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java - 在下拉菜单中拉取 json 文件

转载 作者:太空宇宙 更新时间:2023-11-04 13:54:42 25 4
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我正在尝试在下拉菜单中拉取 json 文件以显示可用的尺寸、颜色和衬衫的图像。我的下拉菜单可以工作,但当我测试它时,它在我的 div 上没有显示任何内容。这就是我到目前为止所得到的。

p {
display: inline;
position: relative;
letter-spacing: 20px; /* This will push it together giving us a nice 3D vibe */
color: rgba(0,0,255,0.5); /* This will give us a blue at 50% opacity */
}
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta charset="utf-8" />
<title>Shirts That you love</title>
<link rel="stylesheet" type="text/css" href="StyleSheet.css" />
<script src="http://code.jquery.com/jquery-1.6.3.min.js"></script>
<script language="javascript" type="text/javascript">

var JSON_Response;

$(document).ready(function () {

$.getJSON('man.json', function (data) {
JSON_Response = data;


var mySelect = document.getElementById("selman");

for (i = 0; i < JSON_Response.man.length; i++) {
var myOption = document.createElement("option");
myOption.text = JSON_Response.man[i].Shirt;
myOption.value = i;
try {

mySelect.add(myOption, mySelect.options[null]);
}
catch (e) {
mySelect.add(myOption, null);
}
}

});

$.getJSON('women.json', function (data) {
JSON_Response = data;


http://www.w3schools.com/jsref/met_select_add.asp
var mySelect = document.getElementById("selwoman");

for (i = 0; i < JSON_Response.women.length; i++) {
var myOption = document.createElement("option");
myOption.text = JSON_Response.women[i].Shirt;
myOption.value = i;
try {

mySelect.add(myOption, mySelect.options[null]);
}
catch (e) {
mySelect.add(myOption, null);
}
}

});

$.getJSON('kids.json', function (data) {
JSON_Response = data;


var mySelect = document.getElementById("selkids");

for (i = 0; i < JSON_Response.kids.length; i++) {
var myOption = document.createElement("option");
myOption.text = JSON_Response.kids[i].Shirt;
myOption.value = i;
try {

mySelect.add(myOption, mySelect.options[null]);
}
catch (e) {
mySelect.add(myOption, null);
}
}

});

$("#selman").change(function () {
var myIndex = $("#selman").val();
$("#shirts").attr("src", JSON_Response.man[myIndex].shirtimage);

var info = JSON_Response.man[myIndex].Shirt + "<br />";
info += JSON_Response.man[myIndex].Color + "<br />";
info += JSON_Response.man[myIndex].Sizes + "<br />";
info += "Man" + "<br />";

$("#divDisplay").html(info);

});

$("#selwoman").change(function () {
var myIndex = $("#selkids").val();
$("#shirts0").attr("src", JSON_Response.women[myIndex].shirtimage);

var info = JSON_Response.women[myIndex].Shirt + "<br />";
info += JSON_Response.women[myIndex].Color + "<br />";
info += JSON_Response.women[myIndex].Sizes + "<br />";
info += "women" + "<br />";

$("#divDisplay0").html(info);

});

$("#selkids").change(function () {
var myIndex = $("#selwomen").val();
$("#shirts1").attr("src", JSON_Response.kids[myIndex].shirtimage);

var info = JSON_Response.kids[myIndex].Shirt + "<br />";
info += JSON_Response.kids[myIndex].Color + "<br />";
info += JSON_Response.kids[myIndex].Sizes + "<br />";
info += "kids" + "<br />";

$("#divDisplay1").html(info);

});
});

</script>

</head>
<body>
<h1>Pick A Genre</h1>

<p>
<br />
Are you a man?
</p>

<table>
<tr>
<td rowspan="2">
<img id="shirts" alt="" src="" /></td>
<td>
<select id="selman">
<option></option>

</select></td>
</tr>
<tr>
<td>
<div id="divDisplay">
</div>
</td>
</tr>

</table>
<p>
<br />
Are you a woman?
</p>

<table>
<tr>
<td rowspan="2">
<img id="shirts0" alt="" src="" /></td>
<td>
<select id="selwoman">
<option></option>

</select></td>
</tr>
<tr>
<td>
<div id="divDisplay0">
</div>
</td>
</tr>

</table>
<p>
<br />
Are you a kid?
</p>

<table>
<tr>
<td rowspan="2">
<img id="shirts1" alt="" src="" /></td>
<td>
<select id="selkids">
<option></option>

</select></td>
</tr>
<tr>
<td>
<div id="divDisplay1">
</div>
</td>
</tr>

</table>
<p>
&nbsp;
</p>
</body>
</html>

最佳答案

有两种错误可能性 -

  1. 选择 Try this- 中的选项插入

    mySelect.add(myOption);

  2. 检查您的 JSONm,可能是您没有提取正确的信息..

关于java - 在下拉菜单中拉取 json 文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29944421/

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