douzhang3822 2015-04-03 09:11
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jQuery Tokeninput插件PHP没有json的结果

I try to use the JQuery Tokeninput. While I introduce on the search fields the text , it doesn't show anything. It stays on searching forever...

From the documentation I see that the format of the resunt of json_encode should looks like :

[{"id":"856","name":"House"},
{"id":"1035","name":"Desperate Housewives"},
...]

but when I access my json localhost/json.php?q=house returns datas like :

["house1","house2" ...] 

so it doesn't contains the "{}" .

I'm sure that the paths to the js,and css are ok. the script is here:

<script type="text/javascript">
                    $(document).ready(function () {
                        $("#textarea").tokenInput('link/json.php');
                    });
                </script>

The json code looks like :

$rows = array();
$q=$_GET['q']; 

$query_get_element = "SELECT DISTINCT `column` FROM `table` WHERE `column` LIKE '%".$q."%'";
$query_get_element = mysql_query($query_get_element);
$row_cen = mysql_fetch_array($query_get_element);

$json=array();

while ($row2 = mysql_fetch_array($query_get_element))
    array_push($json, $row2['column']);

echo json_encode($json);

While I search the console shows localhost/json.php?q=house Any advice to solve this ? Thanks in advance!

on console (google chrome I see:

GET http://localhost/json.php?q=house 404 (Not Found)
  • 写回答

1条回答 默认 最新

  • douyoufan7881 2015-04-03 10:40
    关注

    I solved the issue. The problem was in my JSON . It was : array_push($json, array($row2['column'])); the pluggin search for the name of column and the array of column so it need to be like :

    array_push($json, array("column" => $row2['column']));

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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