jquery ui令牌

问题描述 投票:1回答:2

你好我遵循了本教程,其中使用jquery UI生成令牌Facebook,例如:http://net.tutsplus.com/tutorials/javascript-ajax/how-to-use-the-jquery-ui-autocomplete-widget/

我的问题是,我需要通过Json传递两个值:ID和NAME:服务器端脚本如下所示:

header('Content-Type: text/html; charset=iso-8859-1', true);
include($_SERVER['DOCUMENT_ROOT'].'/inrees/inrees/communaute/includes/_db.php');

$param = $_GET["term"];
$query = mysql_query("SELECT * FROM comm_carnet, in_emails 
                       WHERE carnet_iduser=emails_id 
                         AND emails_id!='".$_COOKIE['INREES_ID']."'  
                         AND emails_nom REGEXP '^$param'");

//build array of results
for ($x = 0, $numrows = mysql_num_rows($query); $x < $numrows; $x++) {
    $row = mysql_fetch_assoc($query);
    $friends[$x] = array("name" = > $row["emails_nom"], "id" = > $row["emails_id"]);
}

//echo JSON to page
$response = $_GET["callback"]."(".json_encode($friends).")";

echo $response;

来自服务器端脚本的回显是:

([{"name":"dupont","id":"34998"},{"name":"castro","id":"34996"},{"name":"castelbajac","id":"34995"}])

(这正是我所需要的)

现在,我传递的是“名称”数组,但不需要传递“ id”,它需要是数据库中具有相应ID的隐藏输入,完成对php的调用的html页面如下所示:] >

//attach autocomplete
$("#to").autocomplete({

    //define callback to format results
    source: function (req, add) {

        //pass request to server
        $.getJSON("messages_ajax.php?callback=?", req, function (data) {

            //create array for response objects
            var suggestions = [];

            //process response
            $.each(data, function (i, val) {
                suggestions.push(val.name);
            });

            //pass array to callback
            add(suggestions);
        });
    },

    //define select handler
    select: function (e, ui) {

        //create formatted friend
        var friend = ui.item.value,
            span = $("<span>").text(friend),
            a = $("<a>").addClass("remove").attr({
                href: "javascript:",
                title: "Remove " + friend
            }).text("x").appendTo(span);
        $("<input />", {
            value: "id",
            type: "hidden",
            name: "id"
        }).appendTo(span);
        //add friend to friend div
        span.insertBefore("#to");
    },

    //define select handler
    change: function () {
        //prevent 'to' field being updated and correct position
        $("#to").val("").css("top", 2);
    }
});

//add click handler to friends div
$("#friends").click(function () {
    //focus 'to' field
    $("#to").focus();
});

//add live handler for clicks on remove links
$(".remove", document.getElementById("friends")).live("click", function () {

    //remove current friend
    $(this).parent().remove();

    //correct 'to' field position
    if ($("#friends span").length === 0) {
        $("#to").css("top", 0);
    }
});

所以基本上就是您看到评论的地方:"//define select handler"需要做些什么,但我不能做!我添加了一行:

$("<input />", {value:"id", type:"hidden", name:"id"}).appendTo(span);但它不会获取我的数组“ id”,因此不提供任何帮助。

问候

您好,我已遵循本教程,其中使用jquery UI生成令牌facebook,例如:http://net.tutsplus.com/tutorials/javascript-ajax/how-to-use-the-jquery-ui-autocomplete-widget/我的问题是...

javascript jquery ajax jquery-ui jsonp
2个回答
3
投票

您的代码应为:


0
投票

因此,您似乎只向建议列表添加了name,而不是将包含dataname成员的整个id对象。而不是这样做:

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