Question

我有一个修改表格＆＃39; wp_postmeta＆＃39;的元内容的表单。我试图找出从表单到INSERT到另一个表的操作的方法。

这可能吗？这就是我所拥有的，并且数据不会插入到事务表中。

CURRENT FORM提交按钮div：

<div class='inventory-management-form-item-holders inventory-management-submit-holder'>
    <input type="submit" name="submit" value="SUBMIT">
</div>

＆＃34;库存同治提交＆＃34; js SECTION：

// submit form
jQuery("#submit").click(function(){
    // Change thw loading Gif url 
    jQuery('#the-results-holder').html("<img src='/wp-content/plugins/mcs-inventory-management/assets/images/loading_icon.gif' />");
    jQuery.ajax({
        type: "POST",
        data: {'im-product-id': theProductId, 'status-quantity': jQuery('#status-quantity').val(), 'status-action': jQuery('#im-action-box').val(), 'status-location': jQuery('#status-location').val(), 'status-move-location': jQuery('#status-move-location').val()},
        dataType: "json",
        url: "/inventory-management-process/",
        processData: true,
        success: function(data) {
            jQuery('#the-results-holder').html(""); 
            jQuery('#status-stock').val(data.stock);
            jQuery('#status-res').val(data.res);
            jQuery('#status-prepro').val(data.prepro);
            jQuery('#status-clean').val(data.clean);
            jQuery('#status-quantity').val("0");
        },
        error: function (xhr, ajaxOptions, thrownError) {
            jQuery('#the-results-holder').html(""); 
            jQuery('#the-results-holder-error').html("There has been an error!"); 
        }
    });
    return false;
});

这是我在哪里尝试执行INSERT表单进入事务表的内容？

<form name='inventory-management' id='inventory-management-form' method="POST" action='<?php $transaction ?>'>


<?php
$product= $_POST['part-number'];
$action= $_POST['status-action'];
$from_location= $_POST['status-location'];
$to_location= $_POST['status-move-location'];
$qty= $_POST['status-quantity'];


if(isset($_POST['submit']))
{
 $transaction = "INSERT INTO mcs_inventory_transactions (id,product,action,from_location,to_location,qty) VALUES ('','$product','$action','$from_location','$to_location','$qty')";
 $result = mysql_query($transaction);
}
?>

Answer 1

将你的js改为

  JS
// submit form

    jQuery("#inventory-management-submit").click(function(){
        // Change thw loading Gif url 
        jQuery('#the-results-holder').html("<img src='/wp-content/plugins/mcs-inventory-management/assets/images/loading_icon.gif' />");
        jQuery.ajax({
            type: "POST",
            data: {'im-product-id': theProductId, 'status-quantity': jQuery('#status-quantity').val(), 'status-action': jQuery('#im-action-box').val(), 'status-location': jQuery('#status-location').val(), 'status-move-location': jQuery('#status-move-location').val(),submit: "test"},
            dataType: "json",
            url: "/inventory-management-process/",
            processData: true,
            success: function(data) {
                jQuery('#the-results-holder').html(""); 
                jQuery('#status-stock').val(data.stock);
                jQuery('#status-res').val(data.res);
                jQuery('#status-prepro').val(data.prepro);
                jQuery('#status-clean').val(data.clean);
                jQuery('#status-quantity').val("0");
            },
            error: function (xhr, ajaxOptions, thrownError) {
                jQuery('#the-results-holder').html(""); 
                jQuery('#the-results-holder-error').html("There has been an error!"); 
            }
        });
        return false;
    });

希望这有助于你

Answer 2

这是错误的：

$transaction= if(isset($_POST['submit']))

应该是

if(isset($_POST['submit']))

提交按钮更新一个表并将详细信息发送到另一个表

2 个答案: