我有一个收集页面,学生将参加哪些课程,并在表格中同时显示Course Title
和Catalog Number
。如果学生选择Course Title
,它将自动填写下一个单元格中的Catalog Number
,反之亦然。
我唯一遇到的问题是,当您填写一个Course Title
时,它会填充每个Catalog Number
单元格(反之亦然)。
<body>
<table>
<tbody>
<tr>
<td>
<input list="courses" name="courseInput" placeholder="Course" oninput="UpdateCatNumbers()">
<datalist id="courses" name="courseDatalist">
<!--Filled in script-->
</datalist>
</td>
<td>
<input list="catalogs" name="catalogInput" placeholder="Catalog Number" id="catalogID" oninput="UpdateCourseNames()">
<datalist id="catalogs" name="catalogDatalist">
<!--Filled in script-->
</datalist>
</td>
</tr>
<tr>
<td><input list="courses" name="courseInput" placeholder="Course" oninput="UpdateCatNumbers()"></td>
<td><input list="catalogs" name="catalogInput" placeholder="Catalog Number" id="catalogID" oninput="UpdateCourseNames()"></td>
</tr>
<tr>
<td><input list="courses" name="courseInput" placeholder="Course" oninput="UpdateCatNumbers()"></td>
<td><input list="catalogs" name="catalogInput" placeholder="Catalog Number" id="catalogID" oninput="UpdateCourseNames()"></td>
</tr>
<tr>
<td><input list="courses" name="courseInput" placeholder="Course" oninput="UpdateCatNumbers()"></td>
<td><input list="catalogs" name="catalogInput" placeholder="Catalog Number" id="catalogID" oninput="UpdateCourseNames()"></td>
</tr>
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<script src="testTemplateJS.js"></script>
<script>
//Course Names
var courseNames = ["Class A", "Class B", "Class C", "Class D", "Class E"];
var list = $('#courses')[0];
courseNames.forEach(function(item) {
var option = document.createElement('option');
option.value = item;
list.appendChild(option);
});
</script>
<script>
//Catalog Numbers
var catNumbers = ["Catalog 1", "Catalog 2", "Catalog 3", "Catalog 4", "Catalog 5"];
var list = $('#catalogs')[0];
catNumbers.forEach(function(item) {
var option = document.createElement('option');
option.value = item;
list.appendChild(option);
});
</script>
<script>
function UpdateCatNumbers() {
//on change of courseInput, run...
$("input[name=courseInput]").change(function() {
//declare a as the array index # of courseNames
var a = courseNames.indexOf($(this).val());
//change catalog input to display the same array index of catNumbers
$("input[name=catalogInput]").val(catNumbers[a]);
});
}
</script>
<script>
function UpdateCourseNames() {
$("input[name=catalogInput]").change(function() {
var b = catNumbers.indexOf($(this).val());
$("input[name=courseInput]").val(courseNames[b]);
});
}
</script>
</body>
如何获取它,以便UpdateCatNumbers()
和UpdateCourseNames()
函数仅填充直接位于它们旁边的单元格?
答案 0 :(得分:0)
因此,您只需要将该值应用于其他输入。在这种情况下,您可以使用jQuery import { FETCH_SONG } from "../actions/types";
export default function(state = null, action) {
switch (action.type) {
case FETCH_SONG:
console.log(action.payload);
return action.payload;
default:
return state;
}
}
-通过它您可以基于索引从jQuery集合中选择单个元素:
eq(index)