在文件之间传递PHP变量并将变量插入SQL表(Passing PHP variable between files and inserting the variable into a SQL table)

 我有一个PHP文件，通过$ _GET通过URL获取值。 代码是$ thisID = $ _GET ['blog_id']; 
 
 现在在同一个文件中，我正在收集用户的条目并将其传递给带有post的HTML表单，该文件将数据输入到SQL表中。 我还想将blog_id输入数据库。 
 
 这是我正在使用的代码： 
 
<form action="comment.php" method="POST">
Your name: <input type="text" name="display_name" size="25"><br>
Your email: <input type="text" name="email" size="30"><br>
Your comment: <input type="text" name="comment" size="40"><br>
<input type="hidden" name="blog_id"
value="<?php echo $_POST['$thisID']; ?>">
<input type="submit" value="Submit">
 
 这就是我将值输入SQL表的方法 
 
 $command = "insert into $table_name
values('','".$db->real_escape_string($_POST['comment'])."',
'".$db->real_escape_string($_POST['blog_id'])."',        
'".$db->real_escape_string($_POST['display_name'])."',
'".$db->real_escape_string($_POST['email'])."',now()
);";
 
 代码执行，我没有得到任何错误，但没有输入变量（blog_id），数据库中的字段为空。 正确输入所有其他字段。 是否有可能交出这样的PHP变量？ 

I have a PHP file that is getting a value through the URL through $_GET. The code is $thisID = $_GET['blog_id'];
 
Now in the same file, I am collecting entries from a user and passing it through an HTML form with post to a file that enters the data into a SQL table. I also want to enter the blog_id into the database.
 
This is the code I am using:
 
<form action="comment.php" method="POST">
Your name: <input type="text" name="display_name" size="25"><br>
Your email: <input type="text" name="email" size="30"><br>
Your comment: <input type="text" name="comment" size="40"><br>
<input type="hidden" name="blog_id"
value="<?php echo $_POST['$thisID']; ?>">
<input type="submit" value="Submit">
 
This is how I am entering the values into the SQL table
 
 $command = "insert into $table_name
values('','".$db->real_escape_string($_POST['comment'])."',
'".$db->real_escape_string($_POST['blog_id'])."',        
'".$db->real_escape_string($_POST['display_name'])."',
'".$db->real_escape_string($_POST['email'])."',now()
);";
 
The code is executed, I don't get any errors but the variable (blog_id) is not entered, the field in the database is null. All other fields are entered correctly. Is it even possible to hand over a PHP variable like that? 

原文：https://stackoverflow.com/questions/27222545