Select2.js错误：无法读取未定义的属性“长度”
[英]Select2.js error: Cannot read property 'length' of undefined
问题描述
I am using Select2 jquery plugin and I can't get results with json. 
我正在使用Select2 jquery插件，我无法使用json获得结果。 When looking json response in browser it looks ok. 在浏览器中查看json响应时看起来没问题。 Like this for example: 像这样例如：
[{
"id" : "50",
"family" : "Portulacaceae "
}, {
"id" : "76",
"family" : "Styracaceae "
}, {
"id" : "137",
"family" : "Dipsacaceae"
}
]
URL called with ajax in this case is: http://localhost/webpage/json_family.php?term=acac&_=1417999511783 but I can't get that results in select2 input, console says: 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输入中得到结果，控制台说：
Uncaught TypeError: Cannot read property 'length' of undefined
Here is code: 这是代码：
html HTML
<input type="hidden" id="select2_family" name="term" style="width:30%" />
js 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 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);
}
Is there error in code? 代码中是否有错误？
3 个解决方案
解决方案1
6 已采纳 2014-12-09 15:12:44
Ok, i have your example working on my test server, please do the following 好的，我的示例在我的测试服务器上工作，请执行以下操作
change your query to this, changed a few names for readability but should be the same functionality, important part is addition of "AS TEXT" in query 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，重要的部分是在查询中添加“AS TEXT”
$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);
second, it looks like you are trying to call a property from the json response called "results" 第二，看起来你试图从名为“结果”的json响应中调用一个属性
if that was the case your json should look like this, note that family is now text due to the change above: 如果是这种情况你的json应该是这样的，请注意由于上面的更改，系列现在是文本：
{
"results":
[
{
"id": "50",
"text": "Portulacaceae "
},
{
"id": "76",
"text": "Styracaceae "
},
{
"id": "137",
"text": "Dipsacaceae"
}
]
}
But your php does not create the property results, so change your results function to remove the .results property call 但是您的php不会创建属性结果，因此请更改结果函数以删除.results属性调用
results: function (data) {
return { results: data };
}
final code i used (note i did not escape/sanitize the $_GET[term] or bind it to the query, recommend you do so ) if you are still having issues i can send you a link to my site example 我使用的最终代码（注意我没有逃避/清理$ _GET [term]或将其绑定到查询，建议你这样做）如果你仍然有问题我可以发送一个链接到我的网站示例
<!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>
php 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);
?>
解决方案2
1 2014-12-08 01:12:57
Note: just a stab at it. 注意：只是刺它。 Just what stuck out. 什么伸出来。
Your json has no property results, so try. 你的json没有财产结果，所以试试吧。
$("#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 };
}
}
});
changed the query -- see if this helps 更改了查询 - 看看这是否有帮助
$myArray = array();
// here
if ($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);
}
解决方案3
0 2014-12-08 01:13:06
you need to define the text property on the results 您需要在结果上定义text属性
and you might need to add formatResult and formatSelection 并且您可能需要添加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; }
}
});
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