我的JavaScript for循环有问题吗

我试图循环返回的JSON数组，并在表中显示值。我的数组包含多个具有多个值的对象，因此每个对象将=一个新行，每个值将=一个新单元格

以下是我尝试过的一个版本：

function refreshUrlArray(urlsRetrieved) {
    $('#response').html('Attempting to update your URLs, please wait...');

    var urlsObj = urlsRetrieved;
    console.log("JSON results returned from DB query: " + urlsObj);

    //Create object holding table
    $table = "<table id='urlTable'><td>Name</td><td>Release Time</td><td>Release Date</td><td>Category</td><td>Genre</td><td>URL</td>";

    //debugging (loop count)
    var c = 0;

    for (var key in urlsObj) {
        if (urlsObj.hasOwnProperty(key)) {

            var val = urlsObj[key];

            $table += "<tr>";

            $table + "<td>" + val.name; + "</td>";
            $table + "<td>" + val.releaseTime; + "</td>";
            $table + "<td>" + val.releaseDate; + "</td>";
            $table + "<td>" + val.category; + "</td>";
            $table + "<td>" + val.genre; + "</td>";
            $table + "<td>" + val.url; + "</td>";

            $table += "</tr>";

            //debugging (loop count)
            c++;
            console.log("Count: " + c);
        }
    }

    $('#response').html('Attempting to display your URLs, please wait...');
    $('#content01').append($table);

};

---

您的问题在于JSON解析

首先这样做：这将把JSON字符串对象转换为Javascript对象，您可以在其中循环

var items = JSON.parse('[{"userId":4,"name":"Doctor Who","releaseTime":"2015-11-22 12:45:24","releaseDate":"Monday","category":"Television","genre":"Action","url":"http:\/\/www.netflix.com\/browse?jbv=70142441&jbp=0&jbr=1"},{"userId":4,"name":"Game Of Thrones","releaseTime":"2015-11-22 13:34:06","releaseDate":"Tuesday","category":"Television","genre":"Drama","url":"http:\/\/www.tvmuse.com\/tv-shows\/Game-of-Thrones_25243\/"}]');

那么，做这个

for(i=0; i<items.length; i++){
    console.log(items[i].url);
    console.log(items[i].name); //etc
}

for（i=0；i将脚本更改为：

function refreshUrlArray(urlsRetrieved) {
    $('#response').html('Attempting to update your URLs, please wait...');  

    var urlsObj = urlsRetrieved;

    var details = JSON.parse(urlsObj);

    //console.log(details.length);

    table = "<table id='urlTable'><td>Name</td><td>Release Time</td><td>Release Date</td><td>Category</td><td>Genre</td><td>URL</td>";
    for(var i = 0; i < details.length; i++){
        records = details[i];
        //Create object holding table
        table += "<tr>";

        table += "<td>" + records.name; + "</td>";
        table += "<td>" + records.releaseTime; + "</td>";
        table += "<td>" + records.releaseDate; + "</td>";
        table += "<td>" + records.category; + "</td>";
        table += "<td>" + records.genre; + "</td>";
        table += "<td>" + records.url; + "</td>";

        table += "</tr>";

    }
    table += "</table>";

    $('#response').html('Attempting to display your URLs, please wait...');  
    $('#content01').html(table);
}

函数refreshUrlArray（urlsRetrieve）{
$（'#response'）.html（'正在尝试更新您的URL，请稍候…）；
var urlsObj=urlsRetrieved；
var details=JSON.parse（urlsObj）；
//控制台。日志（详细信息。长度）；
table=“NameRelease TimeRelease DateCategoryGeneurl”；
对于（变量i=0；i
您需要解析JSON字符串。使用JSON.parse解析JSON后，如下所示

var items = JSON.parse(json_response);

您正在添加字符串，但没有将其分配到
$表

更换回路内的管路，如下所示

$table += "<td>" + urlsObj[i].name; + "</td>";
$table += "<td>" + urlsObj[i].releaseTime; + "</td>";
$table += "<td>" + urlsObj[i].releaseDate; + "</td>";
$table += "<td>" + urlsObj[i].category; + "</td>";
$table += "<td>" + urlsObj[i].genre; + "</td>";
$table += "<td>" + urlsObj[i].url; + "</td>";

$table+=“”+urlsObj[i]。名称；+“”；
$table+=“”+urlsObj[i]。释放时间；+“”；
$table+=“”+urlsObj[i].发布日期；+“”；
$table+=“”+urlsObj[i]。类别；+“”；
$table+=“”+urlsObj[i]。流派；+“”；
$table+=''+urlsObj[i].url；+''；
“据我所知，我认为循环对JSON结果中的每个字符循环一次”在这种情况下，您还没有将JSON解析为JavaScript数组。请参阅“据我所知，我认为…”-为什么要猜测？学习使用调试器并了解问题所在。如果使用数组访问器（如val[“name”]）访问成员，…？与您的问题无关，只是一个小提示。您当然可以随意命名变量，但与jQuery结合使用时，我将仅对实际包含jQuery对象/结果集的变量使用
$
前缀变量，而不是简单的html字符串。谢谢t.niese，我会记住这一点！谢谢mate这很有帮助，正如大多数人指出的那样，我没有将JSON解析为JS数组……仍在学习的家伙们，没有意识到我的错误。
var items = JSON.parse(json_response);

$table += "<td>" + urlsObj[i].name; + "</td>";
$table += "<td>" + urlsObj[i].releaseTime; + "</td>";
$table += "<td>" + urlsObj[i].releaseDate; + "</td>";
$table += "<td>" + urlsObj[i].category; + "</td>";
$table += "<td>" + urlsObj[i].genre; + "</td>";
$table += "<td>" + urlsObj[i].url; + "</td>";