如何从表中的表单提交和处理值

问题描述：

I have a HTML table with an arbitrary amount of rows, each with 4 columns. And the idea is to let the user edit any of these rows and then save the result. These changes should then be saved to a database. But I have several problems.

I have found some posts on * indicating on how you should approach this, but none perfectly useful in my case. This is how it looks when the user is editing my table (although I only have one row now, there can be several).

my table http://oi58.tinypic.com/2gvq9nb.jpg

And of course there's a submit button, that takes my to a php file named something like "handle_edit.php".

Now another post on * about this subject has indicated that I should use some kind of foreach-loop to edit one row in the SQL table per row in my HTML table. My version of this loop in "handle_edit.php" looks like this right now:

foreach( $_POST['appearances'] as $appearances ) {
    $goals = $_POST['goals'];
    $assists = $_POST['assists'];
    $name = ??????? //How to I do this?

    mysqli_query($connection, "UPDATE players
                            SET appearances=$appearances, goals=$goals, assists=$assists
                            WHERE name=$name");
}

First of all. The foreach loop needs to contain an array expression, and the compiler tells me that's not the case right now. But I have no idea on how to count the amount of rows. In the thread I linked above they are doing something similar to this, but I don't think I fully understand what they are doing since my foreach loop isn't working.

Secondly, since the name column in the table isn't an input field, I don't know how to send it to this page with the rest of the information. I need to do it somehow, since I want to determine which row to edit based on the name provided. Each of the players in this SQL table has an unique player id, could I use that somehow?

Am I maybe approaching this the wrong way? Is there a more viable solution?

Please tell me if you need any additional code.

Thanks in advance!

我有一个包含任意行数的HTML表，每行有4列。并且想法是让用户编辑这些行中的任何一行然后保存结果。然后应将这些更改保存到数据库中。但是我有几个问题。 p>

我在*上发现了一些帖子，说明你应该怎么做，但在我的情况下没有一个完全有用。这是用户编辑我的表时的样子（虽然我现在只有一行，但可能有几行）。 p>

我的桌子http://oi58.tinypic.com/2gvq9nb.jpg p>

当然还有提交按钮，将我带到一个名为“handle_edit.php”的php文件。 p>

现在关于*的另一篇文章表明我应该使用某种foreach-loop来编辑SQL中的一行我的HTML表格中每行的表格。我在“handle_edit.php”中的这个循环版本现在看起来像这样： p>

  foreach（$ _POST ['appearances'] as $ appearances）{
 $ goals =  $ _POST ['goal']; 
 $ assists = $ _POST ['assists']; 
 $ name = ???????  //我怎么做？
 
 mysqli_query（$ connection，“UPDATE players 
 SET appearances = $ appearances，goals = $ goals，assists = $ assists 
 WHERE name = $ name”）; 
}  
  code>  pre> 
 
 首先。   foreach  code>循环需要包含一个数组表达式，编译器告诉我现在不是这样。 但我不知道如何计算行数。 在我上面链接的线程中，他们正在做类似的事情，但我不认为我完全理解他们正在做什么，因为我的 foreach  code>循环不起作用。 p> 
 \  n 
其次，由于表中的name列不是输入字段，我不知道如何将其与其余信息一起发送到此页面。 我需要以某种方式执行此操作，因为我想根据提供的名称确定要编辑的行。 这个SQL表中的每个玩家都有一个唯一的玩家ID，我能以某种方式使用它吗？ p> 
 
 

我是否可能以错误的方式接近这个？ 是否有更可行的解决方案？ p> 
 
 
如果您需要任何其他代码，请告诉我。 p> 
 
 
提前致谢！ p>

答

First of all, the name variable. you need to add <input type='hidden' name='name' value='<*YOUR WAY OF GETTING THE NAME*>'/> to your form. if you do so, you can get the value of the name in your foreach loop with $_POST['name']. For more info on how to make a form for php look here: http://www.w3schools.com/php/php_forms.asp

Secondly, the sql string u use is wrong. You cant input the variable names within a string. I suggest declaring the sql string first like this:

$sql = "UPDATE players SET appearances=" . $appearances . ", goals=" . $goals . ", assists=" . $assists . " WHERE name=". $name;

Then make the connection and execute the query like so:

mysqli_query($connection, $sql);

for the foreach problem, i suggest debugging using the var_dump() method and see if you really get an array back with $_POST['appearances'].

I hope this helps.

如何从表中的表单提交和处理值

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