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当使用搜索客户信息时，我有两个 jQuery AJAX 请求。我可以通过 .autocomplete 获取客户名称，也可以从 addressSearch.php 获取 JSON 数据，但是当我无法从第二个 JSON 集中获取所需的值时。

用户应该简单地从下拉列表中选择地址，我应该填写表格。

当我console.log(address)没有得到正确的多维数组时。

如何选择正确的子数组将数据发送到表单域？

HTML

<label>Customer Search:</label><input id="cSearch" type="text" /><br />
<label>Name:</label><input id="name" /><br />
<select id="addresses">
  <option>Please Select Address</option>
</select>
<label>Street:</label><input id="street" type="text" /><br />
<label>City:</label><input id="city" type="text" /><br />
<label>State:</label><input id="state" type="text" /><br />
<label>Zip:</label><input id="zip" type="text" /><br />

JavaScript

$(document).ready(function() {
    $("input#customerSearch").autocomplete({
        source:'inc/customerSearch.php',
        select: function(event, ui) {
                $('#name').val(ui.item.name);
                getAddresses(ui.item.addressid);
            },
            minLength:2
        });
    });
    var address = [];
    function getAddresses(id){
        $.getJSON(
            'inc/addressSearch.php',
            { id: id },
            function(data)  {
                $('#addresses').children().remove().end().append('<option>Please Select Address</option>');
                $.each(data, function(x, val) {
                    $('#addresses').append('<option value="' + val['addressid'] + '">'+ val['street1'] +'</option>');
                    address[x] = [];
                    $.each(val, function(key, value) {
                        address[x][key] = value;
                    });
                });
            }
        );
    $('select#addresses').change(function() {
        var j = $(this).val();
        //fill in fields when select 
    });
}

地址搜索.php

$id = explode(',', $id);//explode address ids
$data = array();//data array
foreach($id as $id) {//evaluate each address
    $query = "SELECT * FROM `address` WHERE address_id = '". $id ."'";//query
    $result = mysql_query($query);//run query
    if (mysql_num_rows($result)) {//results exist
        while($row = mysql_fetch_assoc($result)) {//loop
            $data[$id] = array(//build array
                'addressid' => $row['address_id'],
                'street1' => $row['address_street1'],
                'street2' => $row['address_street2'],
                'city' => $row['address_city'],
                'region' => $row['address_region'],
                'country' => $row['address_country'],
                'code' => $row['address_code']
            );
        }
    }
}
print json_encode($data);// jQuery wants JSON data

JSON

{
    "101":{
        "addressid":"101",
        "street1":"123 ABC St",
        "street2":"",
        "city":"Somewhere",
        "region":"AB",
        "country":"USA",
        "code":"12345"
    },
    "102":{
        "addressid":"102",
        "street1":"123 10th St",
        "street2":"Apt 101",
        "city":"Nowhere",
        "region":"AB",
        "country":"USA",
        "code":"12345"
    }
}
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1 回答 1

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我只好走开再看一遍。上面的代码没有任何问题，只是第二部分不起作用。

这是最终的功能

JavaScript

$('select#addresses').change(function() {
        var j = $(this).val();
        $('#street1').val(address[j]['street1']);
    $('#street2').val(address[j]['street2']);
    $('#city').val(address[j]['city']);
    $('#code').val(address[j]['code']);

    });
于 2012-06-02T02:24:06.723 回答