['name'];
$company = $_POST['company'];
$supervisor = $_POST['supervisor'];
$phone = $_POST['phone'];
$admin = $_POST['admin'];
$exempt = $_POST['exempt'];
$servername = "localhost";
$username = "abc";
$password = "123";
$dbname = "abc";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to create table
$sql = "INSERT INTO users (username, password, name, company, supervisor, phone, admin, exempt)
VALUES($username, $password, $name, $company, $supervisor, $phone, $admin, $exempt)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
</body>
</html>
【问题讨论】:
-
欢迎来到 SO.SE。您应该将代码缩短到演示问题所需的最低限度。见How to Ask。
-
您似乎没有使用
<form>标签。您还编写了函数,但从不调用它们。所以代码永远不会被执行。 -
如果您没有使用任何表单标签,并且您的 html 和 php 代码在同一页面中,请查看此内容是否对您有帮助:stackoverflow.com/questions/4446727/…
-
我并不是说你不应该,而是在深入了解数据库之前先弄清楚 PHP 的一些基本语法和工作原理。
标签: javascript php html