jQuery Tokeninput插件PHP没有json的结果

我尝试使用JQuery Tokeninput。 虽然我在搜索字段中介绍了文本,但它没有显示任何内容。 它一直在寻找... ... </ p>

从文档中我看到json_encode的resunt格式应如下所示:</ p>

   [{“id”:“856”,“name”:“House”},
{“id”:“1035”,“name”:“绝望的主妇”},
...]
</ 代码> </ pre>

但是当我访问我的json localhost / json.php时?q = house返回如下数据:</ p>

  [“house1”  ,“house2”...] 
</ code> </ pre>

因此它不包含“{}”。 </ p>

我确定js和css的路径都可以。 脚本在这里:</ p>

 &lt; script type =“text / javascript”&gt; 
$(document).ready(function(){
$(“# textarea“)。tokenInput('link / json.php');
});
&lt; / script&gt;
</ code> </ pre>

json代码如下所示: </ p>

  $ rows = array(); 
$ q = $ _ GET ['q'];

$ query_get_element =“SELECT DISTINCTcolumn FROMtable WHEREpoint 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']);
\ necho json_encode($ json);
</ code> </ pre>

我搜索控制台时显示localhost / json.php?q = house
任何建议来解决这个问题? 提前致谢!</ p>

在控制台上(google chrome我看到了:</ p>

  GET http://localhost/json.php?q = 房子404(未找到)
</ code> </ pre>
</ div>

展开原文

原文

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)

duanjiu2701
duanjiu2701 我收到以下错误:未捕获的TypeError:无法读取未定义的属性'replace'
5 年多之前 回复
doujindou4356
doujindou4356 array_push($json,array($row2['column']));
5 年多之前 回复
douwen4125
douwen4125 我现在就做了:)
5 年多之前 回复
doumeng06063991
doumeng06063991 所以..postjson.php代码会更好
5 年多之前 回复

1个回答



我解决了这个问题。 问题出在我的JSON中。
它是: array_push($ json,array($ row2 ['column'])); </ code>
插件搜索列名和列数组 所以它需要像:</ p>

array_push($ json,array(“column”=&gt; $ row2 ['column'])); </ code> </ p>
</ div>

展开原文

原文

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']));

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