Question

我想更新表中只有CHARGES的值。我的错误是：

注意：第4行的C：\ xampp \ htdocs \ VCCDMStry \ pages \ save_receipt.php中的未定义索引：id

注意：未定义的索引：第5行的C：\ xampp \ htdocs \ VCCDMStry \ pages \ save_receipt.php收费

致命错误：未捕获的异常＆＃39; PDOException＆＃39; with message＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064 SQL语法中有错误;检查与您的MariaDB服务器版本对应的手册，以便在＃＆lt; WHERE id = NULL＆＃39; WHERE id = NULL＆＃39;附近使用正确的语法。在第1行＆＃39;在C：\ xampp \ htdocs \ VCCDMStry \ pages \ save_receipt.php：8堆栈跟踪：＃0 C：\ xampp \ htdocs \ VCCDMStry \ pages \ save_receipt.php（8）：PDOStatement-＆gt; execute（Array）＃1 {main}在第8行的C：\ xampp \ htdocs \ VCCDMStry \ pages \ save_receipt.php中抛出

这是我的fetch代码：

<?php
    $sql = "SELECT * FROM tenant 
            LEFT JOIN stall ON stall.stall_id = tenant.stall_id
            LEFT JOIN rent ON rent.tenant_id = tenant.tenant_id
            LEFT JOIN receipt ON receipt.tenant_id = tenant.tenant_id
            LEFT JOIN rate ON rate.rate_id = tenant.rate_id
            WHERE rent.rent_status = ? AND rate.rate_id = ?
    ";
    $query = $conn->prepare($sql);
    $query->execute(array(1,1));
    $fetch = $query->fetchAll();

    foreach ($fetch as $key => $value) { ?>                         
    <tr>
        <td>
            <?=$value['tenant_fname']." ".$value['tenant_mname']." ".$value['tenant_lname'];?>
        </td>
        <td>
            <?=$value['stall_number'];?>
        </td>
        <td>
            <?=$value['charges'];?>
        </td>
        <td class = "center" style = "text-align:center;">
            <a href = "#receipt<?php echo $value['tenant_id'];?>" data-toggle="modal" class="btn btn-outline btn-xs btn-primary"> Make Receipt
            </a>
        </td>
    </tr>   
    <?php include 'receipt_modal.php';?>              
<?php }?>

以下是我更新的代码：

<?php    
    include 'dbcon.php';

    $receipt_id = $_POST['id'];
    $charges = $_POST['charges'];

    $query = $conn->prepare("UPDATE receipt SET charges = ? , WHERE id = ? ");
    $query->execute(array($charges,$receipt_id));

    header ('location:elecbills.php');
?>

模态代码：

<div class="form-group">
    <input type = "hidden" class="form-control" name= "id"  value = "<?php echo $value['tenant_id'];?>">
    <label class="col-sm-4 control-label">PRESENT </label>: <input class = "custom-input" id="charges" name="charges"></input> 
</div>

Answer 1

引发致命错误导致您在“之前”处有逗号错误。

UPDATE receipt SET charges = ? , WHERE id = ?

如何在查询中更新仅为LEFT JOIN的表？

1 个答案: