Question

我正在使用Select2 jquery插件，但我无法通过json获得结果。在浏览器中查看json响应时看起来没问题。像这样：例如：

[{
        "id" : "50",
        "family" : "Portulacaceae "
    }, {
        "id" : "76",
        "family" : "Styracaceae "
    }, {
        "id" : "137",
        "family" : "Dipsacaceae"
    }
]

在这种情况下使用ajax调用的URL是：http://localhost/webpage/json_family.php?term=acac&_=1417999511783但是我无法在select2输入中获得结果，控制台说：

未捕获的TypeError：无法读取属性＆＃39;长度＆＃39;未定义的

这是代码：
的 HTML

<input type="hidden" id="select2_family" name="term" style="width:30%" />

JS

$("#select2_family").select2({
  minimumInputLength: 3,
  ajax: {
   url: "json_family.php",
   dataType: 'json',
   data: function (term) {
       return {
         term: term,
       };
   },
   results: function (data) {
     return { results: data.results };
   }

  }
});

PHP

$myArray = array();
if ($result = $mysqli->query("SELECT id,family FROM family WHERE family LIKE '%$term%'")) {
    $tempArray = array();
    while($row = $result->fetch_object()) {
            $tempArray = $row;
            array_push($myArray, $tempArray);
        }
    echo json_encode($myArray);
}

代码中是否有错误？

Answer 1

好的，我的示例是在我的测试服务器上运行，请执行以下操作

将您的查询更改为此，更改了一些名称以便于阅读，但应该是相同的功能，重要的部分是添加＆＃34; AS TEXT＆＃34;在查询中

$query = $mysqli->query("SELECT id, family AS text FROM family WHERE family LIKE '%$term%'"));
    while ($row = mysql_fetch_assoc($query)) {
           $return[] = $row;
         }

    echo json_encode($return);

第二，看起来你试图从json响应中调用一个名为＆＃34;结果＆＃34;

的属性

如果您的json应该是这样的话，请注意由于上述更改，系列现在是文本：

{
"results":
[
    {
        "id": "50",
        "text": "Portulacaceae "
    },
    {
        "id": "76",
        "text": "Styracaceae "
    },
    {
        "id": "137",
        "text": "Dipsacaceae"
    }
]
}

但是你的php不会创建属性结果，所以更改你的结果函数以删除.results属性调用

   results: function (data) {
     return { results: data };
   }

我使用的最终代码（注意我没有逃避/清理$ _GET [term]或将其绑定到查询，建议你这样做）如果你还有问题我可以发送链接到我的网站示例< / p>

<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="//cdnjs.cloudflare.com/ajax/libs/select2/3.5.2/select2.css">
<script type="text/javascript" src="//cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.js"></script>
<script type="text/javascript" src="//cdnjs.cloudflare.com/ajax/libs/select2/3.5.2/select2.js"></script>
</head>
<script>
$(document).ready(function () {

$("#select2_family").select2({
  minimumInputLength: 3,
  ajax: {
   url: "select2.php",
   dataType: 'json',
   data: function (term) {
       return {
         term: term,
       };
   },
   results: function (data) {
     return { results: data };
   }
  }
});

});
</script>

<input type="hidden" id="select2_family" name="term" style="width:30%" />

</html>

<强> PHP

<?

/*** connection strings ***/

// get the database singleton instance
$yog = MySqlDatabase::getInstance();

// connect
try {
    $yog->connect($host, $user, $password, $db_name);
}
catch (Exception $e) {
    die($e->getMessage());
}

$term = $_GET['term'];

if (!$term){
$sub = $yog->query("SELECT id, family AS text FROM family");
} else {
$sub = $yog->query("SELECT id, family AS text FROM family where family like '%$term%'");
}

while ($row = mysql_fetch_assoc($sub)) {
       $return[] = $row;
     }

echo json_encode($return);

?>

Answer 2

注意：只是捅了一下。突然发生了什么。

你的json没有属性结果，所以试试。

$("#select2_family").select2({
  minimumInputLength: 3,
  ajax: {
   url: "json_family.php",
   dataType: 'json',
   data: function (term) {
       return {
         term: term,
       };
   },
   results: function (data) {

     // CHANGED
     return { results: data };

   }

  }
});

更改了查询 - 看看是否有帮助

$myArray = array();

// here
if ($result = $mysqli->query("SELECT id, family AS text FROM family WHERE family LIKE '%$term%'")) {
    $tempArray = array();
    while($row = $result->fetch_object()) {
            $tempArray = $row;
            array_push($myArray, $tempArray);
        }
    echo json_encode($myArray);
}

Answer 3

您需要在结果

上定义text属性

您可能需要添加formatResult和formatSelection

$("#select2_family").select2({
    minimumInputLength: 3,
    ajax: {
        url: "json_family.php",
        dataType: 'json',
        data: function (term) {
            return {
                term: term,
            };
        },
        results: function (data) {return { results: data, text: 'family'}; },
        formatResult: function(item) { return item.family; }, 
        formatSelection: function(item) { return item.family; }
    }
});

Select2.js错误：无法读取未定义的属性“长度”

3 个答案: