Php 显示数据库中的复选框记录

我看过类似的问题和解决方案，但不知何故，我的问题只解决了一半。我试图制作一个表单，从MySQL数据库中检查多条记录，并将检查过的记录显示到另一个页面。不知何故，我设法用复选框来显示页面，但我不知道如何显示选中的记录。无论选中哪个框，它只能显示记录的第一行或所有记录

这是复选框页面

$columns = count($fieldarray); 
//run the query 
$result = mysql_query(
  "SELECT * FROM request_item 
   ORDER BY request_item.IllNo DESC LIMIT 0, 6") or die(mysql_error()); 
$row = mysql_num_rows($result);

while($row=mysql_fetch_array($result)) 
{
  {
    $rows[] = $row['IllNo'];
  }

  foreach($rows as $value); 
  echo "";
  echo " ";
  echo $row['IllNo'];
  echo ""; 
}
echo "";
?>

这是选中的显示记录

$columns = count($fieldarray); 
//run the query 
$result = mysql_query(
  "SELECT * FROM request_item 
   ORDER BY request_item.IllNo DESC LIMIT 0, 6") or die(mysql_error()); 
$row = mysql_num_rows($result);

while($row=mysql_fetch_array($result)) 
{ 
  $rows[]=$row['IllNo'];  
  foreach($rows as $value); 
  if ($rows= 'checked') { 
    echo "";
    echo $value; 
  }

---

欢迎任何帮助。谢谢。

这个脚本实际上有很多问题，包括语法错误、调用错误的变量名、表单没有在应该打开的地方打开、在已经打开之后调用PHP等等

为了得到一个好的答案，您应该共享make$row['IllNo']应该等于多少，以指示是否应该检查它

我重新格式化了一点，这可能会给你一个好的开始

<form NAME ="form1" METHOD ="POST" ACTION ="dari.php">
<table>
<?php
$columns = count($fieldarray);
//run the query
$result = mysql_query("SELECT * FROM request_item ORDER BY request_item.IllNo DESC LIMIT 0, 6") or die(mysql_error()) ;
$row = mysql_num_rows($result);


while($row=mysql_fetch_array($result)) {
  echo "<tr><td>";
  echo "<Input type = 'Checkbox' Name ='ch1' value ='ch1'";
  // check checked if it is. this will be checked if $row['IllNo'] has a value
  // if there were a condition to make it checked, you would put the condition
  // before the ?
  echo $row['IllNo'] ? ' checked' : '';
  echo ' />';
  echo $row['IllNo'];

  echo "</td></tr>";
}
?>
</table>
<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Choose your books">
</FORM>

---

这个脚本实际上有很多问题，包括语法错误、调用错误的变量名、表单没有在应该打开的地方打开、在已经打开的地方调用PHP等等

为了得到一个好的答案，您应该共享make$row['IllNo']应该等于多少，以指示是否应该检查它

我重新格式化了一点，这可能会给你一个好的开始

<form NAME ="form1" METHOD ="POST" ACTION ="dari.php">
<table>
<?php
$columns = count($fieldarray);
//run the query
$result = mysql_query("SELECT * FROM request_item ORDER BY request_item.IllNo DESC LIMIT 0, 6") or die(mysql_error()) ;
$row = mysql_num_rows($result);


while($row=mysql_fetch_array($result)) {
  echo "<tr><td>";
  echo "<Input type = 'Checkbox' Name ='ch1' value ='ch1'";
  // check checked if it is. this will be checked if $row['IllNo'] has a value
  // if there were a condition to make it checked, you would put the condition
  // before the ?
  echo $row['IllNo'] ? ' checked' : '';
  echo ' />';
  echo $row['IllNo'];

  echo "</td></tr>";
}
?>
</table>
<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Choose your books">
</FORM>

是否（$rows='checked'）正确？或者应该是（$rows='checked'）？是否（$rows='checked'）正确？或者应该是（$rows=='checked'）？