Php 使用Jquery获取json数据提要，并在HTML中显示多个json数组对象

我正在使用jquery和getJSON获取由PHP构建的数据提要。访问PHP页面时，提要显示良好。我遇到的问题是，JSON提要在jQuery中被请求时会以多个对象的形式返回，我不知道如何编写jQuery来显示所有对象及其数据

jQuery：

$(document).ready(function () {
    $("#display_results").hide();
    $("#searchButton").click(function (event) {
        $("#display_results").show();
        event.preventDefault();
        search_ajax_way();
    });
    $("#search_results").slideUp();
    $("#searchBox").keyup(function (event) {
        $("#display_results").show();
    });
});

function search_ajax_way() {
    //var search_this=$("#searchBox").val();
    //$.post("http:/url.com/folder/cl/feed.php", {city : search_this},         function(data){
    //$("#display_results").html(data);
    //});
    var search_this = $("#searchBox").val();
    $.getJSON('http://url.com/app/cl/disp_byowner_feed.php', {
        city: search_this
    }, function (result) {
        $.each(result, function (i, r) {
            console.log(r.seller);
            window.title = r.title;
            window.seller = r.seller;
            window.loc = r.location;
            (Plan on listing all keys listed in the data feed below here)
        });
        console.log(result);
        $("#display_results").html(window.seller + < need to list all keys / values here > );
    });
}

示例JSON提要：

$city = 'Kanosh';
$s = "SELECT * FROM `list` WHERE `city` LIKE '%".$city."%'";
$res = $mysqli->query($s) or trigger_error($mysqli->error."[$s]");
$a = array(); 
while($row = $res->fetch_array(MYSQLI_BOTH)) { 
$a[] = array(
'title' => $row['title'],
'price' => $row['price'],
'rooms' => $row['rooms'],
'dimensions' => $row['dimensions'],
'location' => preg_replace('pic\u00a0map', '', $row['location']),
'price' => $row['price'],
'address' => $row['address'],
'seller' => $row['seller'],
'href' => $row['href'],
'date' => $row['date']
); 
}
header('Content-Type: application/json');
echo json_encode($a);
$res->free();
$mysqli->close();

[{
    "title": "Great Ski-In Location with Seller Financing Available ",
    "price": "  (Park City near the Resort)   ",
    "rooms": "",
    "dimensions": "",
    "location": "",
    "address": "Crescent Road at Three Kings Dri",
    "seller": "real estate - by owner",
    "href": "http:\/\/saltlakecity.craigslist.org",
    "date": "20140811223115"
}, {
    "title": "Prospector Building 1 - Tiny Condo, Great Location - Owner Financing",
    "price": "$75000",
    "rooms": false,
    "dimensions": "",
    "location": "  Prospector Square Park City Ut",
    "address": "",
    "seller": "real estate - by owner",
    "href": "http:\/\/saltlakecity.craigslist.org",
    "date": "20140811223115"
}]

---

您的输出是一个JSON对象数组。 幸运的是，JavaScript实际上可以方便地操作JSON，这就是创建JSON的原因，而jQuery可以方便地操作DOM

要解析结果，只需迭代该数组，在数组中构造所需的任何HTML字符串，然后使用jQuery将其插入DOM

下面是一个简单的列表示例：

var html = "";

for (var i = 0; i < result.length; i++) { //assuming result in the JSON array

    html += '<ul id = 'house_' + i>';
    for (var property in result[i]) { //read all the object properties
        html += '<li>' + property + ' : ' + result[i][property] + '</li>';

    }
    html += '</ul>';

};

$("whatever_div").html(html);

如果只想显示某些属性，可以单独读取它们，并对日期进行其他格式设置（例如）。

---

给不同的HTML对象ID与它们所代表的内容相对应也很有用。

完美答案。非常感谢。

var html = "";

for (var i = 0; i < result.length; i++) { //assuming result in the JSON array

    html += '<ul id = 'house_' + i>';
    for (var property in result[i]) { //read all the object properties
        html += '<li>' + property + ' : ' + result[i][property] + '</li>';

    }
    html += '</ul>';

};

$("whatever_div").html(html);