PHP 获取输入、单选、选择数据并插入到 MySQL 表中
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PHP get input , radio , selection data and insert into MySQL table
提问by ETAN
i'm new to php , i have been searching for a tutorial regarding inserting form's input(text) , radio and selection data to MySQL database's table using php. i found some tutorials but most are confusing. So i decided to ask.
我是 php 新手，我一直在寻找有关使用 php 将表单的输入（文本）、收音机和选择数据插入 MySQL 数据库表的教程。我找到了一些教程，但大多数都令人困惑。所以我决定问问。
Okay here's what i want to do. I have a form which have two types of input and a selection 1. input type text 2. input type radio 3. selection
好的，这就是我想要做的。我有一个表单，它有两种类型的输入和一个选择 1. 输入类型文本 2. 输入类型单选 3. 选择
Here's the HTML code :
这是 HTML 代码：
<form action="" method="post" enctype="multipart/form-data">
<strong>Your Name: </strong><br>
<input type="text" name="myname" value="" />
<br /><br/>
<strong>Which class type you want:</strong><br>
<select name="selection">
<option value="A">A</option>
<option value="B">B</option>
<option value="C">C</option>
</select>
<strong>Do you agree?</strong><br>
<input type="radio" name="agree" value="Yes"> or
<input type="radio" name="agree" value="No">
<input type="submit" name="submit" value="Submit">
</form>
I have set the form action to blank because the php code will be in the same file as the HTML (it's a php file btw)
我已将表单操作设置为空白，因为 php 代码将与 HTML 位于同一文件中（顺便说一句，它是一个 php 文件）
MySQL table : info structure : 1. name 2. class 3. agree
MySQL表：信息结构：1.名称2.类3.同意
I want the php code to insert myname into name, selection's selected data into class, radio selected data into agree
我希望 php 代码将 myname 插入name，选择的选定数据插入class，radio selected data 插入同意
P/S Yes i have added a connect to database php script , i just want to know how to get the form data into mysql.
P/S 是的，我添加了一个连接到数据库的 php 脚本，我只想知道如何将表单数据导入 mysql。
Can someone write a php code example on how can i do this?
有人可以写一个关于如何执行此操作的 php 代码示例吗？
Thanks and have a nice day . I hope i have provided enough information. Thanks again if you help.
感谢，并有一个愉快的一天 。我希望我提供了足够的信息。如果你有帮助，再次感谢。
回答by Shef
1. There is a problem with your radioelement. The name should be the same for both options.
1.你的radio元素有问题。两个选项的名称应该相同。
It should be like this:
应该是这样的：
<input type="radio" name="agree" value="Yes"> or
<input type="radio" name="agree" value="No">
2. You can access everything in the $_POSTarray, since you are using the method postfor the form.
2. 您可以访问$_POST数组中的所有内容，因为您使用的post是表单的方法。
$name = $_POST['myname'];
$selection = $_POST['selection'];
$agree = $_POST['agree'];
3. If you are not using parametrized SQL with a library such as PDO, MySQLi, etc... you mustalways escape the data, which will be used in query using mysql_real_escape_string(), in order to protect against SQL injection.
3. 如果您没有使用带有 PDO、MySQLi 等库的参数化 SQL……您必须始终对将在查询中使用的数据进行转义mysql_real_escape_string()，以防止 SQL 注入。
This would be a sample code, to do the escaping and the query.
这将是一个示例代码，用于进行转义和查询。
// write a function somewhere, to use as a shortcut
// for escaping data which will be used in a query
function sql_escape($str){
return "'".mysql_real_escape_string($str)."'";
}
// build the query
$query = sprintf('INSERT INTO table_name(name, class, agree) VALUES(%s, %s, %s)',
sql_escape($_POST['myname']),
sql_escape($_POST['selection']),
sql_escape($_POST['agree']));
// finally run it
$result = mysql_query($query);
回答by flxa
I've taken it a little further here, there is still plenty more that can be done and many way's to do it, for instance you could extend the $errors array to include a field id and then highlight the HTML form field so the user can see exactly where they went wrong. Considering your form is fairly simple you would not need this. @Shef's code would certainly do the job but I thought you might be interested in some more.
