Question

我下载了JTable PHP演示文件并修改了代码。以下是我的表现..

enter code here //Prepare jTable
        $('#PeopleTableContainer').jtable({
            title: 'User Maintenance',
            actions: {
                listAction: 'PersonActions.php?action=list',
                createAction: 'PersonActions.php?action=create',
                updateAction: 'PersonActions.php?action=update',
                deleteAction: 'PersonActions.php?action=delete'
            },
            fields: {
                strUsrName: {
                    key: true,
                    create: false,
                    edit: false,
                    list: true,
                    title: 'UserName'
                },
                strUsrFName: {
                    title: 'FirstName',
                    width: '10%'
                },
                strUsrMName: {
                    title: 'MiddleName',
                    width: '10%'
                },
                dtmUsrCreated: {
                    title: 'Record date',
                    width: '30%',
                    type: 'date',
                    create: false,
                    edit: false
                }
            }
        });

这是我的查询......

//Updating a record (updateAction)
else if($_GET["action"] == "update")
{
    //Update record in database
    $result = mysql_query("UPDATE tbluser SET strUsrFName = '" . $_POST["strUsrFName"] . "', strUsrMName = " . $_POST["strUsrMName"] . " WHERE strUsrName = " . $_POST["strUsrName"] . ";") or die('ERROR: '.mysql_error());

    //Return result to jTable
    $jTableResult = array();
    $jTableResult['Result'] = "OK";
    print json_encode($jTableResult);
}

我的主要问题是：当我删除die（）时，表更新了它上面的数据，但它没有更新数据库中的数据，因此我猜测错误在查询中或者可能在主键中。有人能帮帮我吗？

Answer 1

您的查询编写不正确。您需要使用引号来确保将字符串值传递给非数字的列。您的查询应如下所示：

$result = mysql_query("UPDATE tbluser SET strUsrFName = '" . $_POST["strUsrFName"] . "', strUsrMName = '" . $_POST["strUsrMName"] . "' WHERE strUsrName = '" . $_POST["strUsrName"] . "';") or die('ERROR: '.mysql_error());

正如我在评论中所说，不再使用mysql_query。它被弃用并暴露于sql注入。如果您使用prepared statements，则不再有引用问题。

$sql = "UPDATE tbluser SET strUsrFName = :strUsrFName, strUsrMName = :strUsrMName WHERE strUsrName = :strUsrName;";
$sth = $pdo->prepare($sql, array(PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY));
$sth->execute(array(":strUsrFName" => $_POST["strUsrFName"], ":strUsrMName" => $_POST["strUsrMName"], ":strUsrName" => $_POST["strUsrName"]));

JTable PHP更新无法正常工作

1 个答案: