使用PHP和MySQL在单个查询中更新多个记录

Question

任何人都可以帮助解决这个问题。当我尝试使用catid更新多个记录时（它包含一个或多个相同的catid），它不会更新结果。

以下是代码

在这里输入代码

<?php

//select a database to work with
$selected = mysql_select_db("db_name",$cnn) or die("Could not select examples");

    $sql="SELECT *
                    FROM `plannedbudgetstbl`
                    LEFT JOIN budget_categories ON plannedbudgetstbl.cat_id = budget_categories.cat_id
                    WHERE programid ='".$id."'";
    $result=mysql_query($sql);

    // Count table rows 
    $count=mysql_num_rows($result);
    ?>
<form name="form1" method="post" action="">

    <table width="500" border="0" cellspacing="1" cellpadding="0">
     <tr> 
    <td>
    <table width="500" border="0" cellspacing="1" cellpadding="0">


    <tr>
    <td align="center"><strong>budget</strong></td>

    </tr>
    <?php
    // Check if button name "submit" is active, do this 
    if (isset($_POST['submit'])) {
         $budget = $_POST['budget'];


    for($i=0;$i<$count;$i++){
    $sql1="UPDATE plannedbudgetstbl SET budget='".$budget[$i]."' WHERE programid = '".$id[$![enter image description here][1]i]."'and cat_id='".$id[$i]."'";

    $result1=mysql_query($sql1);
    }
    }
    if(isset($result1)){
    header("location:budget-dispaly1.php");
    }
    ?>
    <?php
    while($rows=mysql_fetch_array($result)){
    ?>
    <tr>
    <td align="center"><?php $id[]=$rows['cat_id']; ?><?php echo $rows['cat_id']; ?></td>
    <td align="center"><input name="budget[]" type="text" id="budget" value="<?php echo $rows['budget']; ?>"></td>

    </tr>
    <?php
    }
    ?>
    <tr>
    <td colspan="4" align="center"><input type="submit" name="submit" value="submit"></td>
    </tr>
    </table>
    </td>
    </tr>

    </table>
        </form>

这是更新输出