我在这里更进一步，还有很多事情可以做，有很多方法可以做到，例如，您可以扩展 $errors 数组以包含字段 id，然后突出显示 HTML 表单字段，以便用户可以准确地看到他们出错的地方。考虑到您的表单相当简单，您不需要这个。@Shef 的代码肯定可以完成这项工作，但我认为您可能对更多内容感兴趣。
<?php
// check the form has been submitted
if (isset($_POST['submit'])) {
// escape the form fields and assign them to variables
// validate myname to ensure the user entered data
if (isset($_POST['myname']) && $_POST['myname']!='') {
$myname = mysql_real_escape_string($_POST['myname']);
} else {
// create an error variable array to store errors to display
$errors[] = 'Please enter your name';
}
// no need to validate selection here as it alway's has a value
$classtype = mysql_real_escape_string($_POST['selection']);
// validate agree unless you want to add 'checked' to one of the values
if (isset($_POST['agree']) && $_POST['agree']!='') {
$agree = mysql_real_escape_string($_POST['agree']);
} else {
$errors[] = 'Please tell us if you agree?';
}
//if errors found tell the user else write and execute the query
if ($errors) {
$message = '<p class="error">We found a problem:</p><ul>';
foreach($error as $msg){
$message .= '<li>'.$msg.'</li>';
}
$message .= '</ul><p>Please fix the error/s to continue.</p>';
} else {
// write the query
$query = "INSERT INTO table (myname, classtype, agree) VALUES ";
$query .= "('$myname','$classtype','$agree')"
// run the query
mysql_query($query);
$message = '<p class="sucessful">Thanks '.htmlspecialchars($myname).'. Your selection has been saved.</p>';
}
}
// print the message
// show the variables in the form field so they don't need re-input
if ($message!='') { echo $message; }
?>
<form action="" method="post" enctype="multipart/form-data">
<strong>Your Name: </strong><br>
<input type="text" name="myname" value="<?php echo htmlspecialchars($myname) ?>" />
<br /><br/>
<strong>Which class type you want:</strong><br>
<select name="selection">
<option value="A"<?php if ($classtype=='A') { echo ' selected'; } ?>>A</option>
<option value="B"<?php if ($classtype=='B') { echo ' selected'; } ?>>B</option>
<option value="C"<?php if ($classtype=='C') { echo ' selected'; } ?>>C</option>
</select>
<strong>Do you agree?</strong><br>
<input type="radio" name="agree" value="Yes"<?php if ($agree=='Yes') { echo ' checked'; } ?>> or
<input type="radio" name="agree" value="No"<?php if ($agree=='No') { echo ' checked'; } ?>>
<input type="submit" name="submit" value="Submit">
</form>
Also: @sqwk, Don't point people towards w3schools, see this: http://w3fools.com/
另外：@sqwk，不要将人们指向 w3schools，请参阅：http://w3fools.com/
回答by bayburt
check this simple example:
检查这个简单的例子：
<form action="welcome.php" method="post">
Name: <input type="text" name="name" />
Sname: <input type="text" name="sname" />
<input type="submit" />
</form>
after you submit form, you can take name and sname.
提交表单后，您可以使用 name 和 sname。
welcome.php::
欢迎.php::
<?php
$name= $_POST["name"];
$sname= $_POST["sname"]; ?>
now you can use this variables as if you want.
现在您可以随意使用此变量。
回答by sqwk
Check whether there is any data in the $_POST array and get the values from it.
检查 $_POST 数组中是否有任何数据并从中获取值。
Have a look here—the second example down is what you need: http://www.w3schools.com/php/php_mysql_insert.asp
看看这里 - 第二个例子是你需要的：http: //www.w3schools.com/php/php_mysql_insert.asp
(You do have to make the changes that Shef suggested, though.)
（不过，您确实必须进行 Shef 建议的更改。）
Also remember to check your data-integrity, otherwise people could use your insert to run malicious code.
还要记住检查您的数据完整性，否则人们可能会使用您的插入来运行恶意代码。
