Question

我有一个PHP代码的工作设置，突然之间没有任何工作，我不太确定我是否在某处意外地在代码中做错了。

基本上我有一个从数据库中读取一些数据的php并将其填充到表中，以便我可以对其进行分组编辑。但是当点击提交按钮时，就不再发生任何事情了。

也许有人会立即发现错误。

这是用于提取数据并创建表格的PHP：

<form method="post" action="fxgroupedit.php">
<table border="1">
 <tr>
  <td>n</td>
  <td>ID</td>
  <td>Year</td>
  <td>Month</td>    
  <td>Day</td>
  <td>Issue</td>
  <td>Revenue</td>
  <td>Category</td>
  <td>Sub Category</td> 
 </tr>
 <?php include("db_fx.php");

 $sql = "SELECT * FROM tbl_data WHERE category=''";
 $result = mysql_query($sql);
 $n=0;

 while ($line = mysql_fetch_array($result)) 
 { 
 $n++;
 ?>
 <tr>
  <td><?PHP echo $n?></td>
  <td><input style="width: 30px" name="id<?echo $n?>" value="<?PHP echo $line[id];?>"</td>
  <td><input style="width: 40px" name="year<?echo $n?>" value="<?PHP echo $line[year];?>"</td>
  <td><input style="width: 30px" name="month<?echo $n?>" value="<?PHP echo $line[month];?>"</td>
  <td><input style="width: 30px" name="day<?echo $n?>" value="<?PHP echo $line[day];?>"</td>
  <td><input style="width: 400px" name="issue<?echo $n?>" value="<?PHP echo $line[issue];?>"</td>
  <td><input style="width: 50px" name="rev<?echo $n?>" value="<?PHP echo $line[rev];?>"</td>
  <td><select name="category<?echo $n?>"
   <?php
    $sql = "SELECT DISTINCT category FROM tbl_data ORDER BY category";
    $rs = mysql_query($sql);
    echo '<option value=""></option>';
    while($row = mysql_fetch_array($rs))
    { 
     echo "<option value=\"".$row['category']."\">".$row['category']."</option>\n  ";
    }
   ?>
  </select></td>
  <td><select name="sub_category<?echo $n?>" 
   <?php
    $sql = "SELECT DISTINCT sub_category FROM tbl_data ORDER BY sub_category";
    $rs = mysql_query($sql);
    echo '<option value=""></option>';
    while($row = mysql_fetch_array($rs))
    { 
     echo "<option value=\"".$row['sub_category']."\">".$row['sub_category']."</option>\n  ";
    }
   ?>
  </select></td>
 </tr>
 <?PHP
 }
 //mysql_free_result($result);
 mysql_close();
 ?>
 </table>
 <input type="hidden" name="rows" value="<?echo $n?>" />
 <input type="submit" />
</form>

这里是fxgroupedit.php脚本，通常可以解决这个问题：

<table border="1">
 <tr>
  <td>n</td>
  <td>ID</td>
  <td>Year</td>
  <td>Month</td>    
  <td>Day</td>
  <td>Issue</td>
  <td>Revenue</td>
  <td>Category</td>
  <td>Sub Category</td> 
 </tr>
<?PHP include("db_fx.php");

$fx_rows=$_POST['rows'.$n];
$n=0;

while($n < $fx_rows)
     {
     $n++;
     $fx_id=$_POST['id'.$n];
     $fx_year=$_POST['year'.$n];
     $fx_month=$_POST['month'.$n];
     $fx_day=$_POST['day'.$n];
     $fx_issue=$_POST['issue'.$n];
     $fx_rev=$_POST['rev'.$n];
     $fx_category=$_POST['category'.$n];
     $fx_sub_category=$_POST['sub_category'.$n];

     $sql="UPDATE tbl_data SET 
     year='$fx_year',
     month='$fx_month', 
     day='$fx_day', 
     issue='$fx_issue', 
     rev='$fx_rev', 
     category='$fx_category', 
     sub_category='$fx_sub_category'
     WHERE
     id='$fx_id'";

     mysql_query($sql);
     print mysql_error();
     echo $fx_rows;
     ?>
     <tr>
      <td><?php echo $n ?></td>
      <td><?php echo $fx_id ?></td>
      <td><?php echo $fx_year ?></td>
      <td><?php echo $fx_month ?></td>
      <td><?php echo $fx_day ?></td>
      <td><?php echo $fx_issue ?></td>
      <td><?php echo $fx_rev ?></td>
      <td><?php echo $fx_category ?></td>
      <td><?php echo $fx_sub_category ?></td>
     </tr>
     <?PHP
     }
     ?>
    </table>

谢谢！

PHP Mysql不会更新

0 个答案: