Javascript Select2.js错误：无法读取属性'；长度'；未定义的
我使用的是Select2 jquery插件，无法使用json获得结果。在浏览器中查看json响应时，它看起来正常。比如说：Javascript Select2.js错误：无法读取属性'；长度'；未定义的,javascript,php,jquery,json,jquery-select2,Javascript,Php,Jquery,Json,Jquery Select2,我使用的是Select2 jquery插件，无法使用json获得结果。在浏览器中查看json响应时，它看起来正常。比如说： [{ "id" : "50", "family" : "Portulacaceae " }, { "id" : "76", "family" : "Styracaceae " }, { "id" : "137", "family" : "Dipsacaceae"
[{
"id" : "50",
"family" : "Portulacaceae "
}, {
"id" : "76",
"family" : "Styracaceae "
}, {
"id" : "137",
"family" : "Dipsacaceae"
}
]
http://localhost/webpage/json_family.php?term=acac&_=1417999511783html
<input type="hidden" id="select2_family" name="term" style="width:30%" />
$("#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 };
}
}
});
$myArray = array();
if ($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);
}
<input type="hidden" id="select2_family" name="term" style="width:30%" />
$("#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 };
}
}
});
$myArray = array();
if ($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);
}
代码中有错误吗？注意：只是尝试一下。就是突出的部分。 您的json没有属性结果，请尝试
$("#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 };
}
}
});
$myArray = array();
// here
if ($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);
}
您需要在结果上定义
文本formatResultformatSelection$("#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; }
}
});
好的，我让您的示例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，为可读性更改了一些名称，但应具有相同的功能，重要的部分是在查询中添加“AS TEXT” 其次，看起来您正试图从名为“results”的json响应调用一个属性 如果是这种情况，您的json应该是这样的，请注意，由于上面的更改，族现在是文本：
{
"results":
[
{
"id": "50",
"text": "Portulacaceae "
},
{
"id": "76",
"text": "Styracaceae "
},
{
"id": "137",
"text": "Dipsacaceae"
}
]
}
results: function (data) {
return { results: data };
}
<!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>
$（文档）.ready（函数（）{
$（“#选择2#U系列”）。选择2({
最小输入长度：3，
阿贾克斯：{
url:“select2.php”，
数据类型：“json”，
数据：功能（术语）{
返回{
任期:任期,，
};
},
结果：功能（数据）{
返回{结果：数据}；
}
}
});
});
<?
/*** 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);
?>
谢谢，但当我这样做时，会显示新的错误：无法读取unfinedi的属性'toUpperCase'，我会得到与我在上述答案中评论的相同的错误。旧的不再显示，但显示了这个“toUpperCase”，并且没有数据显示我更新的php。我对这个插件还不太熟悉，无法直接讨论这个问题，但我浏览了一些文档。希望你能很快把它做好