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javascript - AJAX、JS 和 Select2 的问题

转载 作者:行者123 更新时间:2023-11-30 18:11:43 24 4
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如果我做了一些愚蠢的事情,我深表歉意,但我是 Select2 和 Ajax 的新手,正在努力完成这项工作。我正在尝试创建自己的 json 文件(链接已删除)以与我正在构建的表单(链接已删除)一起使用。在 Chrome 中使用 Firefox Lite,我能够从我的服务器获得响应(200 OK),但选择框不显示搜索结果。我的 json 文件似乎使用 http://jsonformatter.curiousconcept.com 进行了正确验证.您也可以在

有什么想法或提示吗?

json.php 来源:

<?
$myArray = array(
array( "id" => "id1", "text" => "title1" ),
array( "id" => "id2", "text" => "title2" )
);

echo json_encode($myArray);

?>

select.php 来源:

<html>
<head>
<link href="select2/select2.css" rel="stylesheet"/>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script src="select2/select2.js"></script>
<script>
function movieFormatResult(movie) {
var markup = "<table class='movie-result'><tr>";
markup += "<td class='movie-info'><div class='movie-title'>" + movie.title + "</div>";
markup += "</td></tr></table>"
return markup;
}

function movieFormatSelection(movie) {
return movie.title;
}

</script>
<script id="script_e6">
$(document).ready(function() {
$("#e6").select2({
placeholder: "Search for a movie",
minimumInputLength: 1,
ajax: { // instead of writing the function to execute the request we use Select2's convenient helper
url: "http://api.rottentomatoes.com/api/public/v1.0/movies.json",
dataType: 'jsonp',
data: function (term, page) {
return {
q: term, // search term
page_limit: 10,
apikey: "my-api-key" // please do not use so this example keeps working
};
},
results: function (data, page) { // parse the results into the format expected by Select2.
// since we are using custom formatting functions we do not need to alter remote JSON data
return {results: data.movies};
}
},
formatResult: movieFormatResult, // omitted for brevity, see the source of this page
formatSelection: movieFormatSelection, // omitted for brevity, see the source of this page
dropdownCssClass: "bigdrop" // apply css that makes the dropdown taller
});
});
</script>
<script id="script_e6a">
$(document).ready(function() {
$("#e6a").select2({
placeholder: "Search for a movie",
minimumInputLength: 1,
ajax: { // instead of writing the function to execute the request we use Select2's convenient helper
url: "http://willwelch.net/play/nhs/pstudents/json.php",
dataType: 'jsonp',
data: function () {
return;
},
results: function () { // parse the results into the format expected by Select2.
return {results: data};
}
},
formatResult: movieFormatResult, // omitted for brevity, see the source of this page
formatSelection: movieFormatSelection, // omitted for brevity, see the source of this page
dropdownCssClass: "bigdrop" // apply css that makes the dropdown taller
});
});
</script>
</head>

<body style="height:500px;">
<p>original example (works)</p>
<input type="hidden" class="bigdrop" id="e6" style="width: 600px; display: none;"><br><br>
<p>my version (does not work)</p>
<input type="hidden" class="bigdrop" id="e6a" style="width: 600px; display: none;">
</body>

</html>

最佳答案

您正在调用您的 url (willwelch.net) 作为 jsonp 调用。在您的 php 中,您必须捕获 ajax 调用传递的参数 callback,并将结果包装在函数调用中,函数名称 = 回调参数的内容。

因此对您的 php 的调用可能是:

 http://willwelch.net/play/nhs/pstudents/json.php?callback=myfunc

你应该返回的是:

 myfunc([{"id":"id1","text":"title1"},{"id":"id2","text":"title2"}])

注意 ajax 调用会自动添加回调参数。

编辑

您还需要更改 return 方法以将 data 添加为输入参数:

         results: function (data) { // parse the results into the format expected by Select2.
return {results: data};
}

关于javascript - AJAX、JS 和 Select2 的问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14426065/

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