Question

我有以下基本HTML表单：

<form action="" method="post">

Name: <input type="text" name="name" /><br><br>

College: <select name = "colleges">
    <option> ---Select College---</option>
  <option value="option1">DIT</option>
  <option value="option2">University College Dublin</option>
</select><br><br>

Email: <input type="text" name="email" /><br><br>

Password: <input type="text" name="password" /><br><br>
Location: <input type="text" name="location" /><br><br>

 <button type="submit" name="submit" >Submit</button>

</form>

下面是我很新的PHP部分：

if(isset($_POST["submit"])) {
    $name = $_POST["name"];
    $college_name = $_POST["college"];
    $email = $_POST["email"];
    $password = $_POST["password"];
    $location = $_POST["location"];
}
$sql = "INSERT INTO user(name, college, email, password, location) VALUES ($name, $college_name, $email, $password, $location)";
?>

我的数据库的大学类型为int，因此数据库中的DIT为1。谁能告诉我该怎么做，以便它发送1作为college_name而不是用户看到的实际名称？

Answer 1

更改此

College: <select name = "colleges">
    <option> ---Select College---</option>
  <option value="option1">DIT</option>
  <option value="option2">University College Dublin</option>
</select><br><br>

对此

College: <select name = "colleges">
    <option> ---Select College---</option>
  <option value="1">DIT</option>
  <option value="2">University College Dublin</option>
</select><br><br>

还有这个

if(isset($_POST["submit"])) {
    $name = $_POST["name"];
    $college_name = $_POST["college"];
    $email = $_POST["email"];
    $password = $_POST["password"];
    $location = $_POST["location"];
}

对此

if(isset($_POST["submit"])) {
    $name = $_POST["name"];
    $college_name = (int) $_POST["colleges"];
    $email = $_POST["email"];
    $password = $_POST["password"];
    $location = $_POST["location"];
}

Answer 2

只需使用

<option value='1' > DIT </option>

而不是将option1添加为值

Answer 3

在您的选项中输入数值：

College: <select name="colleges">
    <option value="1">DIT</option>
    <option value="2">University College Dublin</option>
</select>

然后更正您的php以使其与$ _POST数组中的select名称匹配：

$college_name = (int) $_POST["colleges"];

Answer 4

为什么不使用option value？

$_POST["college"]返回option1

我认为正确的方法是将值定义为1：

以下内容：

<option value="1">DIT</option>

将返回：1

您还可以使用If else语句

if ($_POST["college"] == 'option1')
{
$college = 1;
}

if ($_POST["college"] == 'option2')
{
$college = 2;
}

在SQL中将字符串从PHP转换为int

4 个答案: