douxian7534 2014-04-08 20:41
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Jquery用Json发送数组

I am using jquery-ui in this snippet:

$(document).ready(function () {
    $(function() {
        function log( message ) {
            $( "<div>" ).text( message ).prependTo( "#log" );
            $( "#log" ).scrollTop( 0 );
        }
    });
    $( "#birds" ).autocomplete({
        source: "search.php",
        minLength: 4,
        select: function( event, ui ) {
            log( ui.item ?
                "Selected: " + ui.item.value + " aka " + ui.item.id :
                "Nothing selected, input was " + this.value );
        }
    });
});



<div class="ui-widget">
    <label for="birds">Birds: </label>
    <input id="birds">
</div>
    <div class="ui-widget" style="margin-top:2em; font-family:Arial">
    Result:
    <div id="log" style="height: 200px; width: 300px; overflow: auto;" class="ui-widget-content">
    </div>
/div>

and my search.php:

$myarray = ["somelabelvalue","somelabelvalue1","somelabelvalue2","somelabelvalue3"];
echo json_encode($myarray);

BUT when i type: "oems" => every option of the array appear´s! why? It should appear only if i type "some". So where is my mistake in search.php?

this one works correctly: http://jqueryui.com/autocomplete/#remote ..

Greetings!

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3条回答 默认 最新

  • drci47425 2014-04-08 20:50
    关注

    The problem is that no matter what the user types, you return the same array. What you should do instead, in your PHP code, is to perform some check based on the value given by the user. Note that the jQuery code only lists what gets returned by your PHP code, so filtering is still up to you.

    To create the array as you mentioned at the comment, you need to do it like this:

    $array = array(
    array("id"=>"the id", "label"=>"the label", "value"=>"the value"),
    array("id"=>"another id", "label"=>"another label", "value"=>"another value"),
    );
    

    After you call echo json_encode($array), you will get the desired result.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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