Question

我正在尝试在自动完成功能中使用jQuery Ajax函数。自动完成功能正常，因为它返回值，但是当我选择一个值时，它不会执行任何操作。我想返回html并在div中显示。

jQuery的：

$(function() {
    $("#store").autocomplete({
        source: "includes/ajax/storenumbers.php",
        select: function(event, ui) {
            var storeid = val(ui.item.id);
            $.ajax({
               url: "includes/ajax/storecontacts.php",
               type: "GET",
               data: {term : storeid},
               dataType: "html",                               
               success: function(msg){
                            //Display html
                                $("#resultsdiv").html(msg);
                           },
                           error: function (request, status, error) {
                                alert(request.responseText);
                            }
                         });
                    }
                });
            });

HTML：

<form action="<?php echo $PHP_SELF;?>"  method="post">

<label for="store">Enter Store #: </label>

<input type="text" id="store"  name="state" /> 

<div id="resultsdiv" style="width: 200px; height: 200px;">
</div>
</form>

storenumbers.php返回“1”：

[{"id":"6","value":"10211","abbrev":"Concord"},{"id":"4","value":"10869","abbrev":"Maplewood"},{"id":"5","value":"16289","abbrev":"Hugo"},{"id":"12","value":"19245","abbrev":"WBL 4th Street"}]

storecontacts.php返回：

Name: Alex E-mail: alex@email.com View Order: yes Print Order: yes<br>
Name: Brooke E-mail: brooke@email.com View Order: no Print Order: no

我希望html返回最终是一系列输入文本字段和复选框。 dataType：html 仅限于某些内容吗？

Answer 1

我明白了。我不得不将var storeid = val（ui.item.id）更改为var storeid = ui.item.id;

jQuery ajax里面自动完成php mysql

1 个答案: