Question

那你好。我有这个问题，我无法弄清楚dateTimePicker的语法错误。请帮助这里是

中的表格

<form name="subExpdate" method="POST" action="" id="subExpdate" >
                            <input type="date" id="expDatePicker" name="expDatePicker" style="margin-bottom: 5px" >
                            <input type="hidden" value="<?php echo $prodLink->valueof('link_id') ?>" id="hidID" name="hidID">
                            <input type="hidden" name="prodInt" id="prodInt" value="<?php echo $selInterface; ?>"/>
                            <button class="btn btn-success" type="button" name="changeExpireDate" id="changeExpireDate" onclick="changeExpireDate23()" >Change expire date</button>
                        </form>

这是我的功能：

function changeExpireDate23() {
        var expTime = new FormData(document.getElementById('subExpdate'));
        console.log('something' + expTime);
        $.ajax({
        url: "my path and it works not allowed to show",
                type: 'POST',
                data:expTime,
                contentType: false,
                processData: false, 
                success: function (data) {
                alert(data);
                        console.log(data);
                }
    });
    }

这是我的ajax页面：

 if (isset($_POST['expDatePicker'])){
            $datePicker = date('d/m/Y  H:i:s' , strtotime($_POST['expDatePicker']));
            $getLinkData = new CGenRs("SELECT * FROM lu_product_link WHERE link_id = '" . $hidid . "' ", $cao_beta2);
            $getLinkData->first();

            $updateExpDate = "UPDATE lu_product_link SET (link_promo_expiry) = '".$datePicker."' WHERE link_id = '".$hidid."'";
            if($cao_beta2->execute($updateExpDate)){
                echo 'SUCCESS date updated';
            } else{
                echo 'Failed sorry mate';
            }

如果你可以帮助我，那将非常感激。提前致谢。

Answer 1

我发现问题就是这一行

$updateExpDate = "UPDATE lu_product_link SET (link_promo_expiry) = '".$datePicker."' WHERE link_id = '".$hidid."'";

（link_promo_expiry）不应该在（）

中

php和js中使用datepicker的语法错误

1 个答案: