验证和保留动态复选框的状态 [英] Validating and retaining state of a dynamic checkboxes

问题描述

新的php ...我一直在战斗在我的动态复选框，在这样一种方式，如果没有检查形式是返回，还需要保留什么检查时，表单回发由于其他无效的输入。

  $ result = mysql_query（SELECT * FROM course）或die（mysql_error（））; 
 if（$ _ POST ['courses']）和$ _POST [
] 
 if（$ result）
 {
 while（$ row = mysql_fetch_array（$ result） 'courses'] == $ row ['cid']）{echo $ row ['cid'];} 
 print< input type = \checkbox\name = ] \value = \$ row [cid] \> $ row [cname] \\\
; 
} 
} 
  

帮助需要纯粹的PHP代码。提前感谢

解决方案

如果复选框出现在您的php页面的HTML中：

 <？php 
 
 $ checked = isset（$ _ POST [checkboxname]）？ checked：''; 
 echo< input type ='checkbox'name ='checkboxname'value ='yes'。 $ checked。 >; 
 
？> 
  

表单发布后将保留复选框状态。

更新：

对于您的代码，只需这样做，我认为：

  $ result = mysql_query（SELECT * FROM course）或die（mysql_error（））; 
 if（$ result）{
 while（$ row = mysql_fetch_array（$ result））{
 $ checked =''; 
 / * ERROR：if（isset（$ _ POST ['courses']）和$ _POST ['courses'] == $ row ['cid']）{* / 
 if（isset _POST ['courses']）{
 if（in_array（$ row ['cid']，$ _POST ['courses']）{
 echo $ row ['cid']; 
 $ checked =checked; 
} 
} 
 echo< input type = \checkbox\name = \courses [] \value = \ $ row [cid] \。$ checked。> $ row ['cname'] \\\
; 
} 
} 
  
 
 
 
编辑：条件也需要改变，我想，正如上面的代码所示。
 
Am new to php... I have been battling on my dynamic checkboxes in such a way that if none is checked the form is return, also I need to retain what was checked when the form postback due to other invalid inputs.  
$result = mysql_query("SELECT * FROM course") or die(mysql_error()); 
if ($result) 
{ 
  while ($row = mysql_fetch_array($result)){
    if (isset($_POST['courses']) and $_POST['courses'] == $row['cid']) {echo $row['cid'];} 
    print "<input type=\"checkbox\" name=\"courses[]\" value=\"$row[cid]\">$row[cname]\n"; 
  }
}
Help needed purely on php codes. Thanks in advance
解决方案
Do this where the checkbox appears in the HTML on your php page:
<?php

   $checked = isset($_POST["checkboxname"]) ? " checked" : '' ;
   echo "<input type='checkbox' name='checkboxname' value='yes'" . $checked . ">";

?>
This will retain the checkbox state after the form has been posted.

UPDATE:

For your code, just do it like this, I think:
$result = mysql_query("SELECT * FROM course") or die(mysql_error()); 
if ($result) {
    while ($row = mysql_fetch_array($result)) {
        $checked = '';
/* ERROR:  if (isset($_POST['courses']) and $_POST['courses'] == $row['cid']) { */
        if (isset($_POST['courses']) {
           if (in_array($row['cid'], $_POST['courses']) {
               echo $row['cid'];
               $checked = " checked";
           }
        }
        echo "<input type=\"checkbox\" name=\"courses[]\"  value=\"$row[cid]\"" . $checked . ">$row['cname']\n"; 
    }
}
EDIT: The condition needs to change too, I think, as I show in the code above.

                        
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